**Unformatted text preview: **3 Higher-Order
Differential Equations EXERCISES 3.1
Preliminary Theory: Linear Equations
1. From y = c1 ex + c2 e−x we ﬁnd y = c1 ex − c2 e−x . Then y(0) = c1 + c2 = 0, y (0) = c1 − c2 = 1 so that c1 =
and c2 = − 1 . The solution is y = 1 ex − 1 e−x .
2
2
2 1
2 2. From y = c1 e4x + c2 e−x we ﬁnd y = 4c1 e4x − c2 e−x . Then y(0) = c1 + c2 = 1, y (0) = 4c1 − c2 = 2 so that
c1 = 3 and c2 = 2 . The solution is y = 3 e4x + 2 e−x .
5
5
5
5
3. From y = c1 x + c2 x ln x we ﬁnd y = c1 + c2 (1 + ln x). Then y(1) = c1 = 3, y (1) = c1 + c2 = −1 so that c1 = 3
and c2 = −4. The solution is y = 3x − 4x ln x.
4. From y = c1 + c2 cos x + c3 sin x we ﬁnd y = −c2 sin x + c3 cos x and y = −c2 cos x − c3 sin x. Then y(π) =
c1 − c2 = 0, y (π) = −c3 = 2, y (π) = c2 = −1 so that c1 = −1, c2 = −1, and c3 = −2. The solution is
y = −1 − cos x − 2 sin x.
5. From y = c1 + c2 x2 we ﬁnd y = 2c2 x. Then y(0) = c1 = 0, y (0) = 2c2 · 0 = 0 and hence y (0) = 1 is not
possible. Since a2 (x) = x is 0 at x = 0, Theorem 3.1 is not violated.
6. In this case we have y(0) = c1 = 0, y (0) = 2c2 · 0 = 0 so c1 = 0 and c2 is arbitrary. Two solutions are y = x2
and y = 2x2 .
7. From x(0) = x0 = c1 we see that x(t) = x0 cos ωt + c2 sin ωt and x (t) = −x0 sin ωt + c2 ω cos ωt. Then
x (0) = x1 = c2 ω implies c2 = x1 /ω. Thus
x1
x(t) = x0 cos ωt +
sin ωt.
ω
8. Solving the system
x(t0 ) = c1 cos ωt0 + c2 sin ωt0 = x0
x (t0 ) = −c1 ω sin ωt0 + c2 ω cos ωt0 = x1
for c1 and c2 gives
c1 =
Thus ωx0 cos ωt0 − x1 sin ωt0
ω and c2 = x1 cos ωt0 + ωx0 sin ωt0
.
ω ωx0 cos ωt0 − x1 sin ωt0
x1 cos ωt0 + ωx0 sin ωt0
cos ωt +
sin ωt
ω
ω
x1
= x0 (cos ωt cos ωt0 + sin ωt sin ωt0 ) + (sin ωt cos ωt0 − cos ωt sin ωt0 )
ω
x1
= x0 cos ω(t − t0 ) +
sin ω(t − t0 ).
ω x(t) = 9. Since a2 (x) = x − 2 and x0 = 0 the problem has a unique solution for −∞ < x < 2. 99 3.1 Preliminary Theory: Linear Equations 10. Since a0 (x) = tan x and x0 = 0 the problem has a unique solution for −π/2 < x < π/2.
11. (a) We have y(0) = c1 + c2 = 0, y (1) = c1 e + c2 e−1 = 1 so that c1 = e/ e2 − 1 and c2 = −e/ e2 − 1 . The
solution is y = e (ex − e−x ) / e2 − 1 .
(b) We have y(0) = c3 cosh 0 + c4 sinh 0 = c3 = 0 and y(1) = c3 cosh 1 + c4 sinh 1 = c4 sinh 1 = 1, so c3 = 0 and
c4 = 1/ sinh 1. The solution is y = (sinh x)/(sinh 1).
(c) Starting with the solution in part (b) we have
y= 1
ex − e−x
2
ex − e−x
e
sinh x = 1
=
= 2
(ex − e−x ).
−1
sinh 1
e −e
2
e − 1/e
e −1 12. In this case we have y(0) = c1 = 1, y (1) = 2c2 = 6 so that c1 = 1 and c2 = 3. The solution is y = 1 + 3x2 .
13. From y = c1 ex cos x + c2 ex sin x we ﬁnd y = c1 ex (− sin x + cos x) + c2 ex (cos x + sin x).
(a) We have y(0) = c1 = 1, y (0) = c1 +c2 = 0 so that c1 = 1 and c2 = −1. The solution is y = ex cos x−ex sin x.
(b) We have y(0) = c1 = 1, y(π) = −eπ = −1, which is not possible.
(c) We have y(0) = c1 = 1, y(π/2) = c2 eπ/2 = 1 so that c1 = 1 and c2 = e−π/2 . The solution is y =
ex cos x + e−π/2 ex sin x.
(d) We have y(0) = c1 = 0, y(π) = c2 eπ sin π = 0 so that c1 = 0 and c2 is arbitrary. Solutions are y = c2 ex sin x,
for any real numbers c2 .
14. (a) We have y(−1) = c1 + c2 + 3 = 0, y(1) = c1 + c2 + 3 = 4, which is not possible.
(b) We have y(0) = c1 · 0 + c2 · 0 + 3 = 1, which is not possible.
(c) We have y(0) = c1 · 0 + c2 · 0 + 3 = 3, y(1) = c1 + c2 + 3 = 0 so that c1 is arbitrary and c2 = −3 − c1 .
Solutions are y = c1 x2 − (c1 + 3)x4 + 3.
(d) We have y(1) = c1 + c2 + 3 = 3, y(2) = 4c1 + 16c2 + 3 = 15 so that c1 = −1 and c2 = 1. The solution is
y = −x2 + x4 + 3.
15. Since (−4)x + (3)x2 + (1)(4x − 3x2 ) = 0 the set of functions is linearly dependent.
16. Since (1)0 + (0)x + (0)ex = 0 the set of functions is linearly dependent. A similar argument shows that any set
of functions containing f (x) = 0 will be linearly dependent.
17. Since (−1/5)5 + (1) cos2 x + (1) sin2 x = 0 the set of functions is linearly dependent.
18. Since (1) cos 2x + (1)1 + (−2) cos2 x = 0 the set of functions is linearly dependent.
19. Since (−4)x + (3)(x − 1) + (1)(x + 3) = 0 the set of functions is linearly dependent.
20. From the graphs of f1 (x) = 2 + x and f2 (x) = 2 + |x| we
see that the set of functions is linearly independent since
they cannot be multiples of each other. 21. Suppose c1 (1 + x) + c2 x + c3 x2 = 0. Then c1 + (c1 + c2 )x + c3 x2 = 0 and so c1 = 0, c1 + c2 = 0, and c3 = 0.
Since c1 = 0 we also have c2 = 0. Thus, the set of functions is linearly independent.
22. Since (−1/2)ex + (1/2)e−x + (1) sinh x = 0 the set of functions is linearly dependent. 100 3.1 Preliminary Theory: Linear Equations 23. The functions satisfy the diﬀerential equation and are linearly independent since
W e−3x , e4x = 7ex = 0
for −∞ < x < ∞. The general solution is
y = c1 e−3x + c2 e4x .
24. The functions satisfy the diﬀerential equation and are linearly independent since
W (cosh 2x, sinh 2x) = 2
for −∞ < x < ∞. The general solution is
y = c1 cosh 2x + c2 sinh 2x.
25. The functions satisfy the diﬀerential equation and are linearly independent since
W (ex cos 2x, ex sin 2x) = 2e2x = 0
for −∞ < x < ∞. The general solution is y = c1 ex cos 2x + c2 ex sin 2x.
26. The functions satisfy the diﬀerential equation and are linearly independent since
W ex/2 , xex/2 = ex = 0
for −∞ < x < ∞. The general solution is
y = c1 ex/2 + c2 xex/2 .
27. The functions satisfy the diﬀerential equation and are linearly independent since
W x3 , x4 = x6 = 0
for 0 < x < ∞. The general solution on this interval is
y = c1 x3 + c2 x4 .
28. The functions satisfy the diﬀerential equation and are linearly independent since
W (cos(ln x), sin(ln x)) = 1/x = 0
for 0 < x < ∞. The general solution on this interval is
y = c1 cos(ln x) + c2 sin(ln x).
29. The functions satisfy the diﬀerential equation and are linearly independent since
W x, x−2 , x−2 ln x = 9x−6 = 0
for 0 < x < ∞. The general solution on this interval is
y = c1 x + c2 x−2 + c3 x−2 ln x.
30. The functions satisfy the diﬀerential equation and are linearly independent since
W (1, x, cos x, sin x) = 1
for −∞ < x < ∞. The general solution on this interval is
y = c1 + c2 x + c3 cos x + c4 sin x. 101 3.1 Preliminary Theory: Linear Equations 31. The functions y1 = e2x and y2 = e5x form a fundamental set of solutions of the associated homogeneous
equation, and yp = 6ex is a particular solution of the nonhomogeneous equation.
32. The functions y1 = cos x and y2 = sin x form a fundamental set of solutions of the associated homogeneous
equation, and yp = x sin x + (cos x) ln(cos x) is a particular solution of the nonhomogeneous equation.
33. The functions y1 = e2x and y2 = xe2x form a fundamental set of solutions of the associated homogeneous
equation, and yp = x2 e2x + x − 2 is a particular solution of the nonhomogeneous equation.
34. The functions y1 = x−1/2 and y2 = x−1 form a fundamental set of solutions of the associated homogeneous
1
equation, and yp = 15 x2 − 1 x is a particular solution of the nonhomogeneous equation.
6
35. (a) We have yp1 = 6e2x and yp1 = 12e2x , so
yp1 − 6yp1 + 5yp1 = 12e2x − 36e2x + 15e2x = −9e2x .
Also, yp2 = 2x + 3 and yp2 = 2, so
yp2 − 6yp2 + 5yp2 = 2 − 6(2x + 3) + 5(x2 + 3x) = 5x2 + 3x − 16.
(b) By the superposition principle for nonhomogeneous equations a particular solution of y − 6y + 5y =
5x2 + 3x − 16 − 9e2x is yp = x2 + 3x + 3e2x . A particular solution of the second equation is
1
1
yp = −2yp2 − yp1 = −2x2 − 6x − e2x .
9
3
36. (a) yp1 = 5
(b) yp2 = −2x
(c) yp = yp1 + yp2 = 5 − 2x
(d) yp = 1 yp1 − 2yp2 =
2 5
2 + 4x 37. (a) Since D2 x = 0, x and 1 are solutions of y = 0. Since they are linearly independent, the general solution
is y = c1 x + c2 .
(b) Since D3 x2 = 0, x2 , x, and 1 are solutions of y
solution is y = c1 x2 + c2 x + c3 . = 0. Since they are linearly independent, the general (c) Since D4 x3 = 0, x3 , x2 , x, and 1 are solutions of y (4) = 0. Since they are linearly independent, the general
solution is y = c1 x3 + c2 x2 + c3 x + c4 .
(d) By part (a), the general solution of y = 0 is yc = c1 x + c2 . Since D2 x2 = 2! = 2, yp = x2 is a particular
solution of y = 2. Thus, the general solution is y = c1 x + c2 + x2 .
(e) By part (b), the general solution of y = 0 is yc = c1 x2 + c2 x + c3 . Since D3 x3 = 3! = 6, yp = x3 is a
particular solution of y = 6. Thus, the general solution is y = c1 x2 + c2 x + c3 + x3 .
(f ) By part (c), the general solution of y (4) = 0 is yc = c1 x3 + c2 x2 + c3 x + c4 . Since D4 x4 = 4! = 24, yp = x4
is a particular solution of y (4) = 24. Thus, the general solution is y = c1 x3 + c2 x2 + c3 x + c4 + x4 .
38. By the superposition principle, if y1 = ex and y2 = e−x are both solutions of a homogeneous linear diﬀerential
equation, then so are
1
ex + e−x
(y1 + y2 ) =
= cosh x and
2
2 102 ex − e−x
1
(y1 − y2 ) =
= sinh x.
2
2 3.2 Reduction of Order 39. (a) From the graphs of y1 = x3 and y2 = |x|3 we see
that the functions are linearly independent since they
cannot be multiples of each other. It is easily shown
that y1 = x3 is a solution of x2 y − 4xy + 6y = 0. To
show that y2 = |x|3 is a solution let y2 = x3 for x ≥ 0
and let y2 = −x3 for x < 0.
(b) If x ≥ 0 then y2 = x3 and
W (y1 , y2 ) =
If x < 0 then y2 = −x3 and
W (y1 , y2 ) = x3
3x2 3x2 x3 −x3 2 3x x3 −3x2 = 0. = 0. This does not violate Theorem 3.3 since a2 (x) = x2 is zero at x = 0.
(c) The functions Y1 = x3 and Y2 = x2 are solutions of x2 y − 4xy + 6y = 0. They are linearly independent
since W x3 , x2 = x4 = 0 for −∞ < x < ∞.
(d) The function y = x3 satisﬁes y(0) = 0 and y (0) = 0.
(e) Neither is the general solution on (−∞, ∞) since we form a general solution on an interval for which
a2 (x) = 0 for every x in the interval.
40. Since ex−3 = e−3 ex = (e−5 e2 )ex = e−5 ex+2 , we see that ex−3 is a constant multiple of ex+2 and the set of
functions is linearly dependent.
41. Since 0y1 + 0y2 + · · · + 0yk + 1yk+1 = 0, the set of solutions is linearly dependent.
42. The set of solutions is linearly dependent. Suppose n of the solutions are linearly independent (if not, then the
set of n + 1 solutions is linearly dependent). Without loss of generality, let this set be y1 , y2 , . . . , yn . Then
y = c1 y1 + c2 y2 + · · · + cn yn is the general solution of the nth-order diﬀerential equation and for some choice,
c∗ , c∗ , . . . , c∗ , of the coeﬃcients yn+1 = c∗ y1 + c∗ y2 + · · · + c∗ yn . But then the set y1 , y2 , . . . , yn , yn+1 is
n
n
1
2
1
2
linearly dependent. EXERCISES 3.2
Reduction of Order In Problems 1-8 we use reduction of order to ﬁnd a second solution. In Problems 9-16 we use formula (5) from the
text.
1. Deﬁne y = u(x)e2x so
y = 2ue2x + u e2x , y = e2x u + 4e2x u + 4e2x u, and y − 4y + 4y = e2x u = 0. Therefore u = 0 and u = c1 x + c2 . Taking c1 = 1 and c2 = 0 we see that a second solution is y2 = xe2x . 103 3.2 Reduction of Order 2. Deﬁne y = u(x)xe−x so
y = (1 − x)e−x u + xe−x u , y = xe−x u + 2(1 − x)e−x u − (2 − x)e−x u, and
y + 2y + y = e−x (xu + 2u ) = 0 or u + If w = u we obtain the linear ﬁrst-order equation w +
e2 dx/x = x2 . Now d 2
[x w] = 0
dx gives 2
u = 0.
x 2
w = 0 which has the integrating factor
x x2 w = c. Therefore w = u = c/x2 and u = c1 /x. A second solution is y2 = 1 −x
xe = e−x .
x 3. Deﬁne y = u(x) cos 4x so
y = −4u sin 4x + u cos 4x, y = u cos 4x − 8u sin 4x − 16u cos 4x and
y + 16y = (cos 4x)u − 8(sin 4x)u = 0 or u − 8(tan 4x)u = 0. If w = u we obtain the linear ﬁrst-order equation w − 8(tan 4x)w = 0 which has the integrating factor
e−8 tan 4x dx = cos2 4x. Now
d
[(cos2 4x)w] = 0
dx gives (cos2 4x)w = c. Therefore w = u = c sec2 4x and u = c1 tan 4x. A second solution is y2 = tan 4x cos 4x = sin 4x.
4. Deﬁne y = u(x) sin 3x so
y = 3u cos 3x + u sin 3x, y = u sin 3x + 6u cos 3x − 9u sin 3x, and
y + 9y = (sin 3x)u + 6(cos 3x)u = 0 or u + 6(cot 3x)u = 0. If w = u we obtain the linear ﬁrst-order equation w + 6(cot 3x)w = 0 which has the integrating factor
e6 cot 3x dx = sin2 3x. Now d
[(sin2 3x)w] = 0 gives (sin2 3x)w = c.
dx
Therefore w = u = c csc2 3x and u = c1 cot 3x. A second solution is y2 = cot 3x sin 3x = cos 3x. 5. Deﬁne y = u(x) cosh x so
y = u sinh x + u cosh x, y = u cosh x + 2u sinh x + u cosh x and
y − y = (cosh x)u + 2(sinh x)u = 0 or u + 2(tanh x)u = 0. If w = u we obtain the linear ﬁrst-order equation w + 2(tanh x)w = 0 which has the integrating factor
e2 tanh x dx = cosh2 x. Now d
[(cosh2 x)w] = 0
dx gives (cosh2 x)w = c. Therefore w = u = c sech2 x and u = c tanh x. A second solution is y2 = tanh x cosh x = sinh x.
6. Deﬁne y = u(x)e5x so
y = 5e5x u + e5x u , y = e5x u + 10e5x u + 25e5x u 104 3.2 Reduction of Order and
y − 25y = e5x (u + 10u ) = 0 or u + 10u = 0. If w = u we obtain the linear ﬁrst-order equation w + 10w = 0 which has the integrating factor
e10 dx = e10x . Now Therefore w = u = ce−10x d 10x
[e w] = 0 gives e10x w = c.
dx
and u = c1 e−10x . A second solution is y2 = e−10x e5x = e−5x . 7. Deﬁne y = u(x)e2x/3 so
y = 2 2x/3
e
u + e2x/3 u ,
3 4
4
y = e2x/3 u + e2x/3 u + e2x/3 u
3
9 and
9y − 12y + 4y = 9e2x/3 u = 0.
Therefore u = 0 and u = c1 x + c2 . Taking c1 = 1 and c2 = 0 we see that a second solution is y2 = xe2x/3 .
8. Deﬁne y = u(x)ex/3 so
y = 1 x/3
e u + ex/3 u ,
3 2
1
y = ex/3 u + ex/3 u + ex/3 u
3
9 and 5
or u + u = 0.
6
5
If w = u we obtain the linear ﬁrst-order equation w + 6 w = 0 which has the integrating factor
6y + y − y = ex/3 (6u + 5u ) = 0 e(5/6) dx = e5x/6 . Now d 5x/6
w] = 0
[e
dx gives e5x/6 w = c. Therefore w = u = ce−5x/6 and u = c1 e−5x/6 . A second solution is y2 = e−5x/6 ex/3 = e−x/2 .
9. Identifying P (x) = −7/x we have
y2 = x4 e− (−7/x) dx dx = x4 x8 1
dx = x4 ln |x|.
x A second solution is y2 = x4 ln |x|.
10. Identifying P (x) = 2/x we have
y2 = x2 e− (2/x) dx x4 dx = x2 1
x−6 dx = − x−3 .
5 A second solution is y2 = x−3 .
11. Identifying P (x) = 1/x we have
y2 = ln x e− dx/x
dx = ln x
(ln x)2 dx
1
= ln x −
2
x(ln x)
ln x = −1. A second solution is y2 = 1.
12. Identifying P (x) = 0 we have
y2 = x1/2 ln x e− 0 dx
1
dx = x1/2 ln x −
2
x(ln x)
ln x A second solution is y2 = x1/2 . 105 = −x1/2 . 3.2 Reduction of Order 13. Identifying P (x) = −1/x we have
e− −dx/x
dx = x sin(ln x)
x2 sin2 (ln x) y2 = x sin(ln x) x
dx
x2 sin2 (ln x) csc2 (ln x)
dx = [x sin(ln x)] [− cot(ln x)] = −x cos(ln x).
x = x sin(ln x)
A second solution is y2 = x cos(ln x).
14. Identifying P (x) = −3/x we have
y2 = x2 cos(ln x)
= x2 cos(ln x) e− −3 dx/x
dx = x2 cos(ln x)
x4 cos2 (ln x) x4 x3
dx
cos2 (ln x) sec2 (ln x)
dx = x2 cos(ln x) tan(ln x) = x2 sin(ln x).
x A second solution is y2 = x2 sin(ln x).
15. Identifying P (x) = 2(1 + x)/ 1 − 2x − x2 we have
y2 = (x + 1)
= (x + 1) e− 2(1+x)dx/(1−2x−x2 ) dx = (x + 1) (x + 1)2 1 − 2x − x2
dx = (x + 1)
(x + 1)2 = (x + 1) − 2
eln(1−2x−x )
dx
(x + 1)2 2
− 1 dx
(x + 1)2 2
− x = −2 − x2 − x.
x+1 A second solution is y2 = x2 + x + 2.
16. Identifying P (x) = −2x/ 1 − x2 we have
y2 = e− −2x dx/(1−x2 ) dx = 2
e− ln(1−x ) dx = 1+x
1
1
.
dx = ln
2
1−x
2
1−x A second solution is y2 = ln |(1 + x)/(1 − x)|.
17. Deﬁne y = u(x)e−2x so
y = −2ue−2x + u e−2x , y = u e−2x − 4u e−2x + 4ue−2x and
y − 4y = e−2x u − 4e−2x u = 0 or u − 4u = 0. If w = u we obtain the linear ﬁrst-order equation w − 4w = 0 which has the integrating factor
e−4 dx = e−4x . Now d −4x
w] = 0 gives e−4x w = c.
[e
dx
Therefore w = u = ce4x and u = c1 e4x . A second solution is y2 = e−2x e4x = e2x . We see by observation that a
particular solution is yp = −1/2. The general solution is
1
y = c1 e−2x + c2 e2x − .
2
18. Deﬁne y = u(x) · 1 so
y =u, y =u and y + y = u + u = 1. 106 3.2 Reduction of Order If w = u we obtain the linear ﬁrst-order equation w + w = 1 which has the integrating factor e
d x
[e w] = ex
dx gives dx = ex . Now ex w = ex + c. Therefore w = u = 1 + ce−x and u = x + c1 e−x + c2 . The general solution is
y = u = x + c1 e−x + c2 .
19. Deﬁne y = u(x)ex so
y = uex + u ex , y = u ex + 2u ex + uex and
y − 3y + 2y = ex u − ex u = 5e3x .
If w = u we obtain the linear ﬁrst-order equation w − w = 5e2x which has the integrating factor e−
Now
d −x
[e w] = 5ex gives e−x w = 5ex + c1 .
dx dx = e−x . Therefore w = u = 5e2x + c1 ex and u = 5 e2x + c1 ex + c2 . The general solution is
2
y = uex = 5 3x
e + c1 e2x + c2 ex .
2 20. Deﬁne y = u(x)ex so
y = uex + u ex , y = u ex + 2u ex + uex and
y − 4y + 3y = ex u − ex u = x.
If w = u we obtain the linear ﬁrst-order equation w − 2w = xe−x which has the integrating factor e−
e−2x . Now
d −2x
1
1
w] = xe−3x gives e−2x w = − xe−3x − e−3x + c1 .
[e
dx
3
9
Therefore w = u = − 1 xe−x − 1 e−x + c1 e2x and u =
3
9
y = uex = 1
3 2dx = xe−x + 4 e−x + c2 e2x + c3 . The general solution is
9 4
1
x + + c2 e3x + c3 ex .
3
9 21. (a) For m1 constant, let y1 = em1 x . Then y1 = m1 em1 x and y1 = m2 em1 x . Substituting into the diﬀerential
1
equation we obtain
ay1 + by1 + cy1 = am2 em1 x + bm1 em1 x + cem1 x
1
= em1 x (am2 + bm1 + c) = 0.
1
Thus, y1 = em1 x will be a solution of the diﬀerential equation whenever am2 +bm1 +c = 0. Since a quadratic
1
equation always has at least one real or complex root, the diﬀerential equation must have a solution of the
form y1 = em1 x .
(b) Write the diﬀerential equation in the form
b
c
y + y + y = 0,
a
a 107 3.2 Reduction of Order
and let y1 = em1 x be a solution. Then a second solution is given by
y2 = em1 x
= em1 x e−bx/a
dx
e2m1 x
e−(b/a+2m1 )x dx 1
em1 x e−(b/a+2m1 )x
b/a + 2m1
1
=−
e−(b/a+m1 )x .
b/a + 2m1 =− (m1 = −b/2a) Thus, when m1 = −b/2a, a second solution is given by y2 = em2 x where m2 = −b/a − m1 . When
m1 = −b/2a a second solution is given by
y2 = em1 x dx = xem1 x . (c) The functions
1 ix
(e − e−ix )
cos x =
2i
1
sinh x = (ex − e−x )
cosh x =
2
are all expressible in terms of exponential functions.
sin x = 1 ix
(e + e−ix )
2
1 x
(e + e−x )
2 22. We have y1 = 1 and y1 = 0, so xy1 − xy1 + y1 = 0 − x + x = 0 and y1 (x) = x is a solution of the diﬀerential
equation. Letting y = u(x)y1 (x) = xu(x) we get
y = xu (x) + u(x) and y = xu (x) + 2u (x). Then xy − xy + y = x2 u + 2xu − x2 u − xu + xu = x2 u − (x2 − 2x)u = 0. If we make the substitution
w = u , the linear ﬁrst-order diﬀerential equation becomes x2 w − (x2 − x)w = 0, which is separable:
dw
1
= 1−
w
dx
x
dw
1
= 1−
dx
w
x
ln w = x − ln x + c
w = c1
Then u = c1 ex /x and u = c1
second solution is ex
.
x ex dx/x. To integrate ex /x we use the series representation for ex . Thus, a y2 = xu(x) = c1 x
= c1 x
= c1 x ex
dx
x
1
1
1
1 + x + x2 + x3 + · · · dx
x
2!
3!
1
1
1
+ 1 + x + x2 + · · · dx
x
2!
3! = c1 x ln x + x + 1 2
1 3
x +
x + ···
2(2!)
3(3!) = c1 x ln x + x2 + 1 3
1 4
x +
x + ··· .
2(2!)
3(3!) An interval of deﬁnition is probably (0, ∞) because of the ln x term. 108 3.3 Homogeneous Linear Equations with Constant Coeﬃcients 23. (a) We have y = y = ex , so
xy − (x + 10)y + 10y = xex − (x + 10)ex + 10ex = 0,
and y = ex is a solution of the diﬀerential equation.
(b) By (5) in the text a second solution is
y2 = y1 e− P (x) dx
2
y1 dx = ex e x+10
x dx dx = ex e2x e (1+10/x)dx e2x dx 10 = ex ex+ln x
e2x dx = ex x10 e−x dx = ex (−3,628,800 − 3,628,800x − 1,814,400x2 − 604,800x3 − 151,200x4
− 30,240x5 − 5,040x6 − 720x7 − 90x8 − 10x9 − x10 )e−x
= −3,628,800 − 3,628,800x − 1,814,400x2 − 604,800x3 − 151,200x4
− 30,240x5 − 5,040x6 − 720x7 − 90x8 − 10x9 − x10 .
10 1
1 n
(c) By Corollary (A) of Theorem 3.2, −
y2 =
x is a solution.
10!
n!
n=0 EXERCISES 3.3 Homogeneous Linear Equations with
Constant Coefﬁcients 1. From 4m2 + m = 0 we obtain m = 0 and m = −1/4 so that y = c1 + c2 e−x/4 .
2. From m2 − 36 = 0 we obtain m = 6 and m = −6 so that y = c1 e6x + c2 e−6x .
3. From m2 − m − 6 = 0 we obtain m = 3 and m = −2 so that y = c1 e3x + c2 e−2x .
4. From m2 − 3m + 2 = 0 we obtain m = 1 and m = 2 so that y = c1 ex + c2 e2x .
5. From m2 + 8m + 16 = 0 we obtain m = −4 and m = −4 so that y = c1 e−4x + c2 xe−4x .
6. From m2 − 10m + 25 = 0 we obtain m = 5 and m = 5 so that y = c1 e5x + c2 xe5x .
7. From 12m2 − 5m − 2 = 0 we obtain m = −1/4 and m = 2/3 so that y = c1 e−x/4 + c2 e2x/3 .
√
√
√
8. From m2 + 4m − 1 = 0 we obtain m = −2 ± 5 so that y = c1 e(−2+ 5 )x + c2 e(−2− 5 )x .
9. From m2 + 9 = 0 we obtain m = 3i and m = −3i so that y = c1 cos 3x + c2 sin 3x.
√
√
√
√
10. From 3m2 + 1 = 0 we obtain m = i/ 3 and m = −i/ 3 so that y = c1 cos(x/ 3 ) + c2 (sin x/ 3 ).
11. From m2 − 4m + 5 = 0 we obtain m = 2 ± i so that y = e2x (c1 cos x + c2 sin x).
12. From 2m2 + 2m + 1 = 0 we obtain m = −1/2 ± i/2 so that
y = e−x/2 [c1 cos(x/2) + c2 sin(x/2)].
13. From 3m2 + 2m + 1 = 0 we obtain m = −1/3 ± √ 2 i/3 so that
√
√
y = e−x/3 [c1 cos( 2x/3) + c2 sin( 2x/3)]. 109 3.3 Homogeneous Linear Equations with Constant Coeﬃcients 14. From 2m2 − 3m + 4 = 0 we obtain m = 3/4 ± √ 23 i/4 so that
√
√
y = e3x/4 [c1 cos( 23x/4) + c2 sin( 23x/4)]. 15. From m3 − 4m2 − 5m = 0 we obtain m = 0, m = 5, and m = −1 so that
y = c1 + c2 e5x + c3 e−x .
√
16. From m3 − 1 = 0 we obtain m = 1 and m = −1/2 ± 3 i/2 so that
√
√
y = c1 ex + e−x/2 [c2 cos( 3x/2) + c3 sin( 3x/2)].
17. From m3 − 5m2 + 3m + 9 = 0 we obtain m = −1, m = 3, and m = 3 so that
y = c1 e−x + c2 e3x + c3 xe3x .
18. From m3 + 3m2 − 4m − 12 = 0 we obtain m = −2, m = 2, and m = −3 so that
y = c1 e−2x + c2 e2x + c3 e−3x .
19. From m3 + m2 − 2 = 0 we obtain m = 1 and m = −1 ± i so that
u = c1 et + e−t (c2 cos t + c3 sin t).
√
20. From m3 − m2 − 4 = 0 we obtain m = 2 and m = −1/2 ± 7 i/2 so that
√
√
x = c1 e2t + e−t/2 [c2 cos( 7t/2) + c3 sin( 7t/2)].
21. From m3 + 3m2 + 3m + 1 = 0 we obtain m = −1, m = −1, and m = −1 so that
y = c1 e−x + c2 xe−x + c3 x2 e−x .
22. From m3 − 6m2 + 12m − 8 = 0 we obtain m = 2, m = 2, and m = 2 so that
y = c1 e2x + c2 xe2x + c3 x2 e2x .
√
23. From m4 + m3 + m2 = 0 we obtain m = 0, m = 0, and m = −1/2 ± 3 i/2 so that
√
√
y = c1 + c2 x + e−x/2 [c3 cos( 3x/2) + c4 sin( 3x/2)].
24. From m4 − 2m2 + 1 = 0 we obtain m = 1, m = 1, m = −1, and m = −1 so that
y = c1 ex + c2 xex + c3 e−x + c4 xe−x .
√
√
25. From 16m4 + 24m2 + 9 = 0 we obtain m = ± 3 i/2 and m = ± 3 i/2 so that
√
√
√
√
y = c1 cos( 3x/2) + c2 sin( 3x/2) + c3 x cos( 3x/2) + c4 x sin( 3x/2).
√
26. From m4 − 7m2 − 18 = 0 we obtain m = 3, m = −3, and m = ± 2 i so that
√
√
y = c1 e3x + c2 e−3x + c3 cos 2x + c4 sin 2x.
27. From m5 + 5m4 − 2m3 − 10m2 + m + 5 = 0 we obtain m = −1, m = −1, m = 1, and m = 1, and m = −5 so
that
u = c1 e−r + c2 re−r + c3 er + c4 rer + c5 e−5r .
28. From 2m5 − 7m4 + 12m3 + 8m2 = 0 we obtain m = 0, m = 0, m = −1/2, and m = 2 ± 2i so that
x = c1 + c2 s + c3 e−s/2 + e2s (c4 cos 2s + c5 sin 2s). 110 3.3 Homogeneous Linear Equations with Constant Coeﬃcients 29. From m2 + 16 = 0 we obtain m = ±4i so that y = c1 cos 4x + c2 sin 4x. If y(0) = 2 and y (0) = −2 then c1 = 2,
c2 = −1/2, and y = 2 cos 4x − 1 sin 4x.
2
30. From m2 + 1 = 0 we obtain m = ±i so that y = c1 cos θ + c2 sin θ. If y(π/3) = 0 and y (π/3) = 2 then
√
1
3
c1 +
c2 = 0
2
2
√
3
1
−
c1 + c2 = 2,
2
2
√
√
so c1 = − 3, c2 = 1, and y = − 3 cos θ + sin θ.
31. From m2 − 4m − 5 = 0 we obtain m = −1 and m = 5, so that y = c1 e−x + c2 e5x . If y(1) = 0 and y (1) = 2,
then c1 e−1 + c2 e5 = 0, −c1 e−1 + 5c2 e5 = 2, so c1 = −e/3, c2 = e−5 /3, and y = − 1 e1−x + 1 e5x−5 .
3
3
32. From 4m2 − 4m − 3 = 0 we obtain m = −1/2 and m = 3/2 so that y = c1 e−x/2 + c2 e3x/2 . If y(0) = 1 and
y (0) = 5 then c1 + c2 = 1, − 1 c1 + 3 c2 = 5, so c1 = −7/4, c2 = 11/4, and y = 7 e−x/2 + 11 e3x/2 .
2
2
4
4
√
√
√
33. From m2 + m + 2 = 0 we obtain m = −1/2 ± 7 i/2 so that y = e−x/2 [c1 cos( 7 x/2) + c2 sin( 7 x/2)]. If
y(0) = 0 and y (0) = 0 then c1 = 0 and c2 = 0 so that y = 0.
34. From m2 − 2m + 1 = 0 we obtain m = 1 and m = 1 so that y = c1 ex + c2 xex . If y(0) = 5 and y (0) = 10 then
c1 = 5, c1 + c2 = 10 so c1 = 5, c2 = 5, and y = 5ex + 5xex .
35. From m3 + 12m2 + 36m = 0 we obtain m = 0, m = −6, and m = −6 so that y = c1 + c2 e−6x + c3 xe−6x . If
y(0) = 0, y (0) = 1, and y (0) = −7 then
−6c2 + c3 = 1, c1 + c2 = 0,
so c1 = 5/36, c2 = −5/36, c3 = 1/6, and y = 5
36 − 5 −6x
36 e 36c2 − 12c3 = −7, + 1 xe−6x .
6 36. From m3 + 2m2 − 5m − 6 = 0 we obtain m = −1, m = 2, and m = −3 so that
y = c1 e−x + c2 e2x + c3 e−3x .
If y(0) = 0, y (0) = 0, and y (0) = 1 then
c1 + c2 + c3 = 0, −c1 + 2c2 − 3c3 = 0, c1 + 4c2 + 9c3 = 1, so c1 = −1/6, c2 = 1/15, c3 = 1/10, and
1
1
1
y = − e−x + e2x + e−3x .
6
15
10
37. From m2 − 10m + 25 = 0 we obtain m = 5 and m = 5 so that y = c1 e5x + c2 xe5x . If y(0) = 1 and y(1) = 0
then c1 = 1, c1 e5 + c2 e5 = 0, so c1 = 1, c2 = −1, and y = e5x − xe5x .
38. From m2 + 4 = 0 we obtain m = ±2i so that y = c1 cos 2x + c2 sin 2x. If y(0) = 0 and y(π) = 0 then c1 = 0 and
y = c2 sin 2x.
39. From m2 + 1 = 0 we obtain m = ±i so that y = c1 cos x + c2 sin x and y = −c1 sin x + c2 cos x. From
y (0) = c1 (0) + c2 (1) = c2 = 0 and y (π/2) = −c1 (1) = 0 we ﬁnd c1 = c2 = 0. A solution of the boundary-value
problem is y = 0.
40. From m2 − 2m + 2 = 0 we obtain m = 1 ± i so that y = ex (c1 cos x + c2 sin x). If y(0) = 1 and y(π) = 1 then
c1 = 1 and y(π) = eπ cos π = −eπ . Since −eπ = 1, the boundary-value problem has no solution.
√
√
41. The auxiliary equation is m2 − 3 = 0 which has roots − 3 and 3 . By (10) the general solution is y =
√
√
√
√
√
√
c1 e 3x + c2 e− 3x . By (11) the general solution is y = c1 cosh 3x + c2 sinh 3x. For y = c1 e 3x + c2 e− 3x the 111 3.3 Homogeneous Linear Equations with Constant Coeﬃcients
√
√
√
initial conditions imply c1 + c2 = 1, 3c1 − 3c2 = 5. Solving for c1 and c2 we ﬁnd c1 = 1 (1 + 5 3 ) and c2 =
2
√
√
√
√
√
√
√
1
3x
y 1
+ 1 (1 − 5 3 )e− 3x . For y = c1 cosh 3x + c2 sinh 3x the initial conditions
2 (1 − 5 3 ) so√ = 2 (1 + 5 3 )e
2
√
√
√
√
imply c1 = 1, 3c2 = 5. Solving for c1 and c2 we ﬁnd c1 = 1 and c2 = 5 3 so y = cosh 3x + 5 3 sinh 3x.
3
3 42. The auxiliary equation is m2 − 1 = 0 which has roots −1 and 1. By (10) the general solution is y = c1 ex + c2 e−x .
By (11) the general solution is y = c1 cosh x + c2 sinh x. For y = c1 ex + c2 e−x the boundary conditions
imply c1 + c2 = 1, c1 e − c2 e−1 = 0. Solving for c1 and c2 we ﬁnd c1 = 1/(1 + e2 ) and c2 = e2 /(1 + e2 )
so y = ex /(1 + e2 ) + e2 e−x /(1 + e2 ). For y = c1 cosh x + c2 sinh x the boundary conditions imply c1 = 1,
c2 = − tanh 1, so y = cosh x − (tanh 1) sinh x.
43. The auxiliary equation should have two positive roots, so that the solution has the form y = c1 ek1 x + c2 ek2 x . Thus, the diﬀerential equation is (f).
44. The auxiliary equation should have one positive and one negative root, so that the solution has the form
y = c1 ek1 x + c2 e−k2 x . Thus, the diﬀerential equation is (a).
45. The auxiliary equation should have a pair of complex roots α ± βi where α < 0, so that the solution has the
form eαx (c1 cos βx + c2 sin βx). Thus, the diﬀerential equation is (e).
46. The auxiliary equation should have a repeated negative root, so that the solution has the form y =
c1 e−x + c2 xe−x . Thus, the diﬀerential equation is (c).
47. The diﬀerential equation should have the form y + k 2 y = 0 where k = 1 so that the period of the solution is
2π. Thus, the diﬀerential equation is (d).
48. The diﬀerential equation should have the form y + k 2 y = 0 where k = 2 so that the period of the solution is
π. Thus, the diﬀerential equation is (b).
49. Since (m − 4)(m + 5)2 = m3 + 6m2 − 15m − 100 the diﬀerential equation is y + 6y − 15y − 100y = 0. The
diﬀerential equation is not unique since any constant multiple of the left-hand side of the diﬀerential equation
would lead to the auxiliary roots.
50. A third root must be m3 = 3 − i and the auxiliary equation is
m+ 1
2 [m − (3 + i)][m − (3 − i)] = m+ 1
2 (m2 − 6x + 10) = m3 − 11 2
m + 7m + 5.
2 The diﬀerential equation is
y − 11
y + 7y + 5y = 0.
2 51. From the solution y1 = e−4x cos x we conclude that m1 = −4 + i and m2 = −4 − i are roots of the auxiliary
equation. Hence another solution must be y2 = e−4x sin x. Now dividing the polynomial m3 + 6m2 + m − 34 by
m − (−4 + i) m − (−4 − i) = m2 + 8m + 17 gives m − 2. Therefore m3 = 2 is the third root of the auxiliary
equation, and the general solution of the diﬀerential equation is
y = c1 e−4x cos x + c2 e−4x sin x + c3 e2x .
52. Factoring the diﬀerence of two squares we obtain
√
√
m4 + 1 = (m2 + 1)2 − 2m2 = (m2 + 1 − 2 m)(m2 + 1 + 2 m) = 0.
√
√
Using the quadratic formula on each factor we get m = ± 2/2± 2 i/2. The solution of the diﬀerential equation
is
√
√
√
√
√
√
2
2
2
2
2 x/2
− 2 x/2
y(x) = e
c1 cos
c3 cos
x + c2 sin
x +e
x + c4 sin
x .
2
2
2
2 112 3.3 Homogeneous Linear Equations with Constant Coeﬃcients 53. Using the deﬁnition of sinh x and the formula for the cosine of the sum of two angles, we have
y = sinh x − 2 cos(x + π/6)
1
1
π
π
= ex − e−x − 2 (cos x) cos
− (sin x) sin
2
2
6
6
√
1
3
1
1
= ex − e−x − 2
cos x − sin x
2
2
2
2
= 1 x 1 −x √
e − e − 3 cos x + sin x.
2
2 This form of the solution can be obtained from the general solution y = c1 ex + c2 e−x + c3 cos x + c4 sin x by
√
choosing c1 = 1 , c2 = − 1 , c3 = − 3 , and c4 = 1.
2
2
54. The auxiliary equation is m2 + α = 0 and we consider three cases where λ = 0, λ = α2 > 0, and λ = −α2 < 0:
Case I When α = 0 the general solution of the diﬀerential equation is y = c1 + c2 x. The boundary conditions
imply 0 = y(0) = c1 and 0 = y(π/2) = c2 π/2, so that c1 = c2 = 0 and the problem possesses only the trivial
solution.
Case II When λ = −α2 < 0 the general solution of the diﬀerential equation is y = c1 eα x + c2 e−α x , or
alternatively, y = c1 cosh α x + c2 sinh α x. Again, y(0) = 0 implies c1 = 0 so y = c2 sinh α x. The second
boundary condition implies 0 = y(π/2) = c2 sinh α π/2 or c2 = 0. In this case also, the problem possesses only
the trivial solution.
When λ = α2 > 0 the general solution of the diﬀerential equation is y = c1 cos α x + c2 sin α x.
In this case also, y(0) = 0 yields c1 = 0, so that y = c2 sin α x. The second boundary condition implies
0 = c2 sin α π/2. When α π/2 is an integer multiple of π, that is, when α = 2k for k a nonzero integer, the Case III problem will have nontrivial solutions. Thus, for λ = α2 = 4k 2 the boundary-value problem will have nontrivial
solutions y = c2 sin 2kx, where k is a nonzero integer. On the other hand, when α is not an even integer, the
boundary-value problem will have only the trivial solution.
55. Applying integration by parts twice we have
1 ax
e f (x) −
a
1
= eax f (x) −
a
1
= eax f (x) −
a eax f (x) dx = 1
eax f (x) dx
a
1 1 ax
1
e f (x) −
eax f (x) dx
a a
a
1 ax
1
eax f (x) dx.
e f (x) + 2
a2
a Collecting the integrals we get
eax f (x) − 1
f (x)
a2 dx = 1 ax
1
e f (x) − 2 eax f (x).
a
a In order for the technique to work we need to have
eax f (x) − 1
f (x)
a2 dx = k eax f (x) dx or 1
f (x) = kf (x),
a2
where k = 0. This is the second-order diﬀerential equation
f (x) − f (x) + a2 (k − 1)f (x) = 0. 113 3.3 Homogeneous Linear Equations with Constant Coeﬃcients If k < 1, k = 0, the solution of the diﬀerential equation is a pair of exponential functions, in which case
the original integrand is an exponential function and does not require integration by parts for its evaluation.
Similarly, if k = 1, f (x) = 0 and f (x) has the form f (x) = ax+b. In this case a single application of integration
by parts will suﬃce. Finally, if k > 1, the solution of the diﬀerential equation is
√
√
f (x) = c1 cos a k − 1 x + c2 sin a k − 1 x,
and we see that the technique will work for linear combinations of cos αx and sin αx.
√
56. (a) The auxiliary equation is m2 − 64/L = 0 which has roots ±8/ L . Thus, the general solution of the
√
√
diﬀerential equation is x = c1 cosh(8t/ L ) + c2 sinh(8t/ L ).
√
(b) Setting x(0) = x0 and x (0) = 0 we have c1 = x0 , 8c2 / L = 0. Solving for c1 and c2 we get c1 = x0 and
√
c2 = 0, so x(t) = x0 cosh(8t/ L ).
√
(c) When L = 20 and x0 = 1, x(t) = cosh(4t/ 5 ). The chain will last touch the peg when x(t) = 10.
√
Solving x(t) = 10 for t we get t1 = 1 5 cosh−1 10 ≈ 1.67326. The velocity of the chain at this instant is
4
x (t1 ) = 12 11/5 ≈ 17.7989 ft/s.
57. Using a CAS to solve the auxiliary equation m3 − 6m2 + 2m + 1 we ﬁnd m1 = −0.270534,
m2 = 0.658675, and m3 = 5.61186. The general solution is
y = c1 e−0.270534x + c2 e0.658675x + c3 e5.61186x .
58. Using a CAS to solve the auxiliary equation 6.11m3 + 8.59m2 + 7.93m + 0.778 = 0 we ﬁnd m1 = −0.110241,
m2 = −0.647826 + 0.857532i, and m3 = −0.647826 − 0.857532i. The general solution is
y = c1 e−0.110241x + e−0.647826x (c2 cos 0.857532x + c3 sin 0.857532x).
59. Using a CAS to solve the auxiliary equation 3.15m4 − 5.34m2 + 6.33m − 2.03 = 0 we ﬁnd m1 = −1.74806,
m2 = 0.501219, m3 = 0.62342 + 0.588965i, and m4 = 0.62342 − 0.588965i. The general solution is
y = c1 e−1.74806x + c2 e0.501219x + e0.62342x (c3 cos 0.588965x + c4 sin 0.588965x).
√
√
60. Using a CAS to solve the auxiliary equation m4 +2m2 −m+2 = 0 we ﬁnd m1 = 1/2+ 3 i/2, m2 = 1/2− 3 i/2,
√
√
m3 = −1/2 + 7 i/2, and m4 = −1/2 − 7 i/2. The general solution is
√
√
√
√
3
3
7
7
x/2
−x/2
y=e
c1 cos
c3 cos
x + c2 sin
x +e
x + c4 sin
x .
2
2
2
2
61. From 2m4 + 3m3 − 16m2 + 15m − 4 = 0 we obtain m = −4, m = 1 , m = 1, and m = 1, so that
2
y = c1 e−4x + c2 ex/2 + c3 ex + c4 xex . If y(0) = −2, y (0) = 6, y (0) = 3, and y (0) = 1 , then
2
c1 + c2 + c3 = −2
1
−4c1 + c2 + c3 + c4 = 6
2
1
16c1 + c2 + c3 + 2c4 = 3
4
1
1
−64c1 + c2 + c3 + 3c4 = ,
8
2
4
so c1 = − 75 , c2 = − 116 , c3 =
3 918
25 , c4 = − 58 , and
5
y=− 4 −4x 116 x/2 918 x 58 x
−
+
e
e
e − xe .
75
3
25
5 114 3.4 Undetermined Coeﬃcients 62. From m4 −3m3 +3m2 −m = 0 we obtain m = 0, m = 1, m = 1, and m = 1 so that y = c1 +c2 ex +c3 xex +c4 x2 ex .
If y(0) = 0, y (0) = 0, y (0) = 1, and y (0) = 1 then
c1 + c2 = 0, c2 + c3 = 0, c2 + 2c3 + 2c4 = 1, c2 + 3c3 + 6c4 = 1, so c1 = 2, c2 = −2, c3 = 2, c4 = −1/2, and
1
y = 2 − 2ex + 2xex − x2 ex .
2 EXERCISES 3.4
Undetermined Coefﬁcients
1. From m2 + 3m + 2 = 0 we ﬁnd m1 = −1 and m2 = −2. Then yc = c1 e−x + c2 e−2x and we assume yp = A.
Substituting into the diﬀerential equation we obtain 2A = 6. Then A = 3, yp = 3 and
y = c1 e−x + c2 e−2x + 3.
2. From 4m2 + 9 = 0 we ﬁnd m1 = − 3 i and m2 = 3 i. Then yc = c1 cos 3 x + c2 sin 3 x and we assume yp = A.
2
2
2
2
Substituting into the diﬀerential equation we obtain 9A = 15. Then A = 5 , yp = 5 and
3
3
3
3
5
y = c1 cos x + c2 sin x + .
2
2
3
3. From m2 − 10m + 25 = 0 we ﬁnd m1 = m2 = 5. Then yc = c1 e5x + c2 xe5x and we assume yp = Ax + B.
Substituting into the diﬀerential equation we obtain 25A = 30 and −10A + 25B = 3. Then A = 6 , B = 6 ,
5
5
yp = 6 x + 6 , and
5
5
6
6
y = c1 e5x + c2 xe5x + x + .
5
5
4. From m2 + m − 6 = 0 we ﬁnd m1 = −3 and m2 = 2. Then yc = c1 e−3x + c2 e2x and we assume yp = Ax + B.
1
Substituting into the diﬀerential equation we obtain −6A = 2 and A − 6B = 0. Then A = − 1 , B = − 18 ,
3
1
yp = − 1 x − 18 , and
3
1
1
y = c1 e−3x + c2 e2x − x −
.
3
18
5. From 1 m2 + m + 1 = 0 we ﬁnd m1 = m2 = −2. Then yc = c1 e−2x + c2 xe−2x and we assume yp = Ax2 + Bx + C.
4
Substituting into the diﬀerential equation we obtain A = 1, 2A + B = −2, and 1 A + B + C = 0. Then A = 1,
2
B = −4, C = 7 , yp = x2 − 4x + 7 , and
2
2
y = c1 e−2x + c2 xe−2x + x2 − 4x + 7
.
2 6. From m2 − 8m + 20 = 0 we ﬁnd m1 = 4 + 2i and m2 = 4 − 2i. Then yc = e4x (c1 cos 2x + c2 sin 2x) and we
assume yp = Ax2 + Bx + C + (Dx + E)ex . Substituting into the diﬀerential equation we obtain
2A − 8B + 20C = 0
−6D + 13E = 0
−16A + 20B = 0
13D = −26
20A = 100. 115 3.4 Undetermined Coeﬃcients Then A = 5, B = 4, C = 11
10 , D = −2, E = − 12 , yp = 5x2 + 4x +
13 11
10 + −2x − 12
13 ex and 11
12 x
+ −2x −
e .
10
13
√
√
√
√
7. From m2 + 3 = 0 we ﬁnd m1 = 3 i and m2 = − 3 i. Then yc = c1 cos 3 x + c2 sin 3 x and we assume
yp = (Ax2 +Bx+C)e3x . Substituting into the diﬀerential equation we obtain 2A+6B +12C = 0, 12A+12B = 0,
and 12A = −48. Then A = −4, B = 4, C = − 4 , yp = −4x2 + 4x − 4 e3x and
3
3
√
√
4 3x
y = c1 cos 3 x + c2 sin 3 x + −4x2 + 4x −
e .
3
y = e4x (c1 cos 2x + c2 sin 2x) + 5x2 + 4x + 8. From 4m2 − 4m − 3 = 0 we ﬁnd m1 = 3 and m2 = − 1 . Then yc = c1 e3x/2 + c2 e−x/2 and we assume
2
2
yp = A cos 2x + B sin 2x. Substituting into the diﬀerential equation we obtain −19 − 8B = 1 and 8A − 19B = 0.
19
8
19
8
Then A = − 425 , B = − 425 , yp = − 425 cos 2x − 425 sin 2x, and
y = c1 e3x/2 + c2 e−x/2 − 19
8
cos 2x −
sin 2x.
425
425 9. From m2 − m = 0 we ﬁnd m1 = 1 and m2 = 0. Then yc = c1 ex + c2 and we assume yp = Ax. Substituting into
the diﬀerential equation we obtain −A = −3. Then A = 3, yp = 3x and y = c1 ex + c2 + 3x.
10. From m2 +2m = 0 we ﬁnd m1 = −2 and m2 = 0. Then yc = c1 e−2x +c2 and we assume yp = Ax2 +Bx+Cxe−2x .
Substituting into the diﬀerential equation we obtain 2A + 2B = 5, 4A = 2, and −2C = −1. Then A = 1 ,
2
B = 2, C = 1
2 , yp = 1 x2 + 2x + 1 xe−2x , and
2
2 1
1
y = c1 e−2x + c2 + x2 + 2x + xe−2x .
2
2
11. From m2 − m + 1
4 = 0 we ﬁnd m1 = m2 = 1 . Then yc = c1 ex/2 + c2 xex/2 and we assume yp = A + Bx2 ex/2 .
2
Substituting into the diﬀerential equation we obtain 1 A = 3 and 2B = 1. Then A = 12, B = 1 , yp =
4
2
12 + 1 x2 ex/2 , and
2
1
y = c1 ex/2 + c2 xex/2 + 12 + x2 ex/2 .
2 12. From m2 − 16 = 0 we ﬁnd m1 = 4 and m2 = −4. Then yc = c1 e4x + c2 e−4x and we assume yp = Axe4x .
Substituting into the diﬀerential equation we obtain 8A = 2. Then A = 1 , yp = 1 xe4x and
4
4
1
y = c1 e4x + c2 e−4x + xe4x .
4
13. From m2 + 4 = 0 we ﬁnd m1 = 2i and m2 = −2i. Then yc = c1 cos 2x + c2 sin 2x and we assume
yp = Ax cos 2x + Bx sin 2x. Substituting into the diﬀerential equation we obtain 4B = 0 and −4A = 3.
Then A = − 3 , B = 0, yp = − 3 x cos 2x, and
4
4
3
y = c1 cos 2x + c2 sin 2x − x cos 2x.
4
14. From m2 − 4 = 0 we ﬁnd m1 = 2 and m2 = −2.
2 Then yc = c1 e2x + c2 e−2x and we assume that 2 yp = (Ax + Bx + C) cos 2x + (Dx + Ex + F ) sin 2x. Substituting into the diﬀerential equation we obtain
−8A = 0
−8B + 8D = 0
2A − 8C + 4E = 0
−8D = 1
−8A − 8E = 0
−4B + 2D − 8F = −3. 116 3.4
Then A = 0, B = − 1 , C = 0, D = − 1 , E = 0, F =
8
8
y = c1 e2x + c2 e−2x − 13
32 Undetermined Coeﬃcients , so yp = − 1 x cos 2x + − 1 x2 +
8
8 1
13
1
x cos 2x + − x2 +
8
8
32 13
32 sin 2x, and sin 2x. 15. From m2 + 1 = 0 we ﬁnd m1 = i and m2 = −i. Then yc = c1 cos x + c2 sin x and we assume
yp = (Ax2 + Bx) cos x + (Cx2 + Dx) sin x. Substituting into the diﬀerential equation we obtain 4C = 0,
2A + 2D = 0, −4A = 2, and −2B + 2C = 0. Then A = − 1 , B = 0, C = 0, D = 1 , yp = − 1 x2 cos x + 1 x sin x,
2
2
2
2
and
1
1
y = c1 cos x + c2 sin x − x2 cos x + x sin x.
2
2
16. From m2 −5m = 0 we ﬁnd m1 = 5 and m2 = 0. Then yc = c1 e5x +c2 and we assume yp = Ax4 +Bx3 +Cx2 +Dx.
Substituting into the diﬀerential equation we obtain −20A = 2, 12A − 15B = −4, 6B − 10C = −1, and
1
53
1
53
2C − 5D = 6. Then A = − 10 , B = 14 , C = 250 , D = − 697 , yp = − 10 x4 + 14 x3 + 250 x2 − 697 x, and
75
625
75
625
y = c1 e5x + c2 − 1 4 14 3
53 2 697
x + x +
x −
x.
10
75
250
625 17. From m2 − 2m + 5 = 0 we ﬁnd m1 = 1 + 2i and m2 = 1 − 2i. Then yc = ex (c1 cos 2x + c2 sin 2x) and we assume
yp = Axex cos 2x + Bxex sin 2x. Substituting into the diﬀerential equation we obtain 4B = 1 and −4A = 0.
Then A = 0, B = 1 , yp = 1 xex sin 2x, and
4
4
1
y = ex (c1 cos 2x + c2 sin 2x) + xex sin 2x.
4
18. From m2 − 2m + 2 = 0 we ﬁnd m1 = 1 + i and m2 = 1 − i. Then yc = ex (c1 cos x + c2 sin x) and we assume
yp = Ae2x cos x+Be2x sin x. Substituting into the diﬀerential equation we obtain A+2B = 1 and −2A+B = −3.
Then A = 7
5 , B = − 1 , yp = 7 e2x cos x − 1 e2x sin x and
5
5
5
7
1
y = ex (c1 cos x + c2 sin x) + e2x cos x − e2x sin x.
5
5 19. From m2 +2m+1 = 0 we ﬁnd m1 = m2 = −1. Then yc = c1 e−x +c2 xe−x and we assume yp = A cos x+B sin x+
C cos 2x + D sin 2x. Substituting into the diﬀerential equation we obtain 2B = 0, −2A = 1, −3C + 4D = 3, and
9
−4C − 3D = 0. Then A = − 1 , B = 0, C = − 25 , D =
2 y = c1 e−x + c2 xe−x − 12
25 , yp = − 1 cos x −
2 9
25 cos 2x + 12
25 sin 2x, and 1
9
12
cos x −
cos 2x +
sin 2x.
2
25
25 20. From m2 + 2m − 24 = 0 we ﬁnd m1 = −6 and m2 = 4. Then yc = c1 e−6x + c2 e4x and we assume
yp = A + (Bx2 + Cx)e4x . Substituting into the diﬀerential equation we obtain −24A = 16, 2B + 10C = −2,
1
19
1
19
and 20B = −1. Then A = − 2 , B = − 20 , C = − 100 , yp = − 2 − 20 x2 + 100 x e4x , and
3
3
y = c1 e−6x + c2 e4x − 2
−
3 19
1 2
x +
x e4x .
20
100 21. From m3 − 6m2 = 0 we ﬁnd m1 = m2 = 0 and m3 = 6. Then yc = c1 + c2 x + c3 e6x and we assume
yp = Ax2 + B cos x + C sin x. Substituting into the diﬀerential equation we obtain −12A = 3, 6B − C = −1,
6
1
6
1
and B + 6C = 0. Then A = − 1 , B = − 37 , C = 37 , yp = − 1 x2 − 37 cos x + 37 sin x, and
4
4
1
6
1
y = c1 + c2 x + c3 e6x − x2 −
cos x +
sin x.
4
37
37 117 3.4 Undetermined Coeﬃcients 22. From m3 − 2m2 − 4m + 8 = 0 we ﬁnd m1 = m2 = 2 and m3 = −2. Then yc = c1 e2x + c2 xe2x + c3 e−2x and we
assume yp = (Ax3 + Bx2 )e2x . Substituting into the diﬀerential equation we obtain 24A = 6 and 6A + 8B = 0.
3
3
Then A = 1 , B = − 16 , yp = 1 x3 − 16 x2 e2x , and
4
4
y = c1 e2x + c2 xe2x + c3 e−2x + 1 3
3
x − x2 e2x .
4
16 23. From m3 − 3m2 + 3m − 1 = 0 we ﬁnd m1 = m2 = m3 = 1. Then yc = c1 ex + c2 xex + c3 x2 ex and we assume
yp = Ax + B + Cx3 ex . Substituting into the diﬀerential equation we obtain −A = 1, 3A − B = 0, and 6C = −4.
Then A = −1, B = −3, C = − 2 , yp = −x − 3 − 2 x3 ex , and
3
3
2
y = c1 ex + c2 xex + c3 x2 ex − x − 3 − x3 ex .
3
24. From m3 − m2 − 4m + 4 = 0 we ﬁnd m1 = 1, m2 = 2, and m3 = −2. Then yc = c1 ex + c2 e2x + c3 e−2x and we
assume yp = A + Bxex + Cxe2x . Substituting into the diﬀerential equation we obtain 4A = 5, −3B = −1, and
4C = 1. Then A = 5 , B = 1 , C = 1 , yp = 5 + 1 xex + 1 xe2x , and
4
3
4
4
3
4
y = c1 ex + c2 e2x + c3 e−2x + 5 1 x 1 2x
+ xe + xe .
4 3
4 25. From m4 +2m2 +1 = 0 we ﬁnd m1 = m3 = i and m2 = m4 = −i. Then yc = c1 cos x+c2 sin x+c3 x cos x+c4 x sin x
and we assume yp = Ax2 + Bx + C. Substituting into the diﬀerential equation we obtain A = 1, B = −2, and
4A + C = 1. Then A = 1, B = −2, C = −3, yp = x2 − 2x − 3, and
y = c1 cos x + c2 sin x + c3 x cos x + c4 x sin x + x2 − 2x − 3.
26. From m4 − m2 = 0 we ﬁnd m1 = m2 = 0, m3 = 1, and m4 = −1. Then yc = c1 + c2 x + c3 ex + c4 e−x and
we assume yp = Ax3 + Bx2 + (Cx2 + Dx)e−x . Substituting into the diﬀerential equation we obtain −6A = 4,
−2B = 0, 10C −2D = 0, and −4C = 2. Then A = − 2 , B = 0, C = − 1 , D = − 5 , yp = − 2 x3 − 1 x2 + 5 x e−x ,
3
2
2
3
2
2
and
2
y = c1 + c2 x + c3 ex + c4 e−x − x3 −
3 1 2 5
x + x e−x .
2
2 27. We have yc = c1 cos 2x + c2 sin 2x and we assume yp = A. Substituting into the diﬀerential equation we ﬁnd
√
A = − 1 . Thus y = c1 cos 2x + c2 sin 2x − 1 . From the initial conditions we obtain c1 = 0 and c2 = 2 , so
2
2
y= √ 2 sin 2x − 1
.
2 28. We have yc = c1 e−2x + c2 ex/2 and we assume yp = Ax2 + Bx + C. Substituting into the diﬀerential equation we
ﬁnd A = −7, B = −19, and C = −37. Thus y = c1 e−2x + c2 ex/2 − 7x2 − 19x − 37. From the initial conditions
we obtain c1 = − 1 and c2 =
5 186
5 , so
1
186 x/2
y = − e−2x +
− 7x2 − 19x − 37.
e
5
5 29. We have yc = c1 e−x/5 + c2 and we assume yp = Ax2 + Bx. Substituting into the diﬀerential equation we ﬁnd
A = −3 and B = 30. Thus y = c1 e−x/5 + c2 − 3x2 + 30x. From the initial conditions we obtain c1 = 200 and
c2 = −200, so
y = 200e−x/5 − 200 − 3x2 + 30x. 118 3.4 Undetermined Coeﬃcients 30. We have yc = c1 e−2x +c2 xe−2x and we assume yp = (Ax3 +Bx2 )e−2x . Substituting into the diﬀerential equation
we ﬁnd A = 1 and B = 3 . Thus y = c1 e−2x + c2 xe−2x + 1 x3 + 3 x2 e−2x . From the initial conditions we
6
2
6
2
obtain c1 = 2 and c2 = 9, so
1 3 3 2 −2x
y = 2e−2x + 9xe−2x +
x + x e
.
6
2
31. We have yc = e−2x (c1 cos x + c2 sin x) and we assume yp = Ae−4x . Substituting into the diﬀerential equation
we ﬁnd A = 7. Thus y = e−2x (c1 cos x + c2 sin x) + 7e−4x . From the initial conditions we obtain c1 = −10 and
c2 = 9, so
y = e−2x (−10 cos x + 9 sin x) + 7e−4x .
32. We have yc = c1 cosh x + c2 sinh x and we assume yp = Ax cosh x + Bx sinh x. Substituting into the diﬀerential
equation we ﬁnd A = 0 and B = 1 . Thus
2
1
y = c1 cosh x + c2 sinh x + x sinh x.
2
From the initial conditions we obtain c1 = 2 and c2 = 12, so
1
y = 2 cosh x + 12 sinh x + x sinh x.
2
33. We have xc = c1 cos ωt + c2 sin ωt and we assume xp = At cos ωt + Bt sin ωt. Substituting into the diﬀerential
equation we ﬁnd A = −F0 /2ω and B = 0. Thus x = c1 cos ωt + c2 sin ωt − (F0 /2ω)t cos ωt. From the initial
conditions we obtain c1 = 0 and c2 = F0 /2ω 2 , so
x = (F0 /2ω 2 ) sin ωt − (F0 /2ω)t cos ωt.
34. We have xc = c1 cos ωt + c2 sin ωt and we assume xp = A cos γt + B sin γt, where γ = ω. Substituting into the
diﬀerential equation we ﬁnd A = F0 /(ω 2 − γ 2 ) and B = 0. Thus
x = c1 cos ωt + c2 sin ωt + F0
cos γt.
ω2 − γ 2 From the initial conditions we obtain c1 = −F0 /(ω 2 − γ 2 ) and c2 = 0, so
x=− F0
F0
cos ωt + 2
cos γt.
ω2 − γ 2
ω − γ2 35. We have yc = c1 + c2 ex + c3 xex and we assume yp = Ax + Bx2 ex + Ce5x . Substituting into the diﬀerential
equation we ﬁnd A = 2, B = −12, and C = 1 . Thus
2
1
y = c1 + c2 ex + c3 xex + 2x − 12x2 ex + e5x .
2
From the initial conditions we obtain c1 = 11, c2 = −11, and c3 = 9, so
1
y = 11 − 11ex + 9xex + 2x − 12x2 ex + e5x .
2
√
√
36. We have yc = c1 e−2x + ex (c2 cos 3 x + c3 sin 3 x) and we assume yp = Ax + B + Cxe−2x . Substituting into
the diﬀerential equation we ﬁnd A = 1 , B = − 5 , and C = 2 . Thus
4
8
3
√
5 2
1
3 x + c3 sin 3 x) + x − + xe−2x .
4
8 3
√
23
59
17
From the initial conditions we obtain c1 = − 12 , c2 = − 24 , and c3 = 72 3 , so
y = c1 e−2x + ex (c2 cos y=− √ √
√
23 −2x
17 √
1
5 2
59
+ ex − cos 3 x +
3 sin 3 x + x − + xe−2x .
e
12
24
72
4
8 3 119 3.4 Undetermined Coeﬃcients 37. We have yc = c1 cos x + c2 sin x and we assume yp = A2 + Bx + C. Substituting into the diﬀerential equation
we ﬁnd A = 1, B = 0, and C = −1. Thus y = c1 cos x + c2 sin x + x2 − 1. From y(0) = 5 and y(1) = 0 we obtain
c1 − 1 = 5
(cos 1)c1 + (sin 1)c2 = 0.
Solving this system we ﬁnd c1 = 6 and c2 = −6 cot 1. The solution of the boundary-value problem is
y = 6 cos x − 6(cot 1) sin x + x2 − 1.
38. We have yc = ex (c1 cos x + c2 sin x) and we assume yp = Ax + B. Substituting into the diﬀerential equation we
ﬁnd A = 1 and B = 0. Thus y = ex (c1 cos x + c2 sin x) + x. From y(0) = 0 and y(π) = π we obtain
c1 = 0
π − e c1 = π.
π Solving this system we ﬁnd c1 = 0 and c2 is any real number. The solution of the boundary-value problem is
y = c2 ex sin x + x.
√
√
39. The general solution of the diﬀerential equation y + 3y = 6x is y = c1 cos 3x + c2 sin 3x + 2x. The
√
condition y(0) = 0 implies c1 = 0 and so y = c2 sin 3x + 2x. The condition y(1) + y (1) = 0 implies
√
√
√
√
√
√
c2 sin 3 + 2 + c2 3 cos 3 + 2 = 0 so c2 = −4/(sin 3 + 3 cos 3 ). The solution is
√
−4 sin 3x
√
√
√ + 2x.
y=
sin 3 + 3 cos 3
√
√
40. Using the general solution y = c1 cos 3x + c2 sin 3x + 2x, the boundary conditions y(0) + y (0) = 0, y(1) = 0
yield the system
√
c1 + 3c2 + 2 = 0
√
√
c1 cos 3 + c2 sin 3 + 2 = 0.
Solving gives Thus, √
√
2(− 3 + sin 3 )
√
√
c1 = √
3 cos 3 − sin 3 and c2 = √ √
2(1 − cos 3 )
√
√ .
3 cos 3 − sin 3 √
√
√
√
√
2(− 3 + sin 3 ) cos 3x 2(1 − cos 3 ) sin 3x
√
√
√
√
√ + 2x.
y=
+ √
3 cos 3 − sin 3
3 cos 3 − sin 3 41. We have yc = c1 cos 2x + c2 sin 2x and we assume yp = A cos x + B sin x on [0, π/2]. Substituting into the
diﬀerential equation we ﬁnd A = 0 and B = 1 . Thus y = c1 cos 2x + c2 sin 2x + 1 sin x on [0, π/2]. On (π/2, ∞)
3
3
we have y = c3 cos 2x + c4 sin 2x. From y(0) = 1 and y (0) = 2 we obtain
c1 = 1
1
+ 2c2 = 2.
3
Solving this system we ﬁnd c1 = 1 and c2 = 5
6 . Thus y = cos 2x + 5
6 sin 2x + 1
3 sin x on [0, π/2]. Now continuity of y at x = π/2 implies
5
1
π
sin π + sin = c3 cos π + c4 sin π
6
3
2
2
= −c3 . Hence c3 = 3 . Continuity of y at x = π/2 implies
cos π + or −1 + 1
3 −2 sin π + 5
1
π
cos π + cos = −2c3 sin π + 2c4 cos π
3
3
2 120 3.4
or − 5 = −2c4 . Then c4 =
3 5
6 Undetermined Coeﬃcients and the solution of the initial-value problem is
cos 2x +
y(x) = 2
3 5
6 cos 2x + sin 2x +
5
6 sin 2x, 1
3 sin x, 0 ≤ x ≤ π/2
x > π/2. 42. We have yc = ex (c1 cos 3x + c2 sin 3x) and we assume yp = A on [0, π]. Substituting into the diﬀerential equation
we ﬁnd A = 2. Thus, y = ex (c1 cos 3x + c2 sin 3x) + 2 on [0, π]. On (π, ∞) we have y = ex (c3 cos 3x + c4 sin 3x).
From y(0) = 0 and y (0) = 0 we obtain
c1 = −2,
Solving this system, we ﬁnd c1 = −2 and c2 =
continuity of y at x = π implies
eπ (−2 cos 3π + 2
3 c1 + 3c2 = 0.
. Thus y = ex (−2 cos 3x + 2
3 sin 3x) + 2 on [0, π]. Now, 2
sin 3π) + 2 = eπ (c3 cos 3π + c4 sin 3π)
3 or 2 + 2eπ = −c3 eπ or c3 = −2e−π (1 + eπ ). Continuity of y at π implies
20 π
e sin 3π = eπ [(c3 + 3c4 ) cos 3π + (−3c3 + c4 ) sin 3π]
3
or −c3 eπ − 3c4 eπ = 0. Since c3 = −2e−π (1 + eπ ) we have c4 = 2 e−π (1 + eπ ). The solution of the initial-value
3
problem is
ex (−2 cos 3x + 2 sin 3x) + 2,
0≤x≤π
3
y(x) =
2
π x−π
(1 + e )e
(−2 cos 3x + 3 sin 3x), x > π.
43. (a) From yp = Aekx we ﬁnd yp = Akekx and yp = Ak 2 ekx . Substituting into the diﬀerential equation we get
aAk 2 ekx + bAkekx + cAekx = (ak 2 + bk + c)Aekx = ekx ,
so (ak 2 + bk + c)A = 1. Since k is not a root of am2 + bm + c = 0, A = 1/(ak 2 + bk + c).
(b) From yp = Axekx we ﬁnd yp = Akxekx +Aekx and yp = Ak 2 xekx +2Akekx . Substituting into the diﬀerential
equation we get
aAk 2 xekx + 2aAkekx + bAkxekx + bAekx + cAxekx
= (ak 2 + bk + c)Axekx + (2ak + b)Aekx
= (0)Axekx + (2ak + b)Aekx = (2ak + b)Aekx = ekx
where ak 2 +bk +c = 0 because k is a root of the auxiliary equation. Now, the roots of the auxiliary equation
√
are −b/2a ± b2 − 4ac /2a, and since k is a root of multiplicity one, k = −b/2a and 2ak + b = 0. Thus
(2ak + b)A = 1 and A = 1/(2ak + b).
(c) If k is a root of multiplicity two, then, as we saw in part (b), k = −b/2a and 2ak + b = 0. From yp = Ax2 ekx
we ﬁnd yp = Akx2 ekx + 2Axekx and yp = Ak 2 x2 ekx + 4Akxekx = 2Aekx . Substituting into the diﬀerential
equation, we get
aAk 2 x2 ekx + 4aAkxekx + 2aAekx + bAkx2 ekx + 2bAxekx + cAx2 ekx
= (ak 2 + bk + c)Ax2 ekx + 2(2ak + b)Axekx + 2aAekx
= (0)Ax2 ekx + 2(0)Axekx + 2aAekx = 2aAekx = ekx .
Since the diﬀerential equation is second order, a = 0 and A = 1/(2a).
44. Using the double-angle formula for the cosine, we have
sin x cos 2x = sin x(cos2 x − sin2 x) = sin x(1 − 2 sin2 x) = sin x − 2 sin3 x. 121 3.4 Undetermined Coeﬃcients Since sin x is a solution of the related homogeneous diﬀerential equation we look for a particular solution of the
form yp = Ax sin x + Bx cos x + C sin3 x. Substituting into the diﬀerential equation we obtain
2A cos x + (6C − 2B) sin x − 8C sin3 x = sin x − 2 sin3 x.
Equating coeﬃcients we ﬁnd A = 0, C = 1
4 , and B =
yp = 1
4 . Thus, a particular solution is 1
1
x cos x + sin3 x.
4
4 45. (a) f (x) = ex sin x. We see that yp → ∞ as x → ∞ and yp → 0 as x → −∞.
(b) f (x) = e−x . We see that yp → ∞ as x → ∞ and yp → ∞ as x → −∞.
(c) f (x) = sin 2x. We see that yp is sinusoidal.
(d) f (x) = 1. We see that yp is constant and simply translates yc vertically.
46. The complementary function is yc = e2x (c1 cos 2x + c2 sin 2x). We assume a particular solution of the form
yp = (Ax3 + Bx2 + Cx)e2x cos 2x + (Dx3 + Ex2 + F )e2x sin 2x. Substituting into the diﬀerential equation and
using a CAS to simplify yields
[12Dx2 + (6A + 8E)x + (2B + 4F )]e2x cos 2x
+ [−12Ax2 + (−8B + 6D)x + (−4C + 2E)]e2x sin 2x
= (2x2 − 3x)e2x cos 2x + (10x2 − x − 1)e2x sin 2x.
This gives the system of equations
6A + 8E = −3, 12D = 2,
−12A = 10,
from which we ﬁnd A = − 5 , B =
6
diﬀerential equation is
yp = 1
4 ,C= 2B + 4F = 0, −8B + 6D = −1,
3
8 ,D= 1
6 ,E= 5
1
3
− x3 + x2 + x e2x cos 2x +
6
4
8 1
4 −4C + 2E = −1, , and F = − 1 . Thus, a particular solution of the
8
1 3 1 2 1
x + x − x e2x sin 2x.
6
4
8 47. The complementary function is yc = c1 cos x + c2 sin x + c3 x cos x + c4 x sin x. We assume a particular solution
of the form yp = Ax2 cos x + Bx3 sin x. Substituting into the diﬀerential equation and using a CAS to simplify
yields
(−8A + 24B) cos x + 3Bx sin x = 2 cos x − 3x sin x.
This implies −8A + 24B = 2 and −24B = −3. Thus B = 122 1
8 ,A= 1
8 , and yp = 1 x2 cos x + 1 x3 sin x.
8
8 3.5 Variation of Parameters EXERCISES 3.5
Variation of Parameters The particular solution, yp = u1 y1 + u2 y2 , in the following problems can take on a variety of forms, especially where
trigonometric functions are involved. The validity of a particular form can best be checked by substituting it back
into the diﬀerential equation.
1. The auxiliary equation is m2 + 1 = 0, so yc = c1 cos x + c2 sin x and
cos x sin x
= 1.
− sin x cos x W =
Identifying f (x) = sec x we obtain sin x sec x
= − tan x
1
cos x sec x
u2 =
= 1.
1
u1 = − Then u1 = ln | cos x|, u2 = x, and
y = c1 cos x + c2 sin x + cos x ln | cos x| + x sin x.
2. The auxiliary equation is m2 + 1 = 0, so yc = c1 cos x + c2 sin x and
W = cos x sin x
= 1.
− sin x cos x Identifying f (x) = tan x we obtain
u1 = − sin x tan x = cos2 x − 1
= cos x − sec x
cos x u2 = sin x.
Then u1 = sin x − ln | sec x + tan x|, u2 = − cos x, and
y = c1 cos x + c2 sin x + cos x (sin x − ln | sec x + tan x|) − cos x sin x
= c1 cos x + c2 sin x − cos x ln | sec x + tan x|.
3. The auxiliary equation is m2 + 1 = 0, so yc = c1 cos x + c2 sin x and
W = cos x sin x
= 1.
− sin x cos x Identifying f (x) = sin x we obtain
u1 = − sin2 x
u2 = cos x sin x.
Then
u1 = 1
1
1
1
sin 2x − x = sin x cos x − x
4
2
2
2 1
u2 = − cos2 x.
2 123 3.5 Variation of Parameters and 1
1
1
sin x cos2 x − x cos x − cos2 x sin x
2
2
2
1
= c1 cos x + c2 sin x − x cos x.
2 y = c1 cos x + c2 sin x + 4. The auxiliary equation is m2 + 1 = 0, so yc = c1 cos x + c2 sin x and
W = cos x sin x
= 1.
− sin x cos x Identifying f (x) = sec x tan x we obtain
u1 = − sin x(sec x tan x) = − tan2 x = 1 − sec2 x
u2 = cos x(sec x tan x) = tan x.
Then u1 = x − tan x, u2 = − ln | cos x|, and
y = c1 cos x + c2 sin x + x cos x − sin x − sin x ln | cos x|
= c1 cos x + c3 sin x + x cos x − sin x ln | cos x|.
5. The auxiliary equation is m2 + 1 = 0, so yc = c1 cos x + c2 sin x and
W = cos x sin x
= 1.
− sin x cos x Identifying f (x) = cos2 x we obtain
u1 = − sin x cos2 x
u2 = cos3 x = cos x 1 − sin2 x .
Then u1 = 1
3 cos3 x, u2 = sin x − 1
3 sin3 x, and 1
1
cos4 x + sin2 x − sin4 x
3
3
1
= c1 cos x + c2 sin x +
cos2 x + sin2 x cos2 x − sin2 x + sin2 x
3
1
2
= c1 cos x + c2 sin x + cos2 x + sin2 x
3
3
1 1
= c1 cos x + c2 sin x + + sin2 x.
3 3 y = c1 cos x + c2 sin x + 6. The auxiliary equation is m2 + 1 = 0, so yc = c1 cos x + c2 sin x and
W = cos x sin x
= 1.
− sin x cos x Identifying f (x) = sec2 x we obtain
u1 = − sin x
cos2 x u2 = sec x.
Then
u1 = − 1
= − sec x
cos x u2 = ln | sec x + tan x| 124 3.5
and y = c1 cos x + c2 sin x − cos x sec x + sin x ln | sec x + tan x|
= c1 cos x + c2 sin x − 1 + sin x ln | sec x + tan x|. 7. The auxiliary equation is m2 − 1 = 0, so yc = c1 ex + c2 e−x and
ex e−x ex W = −e−x = −2. Identifying f (x) = cosh x = 1 (e−x + ex ) we obtain
2
u1 = 1 −2x 1
e
+
4
4 1 1
u2 = − − e2x .
4 4
Then 1
1
u1 = − e−2x + x
8
4
1
1
u2 = − e2x − x
8
4 and 1
1
1
1
y = c1 ex + c2 e−x − e−x + xex − ex − xe−x
8
4
8
4
1
= c3 ex + c4 e−x + x(ex − e−x )
4
1
= c3 ex + c4 e−x + x sinh x.
2 8. The auxiliary equation is m2 − 1 = 0, so yc = c1 ex + c2 e−x and
ex e−x x W = −e−x e = −2. Identifying f (x) = sinh 2x we obtain
1
1
u1 = − e−3x + ex
4
4
u2 =
Then
u1 = 1 −x 1 3x
e − e .
4
4
1 −3x 1 x
+ e
e
12
4 1
1
u2 = − e−x − e3x .
4
12
and
y = c1 ex + c2 e−x + 1 −2x 1 2x 1 −2x
1
+ e − e
− e2x
e
12
4
4
12 = c1 ex + c2 e−x + 1 2x
e − e−2x
6 = c1 ex + c2 e−x + 1
sinh 2x.
3 9. The auxiliary equation is m2 − 4 = 0, so yc = c1 e2x + c2 e−2x and
W = e2x e−2x 2e2x −2e−2x 125 = −4. Variation of Parameters 3.5 Variation of Parameters
Identifying f (x) = e2x /x we obtain u1 = 1/4x and u2 = −e4x /4x. Then
1
ln |x|,
4
1 x e4t
u2 = −
dt
4 x0 t
u1 = and
y = c1 e2x + c2 e−2x + 1
4 x e2x ln |x| − e−2x x0 e4t
dt ,
t x0 > 0. 10. The auxiliary equation is m2 − 9 = 0, so yc = c1 e3x + c2 e−3x and
e3x W = 3x 3e e−3x
= −6.
−3e−3x Identifying f (x) = 9x/e3x we obtain u1 = 3 xe−6x and u2 = − 3 x. Then
2
2
1 −6x 1 −6x
e
− xe
,
24
4
3
u2 = − x2
4 u1 = − and
y = c1 e3x + c2 e−3x − 1 −3x 1 −3x 3 2 −3x
− xe
− x e
e
24
4
4 1
= c1 e3x + c3 e−3x − xe−3x (1 − 3x).
4
11. The auxiliary equation is m2 + 3m + 2 = (m + 1)(m + 2) = 0, so yc = c1 e−x + c2 e−2x and
W = e−x
−x −e e−2x
= −e−3x .
−2e−2x Identifying f (x) = 1/(1 + ex ) we obtain
u1 = ex
1 + ex u2 = − e2x
ex
=
− ex .
x
1+e
1 + ex Then u1 = ln(1 + ex ), u2 = ln(1 + ex ) − ex , and
y = c1 e−x + c2 e−2x + e−x ln(1 + ex ) + e−2x ln(1 + ex ) − e−x
= c3 e−x + c2 e−2x + (1 + e−x )e−x ln(1 + ex ).
12. The auxiliary equation is m2 − 2m + 1 = (m − 1)2 = 0, so yc = c1 ex + c2 xex and
W = ex xex ex xex + ex = e2x . Identifying f (x) = ex / 1 + x2 we obtain
u1 = −
u2 = x
xex ex
=−
2x (1 + x2 )
e
1 + x2 e2x 1
ex ex
=
.
(1 + x2 )
1 + x2 126 3.5 Variation of Parameters Then u1 = − 1 ln 1 + x2 , u2 = tan−1 x, and
2
1
y = c1 ex + c2 xex − ex ln 1 + x2 + xex tan−1 x.
2
13. The auxiliary equation is m2 + 3m + 2 = (m + 1)(m + 2) = 0, so yc = c1 e−x + c2 e−2x and
e−x e−2x −e−x W = −2e−2x = −e−3x . Identifying f (x) = sin ex we obtain
u1 = e−2x sin ex
= ex sin ex
e−3x e−x sin ex
= −e2x sin ex .
−e−3x
Then u1 = − cos ex , u2 = ex cos x − sin ex , and
u2 = y = c1 e−x + c2 e−2x − e−x cos ex + e−x cos ex − e−2x sin ex
= c1 e−x + c2 e−2x − e−2x sin ex .
14. The auxiliary equation is m2 − 2m + 1 = (m − 1)2 = 0, so yc = c1 et + c2 tet and
et tet et W = tet + et = e2t . Identifying f (t) = et tan−1 t we obtain
u1 = −
u2 = tet et tan−1 t
= −t tan−1 t
e2t et et tan−1 t
= tan−1 t.
e2t Then
u1 = − 1 + t2
t
tan−1 t +
2
2 u2 = t tan−1 t − 1
ln 1 + t2
2 and
y = c1 et + c2 tet + −
1
= c1 et + c3 tet + et
2 1 + t2
t
tan−1 t +
2
2 et + t tan−1 t − t2 − 1 tan−1 t − ln 1 + t2 1
ln 1 + t2
2 . 15. The auxiliary equation is m2 + 2m + 1 = (m + 1)2 = 0, so yc = c1 e−t + c2 te−t and
W = e−t
−e−t te−t
−tet + e−t = e−2t . Identifying f (t) = e−t ln t we obtain
u1 = −
u2 = te−t e−t ln t
= −t ln t
e−2t e−t e−t ln t
= ln t.
e−2t 127 tet 3.5 Variation of Parameters Then 1
1
u1 = − t2 ln t + t2
2
4
u2 = t ln t − t and 1
y = c1 e−t + c2 te−t − t2 e−t ln t +
2
1
= c1 e−t + c2 te−t + t2 e−t ln t −
2 1 2 −t
t e + t2 e−t ln t − t2 e−t
4
3 2 −t
t e .
4 16. The auxiliary equation is 2m2 + 2m + 1 = 0, so yc = e−x/2 [c1 cos(x/2) + c2 sin(x/2)] and
e−x/2 cos
W = x
2 e−x/2 sin 1
x 1
x
− e−x/2 cos − e−x/2 sin
2
2 2
2 x
2 x 1
x
1 −x/2
cos − ex/2 sin
e
2
2 2
2 = 1 −x
e .
2 √
Identifying f (x) = 2 x we obtain
√
√
e−x/2 sin(x/2)2 x
x
= −4ex/2 x sin
−x/2
2
e
√
√
e−x/2 cos(x/2)2 x
x
u2 = −
= 4ex/2 x cos .
−x/2
2
e u1 = − Then √
t
et/2 t sin dt
2
x0
x
√
t
u2 = 4
et/2 t cos dt
2
x0
x u1 = −4 and
y = e−x/2 c1 cos x x
x
x
+ c2 sin
− 4e−x/2 cos
2
2
2 x0 √
t
x
et/2 t sin dt + 4e−x/2 sin
2
2 x
x0 17. The auxiliary equation is 3m2 − 6m + 6 = 0, so yc = ex (c1 cos x + c2 sin x) and
ex cos x ex sin x ex cos x − ex sin x W = ex cos x + ex sin x = e2x . Identifying f (x) = 1 ex sec x we obtain
3
u1 = −
u2 =
Then u1 = 1
3 (ex sin x)(ex sec x)/3
1
= − tan x
e2x
3 (ex cos x)(ex sec x)/3
1
= .
e2x
3 ln(cos x), u2 = 1 x, and
3
y = c1 ex cos x + c2 ex cos x + 1
1
ln(cos x)ex cos x + xex sin x.
3
3 18. The auxiliary equation is 4m2 − 4m + 1 = (2m − 1)2 = 0, so yc = c1 ex/2 + c2 xex/2 and
W = ex/2
1 x/2
2e xex/2
1
x/2
2 xe 128 + ex/2 = ex . √
t
et/2 t cos dt.
2 3.5 Variation of Parameters √
Identifying f (x) = 1 ex/2 1 − x2 we obtain
4 √
xex/2 ex/2 1 − x2
1
= − x 1 − x2
4ex
4
√
ex/2 ex/2 1 − x2
1
u2 =
=
1 − x2 .
x
4e
4
To ﬁnd u1 and u2 we use the substitution v = 1 − x2 and the trig substitution x = sin θ, respectively:
1
3/2
u1 =
1 − x2
12
u1 = − u2 =
Thus
y = c1 ex/2 + c2 xex/2 + x
8 1 − x2 + 1 x/2
1 − x2
e
12 1
sin−1 x.
8 1
+ x2 ex/2
8 3/2 1
1 − x2 + xex/2 sin−1 x.
8 19. The auxiliary equation is 4m2 − 1 = (2m − 1)(2m + 1) = 0, so yc = c1 ex/2 + c2 e−x/2 and
W = ex/2 e−x/2 1 x/2
2e − 1 e−x/2
2 = −1. Identifying f (x) = xex/2 /4 we obtain u1 = x/4 and u2 = −xex /4. Then u1 = x2 /8 and
u2 = −xex /4 + ex /4. Thus
1
y = c1 ex/2 + c2 e−x/2 + x2 ex/2 −
8
1
= c3 ex/2 + c2 e−x/2 + x2 ex/2 −
8
and
y = 1 x/2 1 x/2
+ e
xe
4
4
1 x/2
xe
4 1 x/2 1 −x/2
1
1
1
c3 e
− c2 e
+ x2 ex/2 + xex/2 − ex/2 .
2
2
16
8
4 The initial conditions imply
c3 + c2 =1 1
1
1
c3 − c2 − = 0.
2
2
4
Thus c3 = 3/4 and c2 = 1/4, and
3 x/2 1 −x/2 1 2 x/2 1 x/2
+ e
+ x e
− xe .
e
4
4
8
4 y= 20. The auxiliary equation is 2m2 + m − 1 = (2m − 1)(m + 1) = 0, so yc = c1 ex/2 + c2 e−x and
W = ex/2 e−x 1 x/2
2e −x −e 3
= − e−x/2 .
2 Identifying f (x) = (x + 1)/2 we obtain
1 −x/2
(x + 1)
e
3
1
u2 = − ex (x + 1).
3
u1 = Then
u1 = −e−x/2
1
u2 = − xex .
3 129 2
x−2
3 3.5 Variation of Parameters Thus
y = c1 ex/2 + c2 e−x − x − 2
and
y = 1 x/2
− c2 e−x − 1.
c1 e
2 The initial conditions imply
c1 − c2 − 2 = 1
1
c1 − c2 − 1 = 0.
2
Thus c1 = 8/3 and c2 = 1/3, and
y= 8 x/2 1 −x
+ e − x − 2.
e
3
3 21. The auxiliary equation is m2 + 2m − 8 = (m − 2)(m + 4) = 0, so yc = c1 e2x + c2 e−4x and
W = e2x e−4x 2e2x −4e−4x = −6e−2x . Identifying f (x) = 2e−2x − e−x we obtain
1 −4x 1 −3x
− e
e
3
6
1
1
u2 = e3x − e2x .
6
3 u1 = Then 1 −4x
1
e
+ e−3x
12
18
1 3x 1 2x
e − e .
u2 =
18
6
u1 = − Thus
1 −2x
1
1
1
e
+ e−x + e−x − e−2x
12
18
18
6
1 −2x 1 −x
= c1 e2x + c2 e−4x − e
+ e
4
9 y = c1 e2x + c2 e−4x − and 1
1
y = 2c1 e2x − 4c2 e−4x + e−2x − e−x .
2
9 The initial conditions imply
5
=1
36
7
2c1 − 4c2 +
= 0.
18
c1 + c2 − Thus c1 = 25/36 and c2 = 4/9, and
y= 25 2x 4 −4x 1 −2x 1 −x
e + e
− e
+ e .
36
9
4
9 22. The auxiliary equation is m2 − 4m + 4 = (m − 2)2 = 0, so yc = c1 e2x + c2 xe2x and
W = e2x xe2x 2e2x 2xe2x + e2x 130 = e4x . 3.5 Variation of Parameters Identifying f (x) = 12x2 − 6x e2x we obtain
u1 = 6x2 − 12x3
u2 = 12x2 − 6x.
Then u1 = 2x3 − 3x4
u2 = 4x3 − 3x2 . Thus
y = c1 e2x + c2 xe2x + 2x3 − 3x4 e2x + 4x3 − 3x2 xe2x
= c1 e2x + c2 xe2x + e2x x4 − x3
and
y = 2c1 e2x + c2 2xe2x + e2x + e2x 4x3 − 3x2 + 2e2x x4 − x3 .
The initial conditions imply
c1 =1 2c1 + c2 = 0.
Thus c1 = 1 and c2 = −2, and
y = e2x − 2xe2x + e2x x4 − x3 = e2x x4 − x3 − 2x + 1 .
23. Write the equation in the form
y + 1
1
y + 1− 2
x
4x y = x−1/2 and identify f (x) = x−1/2 . From y1 = x−1/2 cos x and y2 = x−1/2 sin x we compute W (y1 , y2 ) = x−1/2 cos x
−x−1/2 sin x − 1 x−3/2 cos x
2 x−1/2 sin x
1
= .
1 −3/2
−1/2
x
x
cos x − 2 x
sin x Now
u1 = − sin x so u1 = cos x, and
u2 = cos x so u2 = sin x. Thus a particular solution is
yp = x−1/2 cos2 x + x−1/2 sin2 x,
and the general solution is
y = c1 x−1/2 cos x + c2 x−1/2 sin x + x−1/2 cos2 x + x−1/2 sin2 x
= c1 x−1/2 cos x + c2 x−1/2 sin x + x−1/2 .
24. Write the equation in the form
1
1
sec(ln x)
y + 2y =
x
x
x2
2
and identify f (x) = sec(ln x)/x . From y1 = cos(ln x) and y2 = sin(ln x) we compute
y + cos(ln x)
W = sin(ln x) sin(ln x)
−
x cos(ln x)
x 131 = 1
.
x 3.5 Variation of Parameters Now
u1 = − tan(ln x)
x so u1 = ln | cos(ln x)|, 1
x so u2 = ln x. and
u2 =
Thus, a particular solution is
yp = cos(ln x) ln | cos(ln x)| + (ln x) sin(ln x),
and the general solution is
y = c1 cos(ln x) + c2 sin(ln x) + cos(ln x) ln | cos(ln x)| + (ln x) sin(ln x).
25. The auxiliary equation is m3 + m = m(m2 + 1) = 0, so yc = c1 + c2 cos x + c3 sin x and
1
W = 0
0 cos x
− sin x
− cos x sin x
cos x = 1.
− sin x Identifying f (x) = tan x we obtain u1 = W 1 = 0
cos x
0
− sin x
tan x − cos x 1
u2 = W2 = 0
0 0
0
tan x sin x
cos x = tan x
− sin x sin x
cos x = − sin x
− sin x 1
cos x
0
cos2 x − 1
= cos x − sec x.
0
= − sin x tan x =
u3 = W3 = 0 − sin x
cos x
0 − cos x tan x
Then
u1 = − ln | cos x|
u2 = cos x
u3 = sin x − ln | sec x + tan x|
and
y = c1 + c2 cos x + c3 sin x − ln | cos x| + cos2 x
+ sin2 x − sin x ln | sec x + tan x|
= c4 + c2 cos x + c3 sin x − ln | cos x| − sin x ln | sec x + tan x|
for −π/2 < x < π/2.
26. The auxiliary equation is m3 + 4m = m m2 + 4 = 0, so yc = c1 + c2 cos 2x + c3 sin 2x and
1
W = 0
0 cos 2x
−2 sin 2x
−4 cos 2x 132 sin 2x
2 cos 2x = 8.
−4 sin 2x 3.5 Variation of Parameters Identifying f (x) = sec 2x we obtain
1
1
u1 = W 1 =
8
8 0
0 cos 2x
−2 sin 2x sec 2x −4 cos 2x 1
1
1
0
u2 = W2 =
8
8
0 0
0
sec 2x 1
1
1
u3 = W3 =
0
8
8
0 cos 2x
−2 sin 2x Then sin 2x
1
2 cos 2x = sec 2x
4
−4 sin 2x sin 2x
1
2 cos 2x = −
4
−4 sin 2x
0
0 −4 cos 2x sec 2x 1
= − tan 2x.
4 1
ln | sec 2x + tan 2x|
8
1
u2 = − x
4
1
u3 = ln | cos 2x|
8
u1 = and
y = c1 + c2 cos 2x + c3 sin 2x + 1
1
1
ln | sec 2x + tan 2x| − x cos 2x + sin 2x ln | cos 2x|
8
4
8 for −π/4 < x < π/4.
27. The auxiliary equation is 3m2 −6m+30 = 0, which has roots 1±3i, so yc = ex (c1 cos 3x+c2 sin 3x). We consider
ﬁrst the diﬀerential equation 3y − 6y + 30y = 15 sin x, which can be solved using undetermined coeﬃcients.
Letting yp1 = A cos x + B sin x and substituting into the diﬀerential equation we get
(27A − 6B) cos x + (6A + 27B) sin x = 15 sin x.
Then
27A − 6B = 0 and 6A + 27B = 15, 9
2
9
and B = 17 . Thus, yp1 = 17 cos x+ 17 sin x. Next, we consider the diﬀerential equation 3y −6y +30y,
for which a particular solution yp2 can be found using variation of parameters. The Wronskian is so A = 2
17 W = ex cos 3x ex sin 3x ex cos 3x − 3ex sin 3x 3ex cos 3x + ex sin 3x = 3e2x . Identifying f (x) = 1 ex tan x we obtain
3
1
1
u1 = − sin 3x tan 3x = −
9
9
so
u1 = −
Next
u2 =
Thus sin2 3x
cos 3x =− 1
9 1 − cos2 3x
cos 3x 1
= − (sec 3x − cos 3x)
9 1
1
ln | sec 3x + tan 3x| +
sin 3x.
27
27
1
sin 3x so
9 u2 = − 1
cos 3x.
27 1 x
1 x
e cos 3x(ln | sec 3x + tan 3x| − sin 3x) −
e sin 3x cos 3x
27
27
1
= − ex (cos 3x) ln | sec 3x + tan 3x|
27 yp2 = − 133 3.5 Variation of Parameters and the general solution of the original diﬀerential equation is
y = ex (c1 cos 3x + c2 sin 3x) + yp1 (x) + yp2 (x).
28. The auxiliary equation is m2 − 2m + 1 = (m − 1)2 = 0, which has repeated root 1, so yc = c1 ex + c2 xex .
We consider ﬁrst the diﬀerential equation y − 2y + y = 4x2 − 3, which can be solved using undetermined
coeﬃcients. Letting yp1 = Ax2 + Bx + C and substituting into the diﬀerential equation we get
Ax2 + (−4A + B)x + (2A − 2B + C) = 4x2 − 3.
Then
A = 4, −4A + B = 0, and 2A − 2B + C = −3, so A = 4, B = 16, and C = 21. Thus, yp1 = 4x2 + 16x + 21. Next we consider the diﬀerential equation
y − 2y + y = x−1 ex , for which a particular solution yp2 can be found using variation of parameters. The
Wronskian is
ex
xex
W = x
= e2x .
x
e
xe + ex
Identifying f (x) = ex /x we obtain u1 = −1 and u2 = 1/x. Then u1 = −x and u2 = ln x, so that
yp2 = −xex + xex ln x,
and the general solution of the original diﬀerential equation is
y = yc + yp1 + yp2 = c1 ex + c2 xex + 4x2 + 16x + 21 − xex + xex ln x
= c1 ex + c3 xex + 4x2 + 16x + 21 + xex ln x . 29. The interval of deﬁnition for Problem 1 is (−π/2, π/2), for Problem 7 is (−∞, ∞), for Problem 9 is (0, ∞), and
for Problem 18 is (−1, 1). In Problem 24 the general solution is
y = c1 cos(ln x) + c2 sin(ln x) + cos(ln x) ln | cos(ln x)| + (ln x) sin(ln x)
for −π/2 < ln x < π/2 or e−π/2 < x < eπ/2 . The bounds on ln x are due to the presence of sec(ln x) in the
diﬀerential equation.
30. We are given that y1 = x2 is a solution of x4 y + x3 y − 4x2 y = 0. To ﬁnd a second solution we use reduction
of order. Let y = x2 u(x). Then the product rule gives
y = x2 u + 2xu and y = x2 u + 4xu + 2u,
so
x4 y + x3 y − 4x2 y = x5 (xu + 5u ) = 0.
Letting w = u , this becomes xw + 5w = 0. Separating variables and integrating we have
dw
5
= − dx
w
x and ln |w| = −5 ln x + c. Thus, w = x−5 and u = − 1 x−4 . A second solution is then y2 = x2 x−4 = 1/x2 , and the general solution of the
4
homogeneous diﬀerential equation is yc = c1 x2 + c2 /x2 . To ﬁnd a particular solution, yp , we use variation of
parameters. The Wronskian is
W =
4 Identifying f (x) = 1/x we obtain u1 =
yp = − 1 −5
4x x2 1/x2 2x −2/x 3 4
=− .
x 1
and u2 = − 1 x−1 . Then u1 = − 16 x−4 and u2 = − 1 ln x, so
4
4 1 −4 2 1
1
1
x x − (ln x)x−2 = − x−2 − x−2 ln x.
16
4
16
4 134 3.6 Cauchy-Euler Equation The general solution is
y = c1 x2 + c2
1
1
−
− 2 ln x.
x2
16x2
4x EXERCISES 3.6
Cauchy-Euler Equation 1. The auxiliary equation is m2 − m − 2 = (m + 1)(m − 2) = 0 so that y = c1 x−1 + c2 x2 .
2. The auxiliary equation is 4m2 − 4m + 1 = (2m − 1)2 = 0 so that y = c1 x1/2 + c2 x1/2 ln x.
3. The auxiliary equation is m2 = 0 so that y = c1 + c2 ln x.
4. The auxiliary equation is m2 − 4m = m(m − 4) = 0 so that y = c1 + c2 x4 .
5. The auxiliary equation is m2 + 4 = 0 so that y = c1 cos(2 ln x) + c2 sin(2 ln x).
6. The auxiliary equation is m2 + 4m + 3 = (m + 1)(m + 3) = 0 so that y = c1 x−1 + c2 x−3 .
√ 7. The auxiliary equation is m2 − 4m − 2 = 0 so that y = c1 x2− 8. The auxiliary equation is m2 + 2m − 4 = 0 so that y = c1 x−1+
9. The auxiliary equation is 25m2 + 1 = 0 so that y = c1 cos 1
5 6 √ + c2 x2+
5 √ 6 .
√ + c2 x−1− ln x + c2 sin 1
5 5 . ln x . 10. The auxiliary equation is 4m2 − 1 = (2m − 1)(2m + 1) = 0 so that y = c1 x1/2 + c2 x−1/2 .
11. The auxiliary equation is m2 + 4m + 4 = (m + 2)2 = 0 so that y = c1 x−2 + c2 x−2 ln x.
12. The auxiliary equation is m2 + 7m + 6 = (m + 1)(m + 6) = 0 so that y = c1 x−1 + c2 x−6 .
13. The auxiliary equation is 3m2 + 3m + 1 = 0 so that
√
3
−1/2
y=x
c1 cos
ln x
6 √
+ c2 sin 3
ln x
6 . 14. The auxiliary equation is m2 − 8m + 41 = 0 so that y = x4 [c1 cos(5 ln x) + c2 sin(5 ln x)].
15. Assuming that y = xm and substituting into the diﬀerential equation we obtain
m(m − 1)(m − 2) − 6 = m3 − 3m2 + 2m − 6 = (m − 3)(m2 + 2) = 0.
Thus
y = c1 x3 + c2 cos √ 2 ln x + c3 sin √ 2 ln x . 16. Assuming that y = xm and substituting into the diﬀerential equation we obtain
m(m − 1)(m − 2) + m − 1 = m3 − 3m2 + 3m − 1 = (m − 1)3 = 0.
Thus
y = c1 x + c2 x ln x + c3 x(ln x)2 . 135 3.6 Cauchy-Euler Equation 17. Assuming that y = xm and substituting into the diﬀerential equation we obtain
m(m − 1)(m − 2)(m − 3) + 6m(m − 1)(m − 2) = m4 − 7m2 + 6m = m(m − 1)(m − 2)(m + 3) = 0.
Thus
y = c1 + c2 x + c3 x2 + c4 x−3 .
18. Assuming that y = xm and substituting into the diﬀerential equation we obtain
m(m − 1)(m − 2)(m − 3) + 6m(m − 1)(m − 2) + 9m(m − 1) + 3m + 1 = m4 + 2m2 + 1 = (m2 + 1)2 = 0.
Thus
y = c1 cos(ln x) + c2 sin(ln x) + c3 (ln x) cos(ln x) + c4 (ln x) sin(ln x).
19. The auxiliary equation is m2 − 5m = m(m − 5) = 0 so that yc = c1 + c2 x5 and
1 x5 0 W (1, x5 ) = 5x4 = 5x4 . 1
Identifying f (x) = x3 we obtain u1 = − 1 x4 and u2 = 1/5x. Then u1 = − 25 x5 , u2 =
5 y = c1 + c2 x5 − 1
5 ln x, and 1 5 1 5
1
x + x ln x = c1 + c3 x5 + x5 ln x.
25
5
5 20. The auxiliary equation is 2m2 + 3m + 1 = (2m + 1)(m + 1) = 0 so that yc = c1 x−1 + c2 x−1/2 and
W (x−1 , x−1/2 ) =
Identifying f (x) = 1
2 − 1
2x x−1 x−1/2 −2 −x − 1 x−3/2
2 = 1 −5/2
x
.
2 we obtain u1 = x − x and u2 = x3/2 − x1/2 . Then u1 = 1 x2 − 1 x3 ,
2
3
2 u2 = 2 x5/2 − 2 x3/2 , and
5
3
1
2
2
1
1
1
y = c1 x−1 + c2 x−1/2 + x − x2 + x2 − x = c1 x−1 + c2 x−1/2 − x + x2 .
2
3
5
3
6
15
21. The auxiliary equation is m2 − 2m + 1 = (m − 1)2 = 0 so that yc = c1 x + c2 x ln x and
x x ln x
= x.
1 1 + ln x W (x, x ln x) = Identifying f (x) = 2/x we obtain u1 = −2 ln x/x and u2 = 2/x. Then u1 = −(ln x)2 , u2 = 2 ln x, and
y = c1 x + c2 x ln x − x(ln x)2 + 2x(ln x)2
= c1 x + c2 x ln x + x(ln x)2 , x > 0. 22. The auxiliary equation is m2 − 3m + 2 = (m − 1)(m − 2) = 0 so that yc = c1 x + c2 x2 and
W (x, x2 ) = x x2 1 2x = x2 . Identifying f (x) = x2 ex we obtain u1 = −x2 ex and u2 = xex . Then u1 = −x2 ex + 2xex − 2ex , u2 = xex − ex ,
and
y = c1 x + c2 x2 − x3 ex + 2x2 ex − 2xex + x3 ex − x2 ex
= c1 x + c2 x2 + x2 ex − 2xex .
23. The auxiliary equation m(m − 1) + m − 1 = m2 − 1 = 0 has roots m1 = −1, m2 = 1, so yc = c1 x−1 + c2 x. With
y1 = x−1 , y2 = x, and the identiﬁcation f (x) = ln x/x2 , we get
W = 2x−1 , W1 = − ln x/x, 136 and W2 = ln x/x3 . 3.6 Cauchy-Euler Equation Then u1 = W1 /W = −(ln x)/2, u2 = W2 /W = (ln x)/2x2 , and integration by parts gives
1
1
x − x ln x
2
2
1 −1
1
u2 = − x ln x − x−1 ,
2
2
u1 = so
yp = u 1 y 1 + u 2 y 2 = 1
1
1
1
x − x ln x x−1 + − x−1 ln x − x−1 x = − ln x
2
2
2
2 and
y = yc + yp = c1 x−1 + c2 x − ln x, x > 0. 24. The auxiliary equation m(m − 1) + m − 1 = m2 − 1 = 0 has roots m1 = −1, m2 = 1, so yc = c1 x−1 + c2 x. With
y1 = x−1 , y2 = x, and the identiﬁcation f (x) = 1/x2 (x + 1), we get
W = 2x−1 ,
Then u1 = W1 /W = −1/2(x + 1),
gives W1 = −1/x(x + 1), W2 = 1/x3 (x + 1). and u2 = W2 /W = 1/2x2 (x + 1), and integration (by partial fractions for u2 )
1
u1 = − ln(x + 1)
2
1
1 −1 1
u2 = − x − ln x + ln(x + 1),
2
2
2 so
1
1
1
1
yp = u1 y1 + u2 y2 = − ln(x + 1) x−1 + − x−1 − ln x + ln(x + 1) x
2
2
2
2
1 1
1
ln(x + 1)
1 1
1
= − − x ln x + x ln(x + 1) −
= − + x ln 1 +
2 2
2
2x
2 2
x − ln(x + 1)
2x and
y = yc + yp = c1 x−1 + c2 x − 1 1
1
+ x ln 1 +
2 2
x − ln(x + 1)
,
2x x > 0.
y 25. The auxiliary equation is m2 + 2m = m(m + 2) = 0, so that y = c1 + c2 x−2
and y = −2c2 x−3 . The initial conditions imply 5x c1 + c2 = 0
−2c2 = 4.
Thus, c1 = 2, c2 = −2, and y = 2 − 2x−2 . The graph is given to the right. -10 -20 137 3.6 Cauchy-Euler Equation 26. The auxiliary equation is m2 − 6m + 8 = (m − 2)(m − 4) = 0, so that
2 4 y = c1 x + c2 x y 3 and y = 2c1 x + 4c2 x . 30 The initial conditions imply 20 4c1 + 16c2 = 32 10 4c1 + 32c2 = 0.
-1 Thus, c1 = 16, c2 = −2, and y = 16x2 − 2x4 . The graph is given to the right. 4 x -4
-20
-30 27. The auxiliary equation is m2 + 1 = 0, so that y
3 y = c1 cos(ln x) + c2 sin(ln x)
and 1
1
sin(ln x) + c2 cos(ln x).
x
x
The initial conditions imply c1 = 1 and c2 = 2. Thus
y = cos(ln x) + 2 sin(ln x). The graph is given to the right.
y = −c1 100 x 50
-3 28. The auxiliary equation is m2 − 4m + 4 = (m − 2)2 = 0, so that y y = c1 x2 + c2 x2 ln x and y = 2c1 x + c2 (x + 2x ln x). 5 The initial conditions imply c1 = 5 and c2 + 10 = 3. Thus y = 5x2 − 7x2 ln x.
The graph is given to the right. x
-10 -20 -30 29. The auxiliary equation is m2 = 0 so that yc = c1 + c2 ln x and
W (1, ln x) = 1
0 y ln x
1
= .
x
1/x 15 Identifying f (x) = 1 we obtain u1 = −x ln x and u2 = x. Then 10 u1 = 1 x2 − 1 x2 ln x, u2 = 1 x2 , and
4
2
2
1
1
1
1
y = c1 + c2 ln x + x2 − x2 ln x + x2 ln x = c1 + c2 ln x + x2 .
4
2
2
4
The initial conditions imply c1 + 1 = 1 and c2 + 1 = − 1 . Thus, c1 =
4
2
2
y = 3 − ln x + 1 x2 . The graph is given to the right.
4
4 3
4 5 , c2 = −1, and
5 138 3.6
30. The auxiliary equation is m2 − 6m + 8 = (m − 2)(m − 4) = 0, so that
yc = c1 x2 + c2 x4 and
W = x2 x4 2x 4x3 Cauchy-Euler Equation y
0.05 = 2x5 . Identifying f (x) = 8x4 we obtain u1 = −4x3 and u2 = 4x. Then
-1
u1 = −x4 , u2 = 2x2 , and y = c1 x2 + c2 x4 + x6 . The initial conditions
imply
1
1
1
c1 + c2 = −
4
16
64
1
3
c1 + c2 = − .
2
16
1
1
Thus c1 = 16 , c2 = − 1 , and y = 16 x2 − 1 x4 + x6 . The graph is given above.
2
2 1 x 31. Substituting x = et into the diﬀerential equation we obtain
d2 y
dy
+8
− 20y = 0.
2
dt
dt
The auxiliary equation is m2 + 8m − 20 = (m + 10)(m − 2) = 0 so that
y = c1 e−10t + c2 e2t = c1 x−10 + c2 x2 .
32. Substituting x = et into the diﬀerential equation we obtain
d2 y
dy
− 10
+ 25y = 0.
2
dt
dt
The auxiliary equation is m2 − 10m + 25 = (m − 5)2 = 0 so that
y = c1 e5t + c2 te5t = c1 x5 + c2 x5 ln x.
33. Substituting x = et into the diﬀerential equation we obtain
d2 y
dy
+9
+ 8y = e2t .
2
dt
dt
The auxiliary equation is m2 + 9m + 8 = (m + 1)(m + 8) = 0 so that yc = c1 e−t + c2 e−8t . Using undetermined
coeﬃcients we try yp = Ae2t . This leads to 30Ae2t = e2t , so that A = 1/30 and
y = c1 e−t + c2 e−8t + 1 2t
1
e = c1 x−1 + c2 x−8 + x2 .
30
30 34. Substituting x = et into the diﬀerential equation we obtain
d2 y
dy
−5
+ 6y = 2t.
2
dt
dt
The auxiliary equation is m2 − 5m + 6 = (m − 2)(m − 3) = 0 so that yc = c1 e2t + c2 e3t . Using undetermined
coeﬃcients we try yp = At + B. This leads to (−5A + 6B) + 6At = 2t, so that A = 1/3, B = 5/18, and
1
1
5
5
y = c1 e2t + c2 e3t + t +
= c1 x2 + c2 x3 + ln x + .
3
18
3
18
35. Substituting x = et into the diﬀerential equation we obtain
d2 y
dy
−4
+ 13y = 4 + 3et .
2
dt
dt 139 3.6 Cauchy-Euler Equation
The auxiliary equation is m2 −4m+13 = 0 so that yc = e2t (c1 cos 3t+c2 sin 3t). Using undetermined coeﬃcients
we try yp = A + Bet . This leads to 13A + 10Bet = 4 + 3et , so that A = 4/13, B = 3/10, and
4
3
+ et
13 10
4
3
= x2 [c1 cos(3 ln x) + c2 sin(3 ln x)] +
+ x.
13 10 y = e2t (c1 cos 3t + c2 sin 3t) + 36. From it follows that d2 y
1
= 2
dx2
x
d3 y
1 d
= 2
dx3
x dx d2 y dy
−
dt2
dt = 1 d
x2 dx d2 y
dt2 = 1 d3 y
x2 dt3 1
x = 1
x3 −
− d2 y dy
−
dt2
dt
− d2 y dy
−
dt2
dt 2
x3 1 d
x2 dx dy
dt 1 d2 y
x2 dt2 1
x d3 y
d2 y
dy
−3 2 +2
dt3
dt
dt −
− 2 d2 y
2 dy
+ 3
3 dt2
x
x dt 2 dy
2 d2 y
+ 3
3 dt2
x
x dt . Substituting into the diﬀerential equation we obtain
d3 y
d2 y
dy
−3 2 +2
−3
3
dt
dt
dt
or d2 y dy
−
dt2
dt +6 dy
− 6y = 3 + 3t
dt d3 y
d2 y
dy
− 6 2 + 11
− 6y = 3 + 3t.
3
dt
dt
dt The auxiliary equation is m3 −6m2 +11m−6 = (m−1)(m−2)(m−3) = 0 so that yc = c1 et +c2 e2t +c3 e3t . Using
undetermined coeﬃcients we try yp = A + Bt. This leads to (11B − 6A) − 6Bt = 3 + 3t, so that A = −17/12,
B = −1/2, and
17 1
17 1
y = c1 et + c2 e2t + c3 e3t −
− t = c1 x + c2 x2 + c3 x3 −
− ln x.
12 2
12 2
In the next two problems we use the substitution t = −x since the initial conditions are on the interval (−∞, 0). In
this case
dy
dy dx
dy
=
=−
dt
dx dt
dx
and
d2 y
d
dy
d
d2 y dx
d2 y
d dy
dy dx
=
−
= − (y ) = −
=− 2
= 2.
=
2
dt
dt dt
dt
dx
dt
dx dt
dx dt
dx 37. The diﬀerential equation and initial conditions become
4t2 d2 y
+ y = 0;
dt2 y(t) = 2, = −4. y (t) t=1 t=1 The auxiliary equation is 4m2 − 4m + 1 = (2m − 1)2 = 0, so that
y = c1 t1/2 + c2 t1/2 ln t and y = 1 −1/2
1
+ c2 t−1/2 + t−1/2 ln t .
c1 t
2
2 140 3.6 Cauchy-Euler Equation The initial conditions imply c1 = 2 and 1 + c2 = −4. Thus
y = 2t1/2 − 5t1/2 ln t = 2(−x)1/2 − 5(−x)1/2 ln(−x), x < 0. 38. The diﬀerential equation and initial conditions become
t2 d2 y
dy
− 4t
+ 6y = 0;
dt2
dt = 8, y(t) y (t) t=2 = 0.
t=2 The auxiliary equation is m2 − 5m + 6 = (m − 2)(m − 3) = 0, so that
y = c1 t2 + c2 t3 and y = 2c1 t + 3c2 t2 . The initial conditions imply
4c1 + 8c2 = 8
4c1 + 12c2 = 0
from which we ﬁnd c1 = 6 and c2 = −2. Thus
y = 6t2 − 2t3 = 6x2 + 2x3 , x < 0. 39. Letting u = x + 2 we obtain dy/dx = dy/du and, using the Chain Rule,
d2 y
d
=
dx2
dx dy
du = d2 y
d2 y du
d2 y
= 2 (1) = 2 .
du2 dx
du
du Substituting into the diﬀerential equation we obtain
u2 d2 y
dy
+u
+ y = 0.
du2
du The auxiliary equation is m2 + 1 = 0 so that
y = c1 cos(ln u) + c2 sin(ln u) = c1 cos[ ln(x + 2)] + c2 sin[ ln(x + 2)].
40. If 1 − i is a root of the auxiliary equation then so is 1 + i, and the auxiliary equation is
(m − 2)[m − (1 + i)][m − (1 − i)] = m3 − 4m2 + 6m − 4 = 0.
We need m3 − 4m2 + 6m − 4 to have the form m(m − 1)(m − 2) + bm(m − 1) + cm + d. Expanding this last
expression and equating coeﬃcients we get b = −1, c = 3, and d = −4. Thus, the diﬀerential equation is
x3 y − x2 y + 3xy − 4y = 0. 41. For x2 y = 0 the auxiliary equation is m(m − 1) = 0 and the general solution is y = c1 + c2 x. The initial
conditions imply c1 = y0 and c2 = y1 , so y = y0 + y1 x. The initial conditions are satisﬁed for all real values of
y0 and y1 .
For x2 y − 2xy + 2y = 0 the auxiliary equation is m2 − 3m + 2 = (m − 1)(m − 2) = 0 and the general solution
is y = c1 x + c2 x2 . The initial condition y(0) = y0 implies 0 = y0 and the condition y (0) = y1 implies c1 = y1 .
Thus, the initial conditions are satisﬁed for y0 = 0 and for all real values of y1 .
For x2 y − 4xy + 6y = 0 the auxiliary equation is m2 − 5m + 6 = (m − 2)(m − 3) = 0 and the general solution
is y = c1 x2 + c2 x3 . The initial conditions imply y(0) = 0 = y0 and y (0) = 0. Thus, the initial conditions are
satisﬁed only for y0 = y1 = 0.
√
42. The function y(x) = − x cos(ln x) is deﬁned for x > 0 and has x-intercepts where ln x = π/2 + kπ for k an
integer or where x = eπ/2+kπ . Solving π/2 + kπ = 0.5 we get k ≈ −0.34, so eπ/2+kπ < 0.5 for all negative
integers and the graph has inﬁnitely many x-intercepts in the interval (0, 0.5). 141 3.6 Cauchy-Euler Equation 43. The auxiliary equation is 2m(m − 1)(m − 2) − 10.98m(m − 1) + 8.5m + 1.3 = 0, so that m1 = −0.053299,
m2 = 1.81164, m3 = 6.73166, and
y = c1 x−0.053299 + c2 x1.81164 + c3 x6.73166 .
44. The auxiliary equation is m(m − 1)(m − 2) + 4m(m − 1) + 5m − 9 = 0, so that m1 = 1.40819 and the two
complex roots are −1.20409 ± 2.22291i. The general solution of the diﬀerential equation is
y = c1 x1.40819 + x−1.20409 [c2 cos(2.22291 ln x) + c3 sin(2.22291 ln x)].
45. The auxiliary equation is m(m − 1)(m − 2)(m − 3) + 6m(m − 1)(m − 2) + 3m(m − 1) − 3m + 4 = 0, so that
√
√
m1 = m2 = 2 and m3 = m4 = − 2 . The general solution of the diﬀerential equation is
√ y = c1 x 2 √ + c2 x 2 √ ln x + c3 x− 2 √ + c4 x− 2 ln x. 46. The auxiliary equation is m(m − 1)(m − 2)(m − 3) − 6m(m − 1)(m − 2) + 33m(m − 1) − 105m + 169 = 0, so
that m1 = m2 = 3 + 2i and m3 = m4 = 3 − 2i. The general solution of the diﬀerential equation is
y = x3 [c1 cos(2 ln x) + c2 sin(2 ln x)] + x3 ln x[c3 cos(2 ln x) + c4 sin(2 ln x)].
47. The auxiliary equation
m(m − 1)(m − 2) − m(m − 1) − 2m + 6 = m3 − 4m2 + m + 6 = 0
has roots m1 = −1, m2 = 2, and m3 = 3, so yc = c1 x−1 + c2 x2 + c3 x3 . With y1 = x−1 , y2 = x2 , y3 = x3 , and
the identiﬁcation f (x) = 1/x, we get from (10) of Section 4.6 in the text
W1 = x3 , W2 = −4, W3 = 3/x, and W = 12x. Then u1 = W1 /W = x2 /12, u2 = W2 /W = −1/3x, u3 = 1/4x2 , and integration gives
u1 = x3
,
36 so
yp = u 1 y 1 + u 2 y 2 + u 3 y3 =
and 1
u2 = − ln x,
3 and u3 = − 1
1
x3 −1
x + x2 − ln x + x3 −
36
3
4x 2
1
y = yc + yp = c1 x−1 + c2 x2 + c3 x3 − x2 − x2 ln x,
9
3 142 1
,
4x
2
1
= − x2 − x2 ln x,
9
3
x > 0. 3.7 Nonlinear Equations EXERCISES 3.7
Nonlinear Equations 1. We have y1 = y1 = ex , so
2
(y1 )2 = (ex )2 = e2x = y1 . Also, y2 = − sin x and y2 = − cos x, so
2
(y2 )2 = (− cos x)2 = cos2 x = y2 . However, if y = c1 y1 + c2 y2 , we have (y )2 = (c1 ex − c2 cos x)2 and y 2 = (c1 ex + c2 cos x)2 . Thus (y )2 = y 2 .
2. We have y1 = y1 = 0, so
y1 y1 = 1 · 0 = 0 =
Also, y2 = 2x and y2 = 2, so 1 2
1
(0) = (y1 )2 .
2
2 1
1
(2x)2 = (y2 )2 .
2
2
2
= (c1 · 1 + c2 x )(c1 · 0 + 2c2 ) = 2c2 (c1 + c2 x2 ) and y2 y2 = x2 (2) = 2x2 =
However, if y = c1 y1 + c2 y2 , we have yy
1
2 [c1 · 0 + c2 (2x)] =
2 2c2 x2 .
2 Thus yy = 1
2 (y 1
2 (y )2 = 2 ) . 3. Let u = y so that u = y . The equation becomes u = −u − 1 which is separable. Thus
du
= −dx =⇒ tan−1 u = −x + c1 =⇒ y = tan(c1 − x) =⇒ y = ln | cos(c1 − x)| + c2 .
u2 + 1
4. Let u = y so that u = y . The equation becomes u = 1 + u2 . Separating variables we obtain
du
= dx =⇒ tan−1 u = x + c1 =⇒ u = tan(x + c1 ) =⇒ y = − ln | cos(x + c1 )| + c2 .
1 + u2
5. Let u = y so that u = y . The equation becomes x2 u + u2 = 0. Separating variables we obtain
du
1
1
dx
1
c1 x + 1
=⇒ u = −
= − 2 =⇒ − = + c1 =
2
u
x
u
x
x
c1
1
1
=⇒ y = 2 ln |c1 x + 1| − x + c2 .
c1
c1 x
x + 1/c1 = 1
c1 1
−1
c1 x + 1 6. Let u = y so that y = u du/dy. The equation becomes (y + 1)u du/dy = u2 . Separating variables we obtain
du
dy
=
=⇒ ln |u| = ln |y + 1| + ln c1 =⇒ u = c1 (y + 1)
u
y+1
dy
dy
=⇒
= c1 (y + 1) =⇒
= c1 dx
dx
y+1
=⇒ ln |y + 1| = c1 x + c2 =⇒ y + 1 = c3 ec1 x .
7. Let u = y so that y = u du/dy. The equation becomes u du/dy + 2yu3 = 0. Separating variables we obtain
du
1
1
1
+ 2y dy = 0 =⇒ − + y 2 = c =⇒ u = 2
=⇒ y = 2
u2
u
y + c1
y + c1
1
=⇒ y 2 + c1 dy = dx =⇒ y 3 + c1 y = x + c2 .
3 143 3.7 Nonlinear Equations 8. Let u = y so that y = u du/dy. The equation becomes y 2 u du/dy = u. Separating variables we obtain
du = dy
y
1
c1 y − 1
=⇒
dy = dx
=⇒ u = − + c1 =⇒ y =
y2
y
y
c1 y − 1
1
1
1
1
=⇒
1+
y + 2 ln |y − 1| = x + c2 .
dy = dx (for c1 = 0) =⇒
c1
c1 y − 1
c1
c1 If c1 = 0, then y dy = −dx and another solution is 1 y 2 = −x + c2 .
2
y 9. (a) 10 −π/2 x 3π/2 −10
(b) Let u = y so that y = u du/dy. The equation becomes u du/dy + yu = 0. Separating variables we obtain
1
1
du = −y dy =⇒ u = − y 2 + c1 =⇒ y = − y 2 + c1 .
2
2
When x = 0, y = 1 and y = −1 so −1 = −1/2 + c1 and c1 = −1/2. Then
dy
1
1
1
dy
1
= − y2 −
=⇒ 2
= − dx =⇒ tan−1 y = − x + c2
dx
2
2
y +1
2
2
1
=⇒ y = tan − x + c2 .
2
When x = 0, y = 1 so 1 = tan c2 and c2 = π/4. The solution of the initial-value problem is
y = tan π 1
− x .
4
2 The graph is shown in part (a).
(c) The interval of deﬁnition is −π/2 < π/4 − x/2 < π/2 or −π/2 < x < 3π/2.
10. Let u = y so that u = y . The equation becomes (u )2 + u2 = 1
√
√
which results in u = ± 1 − u2 . To solve u = 1 − u2 we separate y 2 variables:
du
√
= dx =⇒ sin−1 u = x + c1 =⇒ u = sin(x + c1 )
1 − u2
−2π
=⇒ y = sin(x + c1 ).
√
√
When x = π/2, y = 3/2, so 3/2 = sin(π/2 + c1 ) and c1 = −π/6. Thus
y = sin x − π
6 =⇒ y = − cos x − 144 π
+ c2 .
6 x 2π 3.7 Nonlinear Equations
When x = π/2, y = 1/2, so 1/2 = − cos(π/2−π/6)+c2 = −1/2+c2 and c2 = 1. The solution of the initial-value
problem is y = 1 − cos(x − π/6).
√
To solve u = − 1 − u2 we separate variables:
du
√
= −dx =⇒ cos−1 u = x + c1
1 − u2
=⇒ u = cos(x + c1 ) =⇒ y = cos(x + c1 ).
√
√
When x = π/2, y = 3/2, so 3/2 = cos(π/2 + c1 ) and c1 = −π/3. Thus
π
y = cos x −
3 y 1
x −2π 2π
−1 π
=⇒ y = sin x −
+ c2 .
3 When x = π/2, y = 1/2, so 1/2 = sin(π/2 − π/3) + c2 = 1/2 + c2 and c2 = 0. The solution of the initial-value
problem is y = sin(x − π/3).
11. Let u = y so that u = y . The equation becomes u − (1/x)u = (1/x)u3 , which is Bernoulli. Using w = u−2
we obtain dw/dx + (2/x)w = −2/x. An integrating factor is x2 , so
d 2
c1
[x w] = −2x =⇒ x2 w = −x2 + c1 =⇒ w = −1 + 2
dx
x
c1
x
=⇒ u−2 = −1 + 2 =⇒ u = √
x
c1 − x2
dy
x
=⇒
=⇒ y = − c1 − x2 + c2
=√
dx
c1 − x2
=⇒ c1 − x2 = (c2 − y)2 =⇒ x2 + (c2 − y)2 = c1 .
12. Let u = y so that u = y . The equation becomes u − (1/x)u = u2 , which is a Bernoulli diﬀerential equation.
Using the substitution w = u−1 we obtain dw/dx + (1/x)w = −1. An integrating factor is x, so
d
1
1
1
c1 − x2
2x
=⇒ y = − ln c1 − x2 + c2 .
[xw] = −x =⇒ w = − x + c =⇒
=
=⇒ u =
dx
2
x
u
2x
c1 − x2
In Problems 13-16 the thinner curve is obtained using a numerical solver, while the thicker curve is the graph of the
Taylor polynomial.
13. We look for a solution of the form
1
1
1
1
y(x) = y(0) + y (0)x + y (0)x2 + y (0)x3 + y (4) (0)x4 + y (5) (0)x5 .
2!
3!
4!
5! y
40 From y (x) = x + y 2 we compute
y (x) = 1 + 2yy 30 y (4) (x) = 2yy + 2(y )2
y (5) (x) = 2yy + 6y y . 20 Using y(0) = 1 and y (0) = 1 we ﬁnd
y (0) = 1, y (0) = 3, y (4) (0) = 4, y (5) (0) = 12.
10 An approximate solution is
1
1
1
1
y(x) = 1 + x + x2 + x3 + x4 + x5 .
2
2
6
10 145 0.5 1 1.5 2 2.5 3 x 3.7 Nonlinear Equations 14. We look for a solution of the form
1
1
1
1
y(x) = y(0) + y (0)x + y (0)x2 + y (0)x3 + y (4) (0)x4 + y (5) (0)x5 .
2!
3!
4!
5! y
10 From y (x) = 1 − y 2 we compute
y (x) = −2yy
y (4) 5 (x) = −2yy − 2(y ) 2 y (5) (x) = −2yy − 6y y .
0.5 1 1.5 2 2.5 3 x Using y(0) = 2 and y (0) = 3 we ﬁnd
y (0) = −3, y (0) = −12, y (4) (0) = −6, y (5) (0) = 102.
-5 An approximate solution is
3
1
17
y(x) = 2 + 3x − x2 − 2x3 − x4 + x5 .
2
4
20 -10 15. We look for a solution of the form
1
1
1
1
y(x) = y(0) + y (0)x + y (0)x2 + y (0)x3 + y (4) (0)x4 + y (5) (0)x5 .
2!
3!
4!
5! y
40 From y (x) = x2 + y 2 − 2y we compute
30 y (x) = 2x + 2yy − 2y
y (4) (x) = 2 + 2(y )2 + 2yy − 2y
y (5) (x) = 6y y + 2yy − 2y (4) . 20 Using y(0) = 1 and y (0) = 1 we ﬁnd
y (0) = −1, y (0) = 4, y (4) (0) = −6, y (5) (0) = 14.
10 An approximate solution is
1
2
1
7
y(x) = 1 + x − x2 + x3 − x4 + x5 .
2
3
4
60 0.5 1 1.5 2 2.5 3 3.5x 16. We look for a solution of the form
1
1
1
y (0)x2 + y (0)x3 + y (4) (0)x4
2!
3!
4!
1 (5)
1 (6)
5
6
+ y (0)x + y (0)x .
5!
6! y(x) = y(0) + y (0)x + From y (x) = ey we compute y
10
8
6 y (x) = ey y 4 y (4) (x) = ey (y )2 + ey y
2 y (5) (x) = ey (y )3 + 3ey y y + ey y
y (6) (x) = ey (y )4 + 6ey (y )2 y + 3ey (y )2 + 4ey y y + ey y (4) . 1
-2 146 2 3 4 5x 3.7 Nonlinear Equations
Using y(0) = 0 and y (0) = −1 we ﬁnd
y (0) = 1, y (0) = −1, y (5) (0) = −5, y (4) (0) = 2, y (6) (0) = 16. An approximate solution is
1
1
1
1
1
y(x) = −x + x2 − x3 + x4 + x5 + x6 .
2
6
12
24
45
17. We need to solve [1 + (y )2 ]3/2 = y . Let u = y so that u = y . The equation becomes (1 + u2 )3/2 = u or
(1 + u2 )3/2 = du/dx. Separating variables and using the substitution u = tan θ we have
du
3/2 (1 + u2 ) sec2 θ = dx =⇒ 1 + tan2 θ 3/2 dθ = x =⇒ sec2 θ
dθ = x
sec3 θ
u
=x
1 + u2
x2
1 + (y )2 =
1 − x2 cos θ dθ = x =⇒ sin θ = x =⇒ √ =⇒ y = x =⇒ (y )2 = x2
1 + (y )2
x
=⇒ y = √
(for x > 0) =⇒ y = − 1 − x2 .
1 − x2
=⇒ 18. When y = sin x, y = cos x, y = sin x, and
(y )2 − y 2 = sin2 x − sin2 x = 0.
When y = e−x , y = −e−x , y = e−x , and
(y )2 − y 2 = e−2x − e−2x = 0.
From (y )2 − y 2 = 0 we have y = ±y, which can be treated as two linear equations. Since linear combinations
of solutions of linear homogeneous diﬀerential equations are also solutions, we see that y = c1 ex + c2 e−x and
y = c3 cos x + c4 sin x must satisfy the diﬀerential equation. However, linear combinations that involve both
exponential and trigonometric functions will not be solutions since the diﬀerential equation is not linear and each
type of function satisﬁes a diﬀerent linear diﬀerential equation that is part of the original diﬀerential equation.
19. Letting u = y , separating variables, and integrating we have
du
=
dx √ 1 + u2 , du
= dx,
1 + u2 and sinh−1 u = x + c1 . Then
u = y = sinh(x + c1 ), y = cosh(x + c1 ) + c2 , and y = sinh(x + c1 ) + c2 x + c3 . 20. If the constant −c2 is used instead of c2 , then, using partial fractions,
1
1
y=− dx
1
=−
x2 − c2
2c1
1 1
1
−
x − c1
x + c1 dx = 1
ln
2c1 x + c1
x − c1 +c2 . Alternatively, the inverse hyperbolic tangent can be used.
21. Let u = dx/dt so that d2 x/dt2 = u du/dx. The equation becomes u du/dx = −k 2 /x2 . Separating variables we
obtain k2
1
1
k2
k2
+ c =⇒ v 2 =
+ c.
dx =⇒ u2 =
x2
2
x
2
x
When t = 0, x = x0 and v = 0 so 0 = (k 2 /x0 ) + c and c = −k 2 /x0 . Then
u du = − 1 2
v = k2
2 1
1
−
x x0 and 147 √
dx
= −k 2
dt x0 − x
.
xx0 3.7 Nonlinear Equations Separating variables we have
√
xx0
1
dx = k 2 dt =⇒ t = −
x0 − x
k − x0
2 x
dx.
x0 − x Using Mathematica to integrate we obtain
t=−
=
22. x0
(x0 − 2x)
x0
− x(x0 − x) −
tan−1
2
2
2x 1
k 1
k x0
2 x(x0 − x) + x x0 − 2x
x0
tan−1
2
2 x(x0 − x) . x 2 x 2 10
-2 x
x0 − x 20 2 t 10
-2 x1 = 0 20 t 10
-2 x1 = 1 20 t x1 = -1.5 For d2 x/dt2 + sin x = 0 the motion appears to be periodic with amplitude 1 when x1 = 0. The amplitude and
period are larger for larger magnitudes of x1 .
x x 1 -1 x 1 x1 = 0 10 1 t -1 x1 = 1 10 t -1 x1 = -2.5 10 t For d2 x/dt2 + dx/dt + sin x = 0 the motion appears to be periodic with decreasing amplitude. The dx/dt term
could be said to have a damping eﬀect. EXERCISES 3.8
Linear Models: Initial-Value Problems 1. From 1 x + 16x = 0 we obtain
8 √
√
x = c1 cos 8 2 t + c2 sin 8 2 t
√
√
so that the period of motion is 2π/8 2 = 2 π/8 seconds. 2. From 20x + kx = 0 we obtain
x = c1 cos 1
2 k
1
t + c2 sin
5
2 k
t
5 so that the frequency 2/π = 1 k/5 π and k = 320 N/m. If 80x + 320x = 0 then x = c1 cos 2t + c2 sin 2t so
4
that the frequency is 2/2π = 1/π cycles/s.
√
3. From 3 x + 72x = 0, x(0) = −1/4, and x (0) = 0 we obtain x = − 1 cos 4 6 t.
4
4 148 3.8
4. From 3 x + 72x = 0, x(0) = 0, and x (0) = 2 we obtain x =
4 √ 6
12 5. From 5 x + 40x = 0, x(0) = 1/2, and x (0) = 0 we obtain x =
8 Linear Models: Initial-Value Problems √
sin 4 6 t. 1
2 cos 8t. (a) x(π/12) = −1/4, x(π/8) = −1/2, x(π/6) = −1/4, x(π/4) = 1/2, x(9π/32) = √ 2/4. (b) x = −4 sin 8t so that x (3π/16) = 4 ft/s directed downward.
(c) If x = 1
2 cos 8t = 0 then t = (2n + 1)π/16 for n = 0, 1, 2, . . . . 6. From 50x + 200x = 0, x(0) = 0, and x (0) = −10 we obtain x = −5 sin 2t and x = −10 cos 2t.
7. From 20x + 20x = 0, x(0) = 0, and x (0) = −10 we obtain x = −10 sin t and x = −10 cos t.
(a) The 20 kg mass has the larger amplitude.
√
(b) 20 kg: x (π/4) = −5 2 m/s, x (π/2) = 0 m/s; 50 kg: x (π/4) = 0 m/s, x (π/2) = 10 m/s (c) If −5 sin 2t = −10 sin t then 2 sin t(cos t − 1) = 0 so that t = nπ for n = 0, 1, 2, . . ., placing both masses
at the equilibrium position. The 50 kg mass is moving upward; the 20 kg mass is moving upward when n
is even and downward when n is odd.
8. From x + 16x = 0, x(0) = −1, and x (0) = −2 we obtain √
1
5
x = − cos 4t − sin 4t =
cos(4t − 3.605).
2
2
√
The period is π/2 seconds and the amplitude is 5/2 feet. In 4 seconds it will make 8 complete cycles. 9. From 1 x + x = 0, x(0) = 1/2, and x (0) = 3/2 we obtain
4
√
1
13
3
x = cos 2t + sin 2t =
sin(2t + 0.588).
2
4
4
10. From 1.6x + 40x = 0, x(0) = −1/3, and x (0) = 5/4 we obtain
1
1
5
x = − cos 5t + sin 5t =
sin(5t − 0.927).
3
4
12
If x = 5/24 then t = 1
5 π
6 + 0.927 + 2nπ and t = 1
5 5π
6 + 0.927 + 2nπ for n = 0, 1, 2, . . . . 11. From 2x + 200x = 0, x(0) = −2/3, and x (0) = 5 we obtain
(a) x = − 2 cos 10t +
3 1
2 sin 10t = 5
6 sin(10t − 0.927). (b) The amplitude is 5/6 ft and the period is 2π/10 = π/5
(c) 3π = πk/5 and k = 15 cycles.
(d) If x = 0 and the weight is moving downward for the second time, then 10t − 0.927 = 2π or t = 0.721 s.
(e) If x = 25
3 cos(10t − 0.927) = 0 then 10t − 0.927 = π/2 + nπ or t = (2n + 1)π/20 + 0.0927 for n = 0, 1, 2, . . . .
(f ) x(3) = −0.597 ft
(g) x (3) = −5.814 ft/s
(h) x (3) = 59.702 ft/s2
(i) If x = 0 then t =
(j) If x = 5/12 then
(k) If x = 5/12 and 1
10 (0.927 + nπ) for n = 0, 1, 2, . . .. The velocity at these times is x = ±8.33 ft/s.
1
1
t = 10 (π/6 + 0.927 + 2nπ) and t = 10 (5π/6 + 0.927 + 2nπ) for n = 0, 1, 2, . . . .
1
x < 0 then t = 10 (5π/6 + 0.927 + 2nπ) for n = 0, 1, 2, . . . . √
12. From x + 9x = 0, x(0) = −1, and x (0) = − 3 we obtain
√
3
4π
2
x = − cos 3t −
sin 3t = √ sin 3t +
3
3
3
√
and x = 2 3 cos(3t + 4π/3). If x = 3 then t = −7π/18 + 2nπ/3 and t = −π/2 + 2nπ/3 for n = 1, 2, 3, . . . . 149 3.8 Linear Models: Initial-Value Problems 13. From k1 = 40 and k2 = 120 we compute the eﬀective spring constant k = 4(40)(120)/160 = 120. Now,
m = 20/32 so k/m = 120(32)/20 = 192 and x + 192x = 0. Using x(0) = 0 and x (0) = 2 we obtain
√
√
3
x(t) = 12 sin 8 3 t.
14. Let m be the mass and k1 and k2 the spring constants. Then k = 4k1 k2 /(k1 + k2 ) is the eﬀective spring constant
of the system. Since the initial mass stretches one spring 1 foot and another spring 1 foot, using F = ks, we
3
2
= 1 k2 or 2k1 = 3k2 . The given period of the combined system is 2π/ω = π/15, so ω = 30. Since a
2
mass weighing 8 pounds is 1 slug, we have from w2 = k/m
4
have 1
3 k1 302 = k
= 4k
1/4 or k = 225. We now have the system of equations
4k1 k2
= 225
k1 + k2
2k1 = 3k2 .
Solving the second equation for k1 and substituting in the ﬁrst equation, we obtain
2
4(3k2 /2)k2
12k2
12k2
=
=
= 225.
3k2 /2 + k2
5k2
5 Thus, k2 = 375/4 and k1 = 1125/8. Finally, the weight of the ﬁrst mass is
32m = k1
1125/8
375
=
=
≈ 46.88 lb.
3
3
8 15. For large values of t the diﬀerential equation is approximated by x = 0. The solution of this equation is the
linear function x = c1 t + c2 . Thus, for large time, the restoring force will have decayed to the point where the
spring is incapable of returning the mass, and the spring will simply keep on stretching.
16. As t becomes larger the spring constant increases; that is, the spring is stiﬀening. It would seem that the
oscillations would become periodic and the spring would oscillate more rapidly. It is likely that the amplitudes
of the oscillations would decrease as t increases.
17. (a) above (b) heading upward 18. (a) below (b) from rest 19. (a) below (b) heading upward 20. (a) above (b) heading downward + x + 2x = 0, x(0) = −1, and x (0) = 8 we obtain x = 4te−4t − e−4t and x = 8e−4t − 16te−4t . If
x = 0 then t = 1/4 second. If x = 0 then t = 1/2 second and the extreme displacement is x = e−2 feet.
√
√
√
√
22. From 1 x + 2 x + 2x = 0, x(0) = 0, and x (0) = 5 we obtain x = 5te−2 2 t and x = 5e−2 2 t 1 − 2 2 t . If
4
√
√
x = 0 then t = 2/4 second and the extreme displacement is x = 5 2 e−1 /4 feet. 21. From 1
8x 23. (a) From x + 10x + 16x = 0, x(0) = 1, and x (0) = 0 we obtain x = 4 e−2t − 1 e−8t .
3
3
(b) From x + x + 16x = 0, x(0) = 1, and x (0) = −12 then x = − 2 e−2t + 5 e−8t .
3
3
24. (a) x = 1 e−8t 4e6t − 1 is not zero for t ≥ 0; the extreme displacement is x(0) = 1 meter.
3
(b) x = 1 e−8t 5 − 2e6t = 0 when t = 1 ln 5 ≈ 0.153 second; if x = 4 e−8t e6t − 10 = 0 then t =
3
6
2
3
0.384 second and the extreme displacement is x = −0.232 meter. 150 1
6 ln 10 ≈ 3.8 Linear Models: Initial-Value Problems 25. (a) From 0.1x + 0.4x + 2x = 0, x(0) = −1, and x (0) = 0 we obtain x = e−2t − cos 4t −
√
5 −2t
(b) x =
e
sin(4t + 4.25)
2 1
2 sin 4t . (c) If x = 0 then 4t + 4.25 = 2π, 3π, 4π, . . . so that the ﬁrst time heading upward is t = 1.294 seconds.
26. (a) From 1 x + x + 5x = 0, x(0) = 1/2, and x (0) = 1 we obtain x = e−2t
4 1
2 cos 4t + 1
2 sin 4t . 1
π
(b) x = √ e−2t sin 4t +
.
4
2
(c) If x = 0 then 4t + π/4 = π, 2π, 3π, . . . so that the times heading downward are t = (7 + 8n)π/16 for
n = 0, 1, 2, . . . .
(d) x
1 .5 .5 2 t -.5 -1 27. From 5
16 x + βx + 5x = 0 we ﬁnd that the roots of the auxiliary equation are m = − 8 β ±
5 4
5 4β 2 − 25 . (a) If 4β 2 − 25 > 0 then β > 5/2.
(b) If 4β 2 − 25 = 0 then β = 5/2.
(c) If 4β 2 − 25 < 0 then 0 < β < 5/2. √
28. From 0.75x + βx + 6x = 0 and β > 3 2 we ﬁnd that the roots of the auxiliary equation are
m = −2β ±
3 2
3 β 2 − 18 and
x = e−2βt/3 c1 cosh 2
3 β 2 − 18 t + c2 sinh If x(0) = 0 and x (0) = −2 then c1 = 0 and c2 = −3/ 2
3 β 2 − 18 t . β 2 − 18. 29. If 1 x + 1 x + 6x = 10 cos 3t, x(0) = −2, and x (0) = 0 then
2
2
√
√
47
47
−t/2
c1 cos
xc = e
t + c2 sin
t
2
2
and xp = 10
3 (cos 3t + sin 3t) so that the equation of motion is
√
√
4
47
47
64
x = e−t/2 − cos
t − √ sin
t
3
2
2
3 47 + 10
(cos 3t + sin 3t).
3 30. (a) If x + 2x + 5x = 12 cos 2t + 3 sin 2t, x(0) = 1, and x (0) = 5 then xc = e−t (c1 cos 2t + c2 sin 2t) and
xp = 3 sin 2t so that the equation of motion is
x = e−t cos 2t + 3 sin 2t. 151 3.8 Linear Models: Initial-Value Problems
x (b) (c) steady-state 3 2
-3 4 6 x
3 x=xc+xp t 2 4 6 t -3 transient 31. From x + 8x + 16x = 8 sin 4t, x(0) = 0, and x (0) = 0 we obtain xc = c1 e−4t + c2 te−4t and xp = − 1 cos 4t so
4
that the equation of motion is
1
1
x = e−4t + te−4t − cos 4t.
4
4
32. From x + 8x + 16x = e−t sin 4t, x(0) = 0, and x (0) = 0 we obtain xc = c1 e−4t + c2 te−4t and xp =
24
7
− 625 e−t cos 4t − 625 e−t sin 4t so that
x= 1 −t
1 −4t
e (24 + 100t) −
e (24 cos 4t + 7 sin 4t).
625
625 As t → ∞ the displacement x → 0.
33. From 2x + 32x = 68e−2t cos 4t, x(0) = 0, and x (0) = 0 we obtain xc = c1 cos 4t + c2 sin 4t and xp =
1 −2t
2e cos 4t − 2e−2t sin 4t so that
1
9
1
x = − cos 4t + sin 4t + e−2t cos 4t − 2e−2t sin 4t.
2
4
2
√ 85
4 34. Since x = sin(4t − 0.219) − √ 17 −2t
2 e sin(4t − 2.897), the amplitude approaches √ 85/4 as t → ∞. 35. (a) By Hooke’s law the external force is F (t) = kh(t) so that mx + βx + kx = kh(t).
(b) From
xp = 1
2 x + 2x + 4x
56
32
13 cos t + 13 sin t = 20 cos t, x(0) = 0, and x (0) = 0 we obtain xc = e−2t (c1 cos 2t + c2 sin 2t) and
so that
x = e−2t − 72
56
32
56
cos 2t −
sin 2t +
cos t +
sin t.
13
13
13
13 36. (a) From 100x + 1600x = 1600 sin 8t, x(0) = 0, and x (0) = 0 we obtain xc = c1 cos 4t + c2 sin 4t and
xp = − 1 sin 8t so that by a trig identity
3
x=
(b) If x = 1
3 2
1
2
2
sin 4t − sin 8t = sin 4t − sin 4t cos 4t.
3
3
3
3 sin 4t(2 − 2 cos 4t) = 0 then t = nπ/4 for n = 0, 1, 2, . . . . − cos 4t)(1 + 2 cos 4t) = 0 then t = π/3 + nπ/2 and t = π/6 + nπ/2 for
n = 0, 1, 2, . . . at the extreme values. Note: There are many other values of t for which x = 0.
√
√
(d) x(π/6 + nπ/2) = 3/2 cm and x(π/3 + nπ/2) = − 3/2 cm.
(c) If x = (e) 8
3 cos 4t − 8
3 cos 8t = 8
3 (1 x
1 1 2 3 t -1 152 3.8 Linear Models: Initial-Value Problems 37. From x + 4x = −5 sin 2t + 3 cos 2t, x(0) = −1, and x (0) = 1 we obtain xc = c1 cos 2t + c2 sin 2t, xp =
3
5
4 t sin 2t + 4 t cos 2t, and
1
3
5
x = − cos 2t − sin 2t + t sin 2t + t cos 2t.
8
4
4
38. From x + 9x = 5 sin 3t, x(0) = 2, and x (0) = 0 we obtain xc = c1 cos 3t + c2 sin 3t, xp = − 5 t cos 3t, and
6
x = 2 cos 3t + 5
5
sin 3t − t cos 3t.
18
6 39. (a) From x + ω 2 x = F0 cos γt, x(0) = 0, and x (0) = 0 we obtain xc = c1 cos ωt + c2 sin ωt and xp =
(F0 cos γt)/ ω 2 − γ 2 so that
x=−
(b) lim γ→ω ω2 F0
F0
cos ωt + 2
cos γt.
2
−γ
ω − γ2 F0
−F0 t sin γt
F0
(cos γt − cos ωt) = lim
=
t sin ωt.
γ→ω
ω2 − γ 2
−2γ
2ω 40. From x + ω 2 x = F0 cos ωt, x(0) = 0, and x (0) = 0 we obtain xc = c1 cos ωt + c2 sin ωt and xp = (F0 t/2ω) sin ωt
so that x = (F0 t/2ω) sin ωt.
41. (a) From cos(u − v) = cos u cos v + sin u sin v and cos(u + v) = cos u cos v − sin u sin v we obtain sin u sin v =
1
2 [cos(u (b) If − v) − cos(u + v)]. Letting u = 1 (γ − ω)t and v = 1 (γ + ω)t, the result follows.
2
2 = 1 (γ − ω) then γ ≈ ω so that x = (F0 /2 γ) sin t sin γt.
2 42. See the article “Distinguished Oscillations of a Forced Harmonic Oscillator” by T.G. Procter in The College
Mathematics Journal, March, 1995. In this article the author illustrates that for F0 = 1, λ = 0.01, γ = 22/9,
and ω = 2 the system exhibits beats oscillations on the interval [0, 9π], but that this phenomenon is transient
as t → ∞.
x 1
t π −1 9π 43. (a) The general solution of the homogeneous equation is
xc (t) = c1 e−λt cos( ω 2 − λ2 t) + c2 e−λt sin( ω 2 − λ2 t) = Ae−λt sin[ ω 2 − λ2 t + φ],
where A = c2 + c2 , sin φ = c1 /A, and cos φ = c2 /A. Now
1
2
xp (t) = (ω 2 F0 (ω 2 − γ 2 )
F0 (−2λγ)
sin γt + 2
cos γt = A sin(γt + θ),
− γ 2 )2 + 4λ2 γ 2
(ω − γ 2 )2 + 4λ2 γ 2 where sin θ = (ω 2 F0 (−2λγ)
− γ 2 )2 + 4λ2 γ 2 ω2 F0
− γ 2 + 4λ2 γ 2 = and 153 −2λγ
(ω 2 − γ 2 )2 + 4λ2 γ 2 3.8 Linear Models: Initial-Value Problems cos θ = (ω 2 F0 (ω 2 − γ 2 )
− γ 2 )2 + 4λ2 γ 2 ω2 − γ 2 = F0
(ω 2 − γ 2 )2 + 4λ2 γ 2 (ω 2 − γ 2 )2 + 4λ2 γ 2 . √
(b) If g (γ) = 0 then γ γ 2 + 2λ2 − ω 2 = 0 so that γ = 0 or γ = ω 2 − 2λ2 . The ﬁrst derivative test shows
√
that g has a maximum value at γ = ω 2 − 2λ2 . The maximum value of g is
ω 2 − 2λ2 g = F0 /2λ ω 2 − λ2 . √
(c) We identify ω 2 = k/m = 4, λ = β/2, and γ1 = ω 2 − 2λ2 = 4 − β 2 /2 . As β → 0, γ1 → 2 and the
resonance curve grows without bound at γ1 = 2. That is, the system approaches pure resonance. g
β
2.00
1.00
0.75
0.50
0.25 γ1
1.41
1.87
1.93
1.97
1.99 g
0.58
1.03
1.36
2.02
4.01 β=1/4 4 β=0 3
2
1
1 44. (a) For n = 2, sin2 γt =
γ1 = ω/2. 1
2 (1 2 β=1/2
β=0
β=3/4
β=0
β=1
β=1
β=2
β=2
γ 3 4 − cos 2γt). The system is in pure resonance when 2γ1 /2π = ω/2π, or when (b) Note that
sin3 γt = sin γt sin2 γt = 1
[sin γt − sin γt cos 2γt].
2 Now
sin(A + B) + sin(A − B) = 2 sin A cos B
so
sin γt cos 2γt = 1
[sin 3γt − sin γt]
2 and
sin3 γt = 1
3
sin γt − sin 3γt.
4
4 Thus
x + ω2 x = 1
3
sin γt − sin 3γt.
4
4 The frequency of free vibration is ω/2π. Thus, when γ1 /2π = ω/2π or γ1 = ω, and when 3γ2 /2π = ω/2π
or 3γ2 = ω or γ3 = ω/3, the system will be in pure resonance.
(c) γ1=1/2 x
10 γ1=1 x
10 n=2 x n=3 10 5 5
10 20 30 t γ2=1/3
n=3 5
10 20 30 t 20 -5 -5 -5 -10 -10 -10 154 40 t 3.8 Linear Models: Initial-Value Problems + 2q + 100q = 0 we obtain q(t) = e−20t (c1 cos 40t + c2 sin 40t). The initial conditions q(0) = 5 and
q (0) = 0 imply c1 = 5 and c2 = 5/2. Thus 45. Solving 1
20 q q(t) = e−20t 5 cos 40t +
and q(0.01) ≈ 4.5676 coulombs.
0.0509 second. 5
sin 40t
2 = 25 + 25/4 e−20t sin(40t + 1.1071) The charge is zero for the ﬁrst time when 40t + 1.1071 = π or t ≈ 46. Solving 1 q + 20q + 300q = 0 we obtain q(t) = c1 e−20t + c2 e−60t . The initial conditions q(0) = 4 and q (0) = 0
4
imply c1 = 6 and c2 = −2. Thus
q(t) = 6e−20t − 2e−60t .
Setting q = 0 we ﬁnd e40t = 1/3 which implies t < 0. Therefore the charge is not 0 for t ≥ 0.
47. Solving 5
3q + 10q + 30q = 300 we obtain q(t) = e−3t (c1 cos 3t + c2 sin 3t) + 10. The initial conditions q(0) = q (0) = 0 imply c1 = c2 = −10. Thus
q(t) = 10 − 10e−3t (cos 3t + sin 3t) and i(t) = 60e−3t sin 3t. Solving i(t) = 0 we see that the maximum charge occurs when t = π/3 and q(π/3) ≈ 10.432.
48. Solving q + 100q + 2500q = 30 we obtain q(t) = c1 e−50t + c2 te−50t + 0.012. The initial conditions q(0) = 0
and q (0) = 2 imply c1 = −0.012 and c2 = 1.4. Thus, using i(t) = q (t) we get
q(t) = −0.012e−50t + 1.4te−50t + 0.012 and i(t) = 2e−50t − 70te−50t . Solving i(t) = 0 we see that the maximum charge occurs when t = 1/35 second and q(1/35) ≈ 0.01871 coulomb.
√
√
49. Solving q + 2q + 4q = 0 we obtain qc = e−t cos 3 t + sin 3 t . The steady-state charge has the form
qp = A cos t + B sin t. Substituting into the diﬀerential equation we ﬁnd
(3A + 2B) cos t + (3B − 2A) sin t = 50 cos t.
Thus, A = 150/13 and B = 100/13. The steady-state charge is
qp (t) = 150
100
cos t +
sin t
13
13 and the steady-state current is
ip (t) = − 100
150
sin t +
cos t.
13
13 50. From
ip (t) =
and Z = √ E0
Z R
X
sin γt −
cos γt
Z
Z X 2 + R2 we see that the amplitude of ip (t) is
A= 2
E0 R 2
E2X 2
E0
+ 04 = 2
4
Z
Z
Z R2 + X 2 = E0
.
Z 51. The diﬀerential equation is 1 q +20q +1000q = 100 sin 60t. To use Example 10 in the text we identify E0 = 100
2
and γ = 60. Then
X = Lγ −
Z= 1
1
1
= (60) −
≈ 13.3333,
cγ
2
0.001(60) X 2 + R2 = X 2 + 400 ≈ 24.0370, and 155 3.8 Linear Models: Initial-Value Problems
E0
100
=
≈ 4.1603.
Z
Z
From Problem 50, then
ip (t) ≈ 4.1603 sin(60t + φ)
where sin φ = −X/Z and cos φ = R/Z. Thus tan φ = −X/R ≈ −0.6667 and φ is a fourth quadrant angle. Now
φ ≈ −0.5880 and
ip (t) = 4.1603 sin(60t − 0.5880). 52. Solving 1 q + 20q + 1000q = 0 we obtain qc (t) = e−20t (c1 cos 40t + c2 sin 40t). The steady-state charge has the
2
form qp (t) = A sin 60t + B cos 60t + C sin 40t + D cos 40t. Substituting into the diﬀerential equation we ﬁnd
(−1600A − 2400B) sin 60t + (2400A − 1600B) cos 60t
+ (400C − 1600D) sin 40t + (1600C + 400D) cos 40t
= 200 sin 60t + 400 cos 40t.
Equating coeﬃcients we obtain A = −1/26, B = −3/52, C = 4/17, and D = 1/17. The steady-state charge is
qp (t) = − 1
3
4
1
sin 60t −
cos 60t +
sin 40t +
cos 40t
26
52
17
17 and the steady-state current is
ip (t) = − 30
45
160
40
cos 60t +
sin 60t +
cos 40t −
sin 40t.
13
13
17
17 + 10q + 100q = 150 we obtain q(t) = e−10t (c1 cos 10t + c2 sin 10t) + 3/2. The initial conditions
q(0) = 1 and q (0) = 0 imply c1 = c2 = −1/2. Thus 53. Solving 1
2q 1
3
q(t) = − e−10t (cos 10t + sin 10t) + .
2
2
As t → ∞, q(t) → 3/2.
54. In Problem 50 it is shown that the amplitude of the steady-state current is E0 /Z, where
√
Z = X 2 + R2 and X = Lγ − 1/Cγ. Since E0 is constant the amplitude will be a maximum when Z is
a minimum. Since R is constant, Z will be a minimum when X = 0. Solving Lγ − 1/Cγ = 0 for γ we obtain
√
γ = 1/ LC . The maximum amplitude will be E0 /R.
√
55. By Problem 50 the amplitude of the steady-state current is E0 /Z, where Z = X 2 + R2 and X = Lγ − 1/Cγ.
Since E0 is constant the amplitude will be a maximum when Z is a minimum. Since R is constant, Z will be a
minimum when X = 0. Solving Lγ − 1/Cγ = 0 for C we obtain C = 1/Lγ 2 .
56. Solving 0.1q + 10q = 100 sin γt we obtain
q(t) = c1 cos 10t + c2 sin 10t + qp (t)
where qp (t) = A sin γt + B cos γt. Substituting qp (t) into the diﬀerential equation we ﬁnd
(100 − γ 2 )A sin γt + (100 − γ 2 )B cos γt = 100 sin γt.
Equating coeﬃcients we obtain A = 100/(100 − γ 2 ) and B = 0. Thus, qp (t) =
conditions q(0) = q (0) = 0 imply c1 = 0 and c2 = −10γ/(100 − γ 2 ). The charge is
q(t) = 10
(10 sin γt − γ sin 10t)
100 − γ 2 156 100
sin γt. The initial
100 − γ 2 3.9
and the current is
i(t) = Linear Models: Boundary-Value Problems 100γ
(cos γt − cos 10t).
100 − γ 2 57. In an LC-series circuit there is no resistor, so the diﬀerential equation is
L d2 q
1
+ q = E(t).
dt2
C √
√
Then q(t) = c1 cos t/ LC + c2 sin t/ LC + qp (t) where qp (t) = A sin γt + B cos γt. Substituting qp (t) into
the diﬀerential equation we ﬁnd
1
− Lγ 2 A sin γt +
C 1
− Lγ 2 B cos γt = E0 cos γt.
C Equating coeﬃcients we obtain A = 0 and B = E0 C/(1 − LCγ 2 ). Thus, the charge is
q(t) = c1 cos √ 1
1
E0 C
cos γt.
t + c2 sin √
t+
1 − LCγ 2
LC
LC √
The initial conditions q(0) = q0 and q (0) = i0 imply c1 = q0 − E0 C/(1 − LCγ 2 ) and c2 = i0 LC . The current
is i(t) = q (t) or
c1
1
c2
1
E0 Cγ
i(t) = − √
sin γt
sin √
t+ √
cos √
t−
1 − LCγ 2
LC
LC
LC
LC
1
E0 Cγ
sin γt.
t−
1 − LCγ 2
LC
√
58. When the circuit is in resonance the form of qp (t) is qp (t) = At cos kt+Bt sin kt where k = 1/ LC . Substituting
qp (t) into the diﬀerential equation we ﬁnd
= i0 cos √ 1
1
t− √
LC
LC q0 − E0 C
1 − LCγ 2 sin √ qp + k 2 qp = −2kA sin kt + 2kB cos kt = E0
cos kt.
L Equating coeﬃcients we obtain A = 0 and B = E0 /2kL. The charge is
q(t) = c1 cos kt + c2 sin kt + E0
t sin kt.
2kL The initial conditions q(0) = q0 and q (0) = i0 imply c1 = q0 and c2 = i0 /k. The current is
i(t) = −c1 k sin kt + c2 k cos kt + E0
(kt cos kt + sin kt)
2kL E0
E0
− q0 k sin kt + i0 cos kt +
t cos kt.
2kL
2L = EXERCISES 3.9
Linear Models: Boundary-Value Problems 1. (a) The general solution is
y(x) = c1 + c2 x + c3 x2 + c4 x3 + 157 w0 4
x .
24EI 3.9 Linear Models: Boundary-Value Problems The boundary conditions are y(0) = 0, y (0) = 0, y (L) = 0, y (L) = 0. The ﬁrst two conditions give
c1 = 0 and c2 = 0. The conditions at x = L give the system
w0 2
2c3 + 6c4 L +
L =0
2EI
w0
6c4 +
L = 0.
EI
Solving, we obtain c3 = w0 L2 /4EI and c4 = −w0 L/6EI. The deﬂection is
w0
y(x) =
(6L2 x2 − 4Lx3 + x4 ).
24EI
x
(b)
0.2 0.4 0.6 0.8 1 1 2 3 y
2. (a) The general solution is w0 4
x .
24EI
The boundary conditions are y(0) = 0, y (0) = 0, y(L) = 0, y (L) = 0. The ﬁrst two conditions give c1 = 0
and c3 = 0. The conditions at x = L give the system
w0 4
c2 L + c4 L3 +
L =0
24EI
w0 2
6c4 L +
L = 0.
2EI
y(x) = c1 + c2 x + c3 x2 + c4 x3 + Solving, we obtain c2 = w0 L3 /24EI and c4 = −w0 L/12EI. The deﬂection is
w0
y(x) =
(L3 x − 2Lx3 + x4 ).
24EI
(b) 0.2 0.4 0.6 0.8 1 x 1 y
3. (a) The general solution is w0 4
x .
24EI
The boundary conditions are y(0) = 0, y (0) = 0, y(L) = 0, y (L) = 0. The ﬁrst two conditions give c1 = 0
and c2 = 0. The conditions at x = L give the system
w0 4
c3 L2 + c4 L3 +
L =0
24EI
w0 2
2c3 + 6c4 L +
L = 0.
2EI
y(x) = c1 + c2 x + c3 x2 + c4 x3 + 158 3.9 Linear Models: Boundary-Value Problems Solving, we obtain c3 = w0 L2 /16EI and c4 = −5w0 L/48EI. The deﬂection is
w0
y(x) =
(3L2 x2 − 5Lx3 + 2x4 ).
48EI
(b) 0.2 0.4 0.6 0.8 1 1 x y 4. (a) The general solution is
w0 L4
π
sin x.
4
EIπ
L
The boundary conditions are y(0) = 0, y (0) = 0, y(L) = 0, y (L) = 0. The ﬁrst two conditions give c1 = 0
and c2 = −w0 L3 /EIπ 3 . The conditions at x = L give the system
w0 4
c3 L2 + c4 L3 +
L =0
EIπ 3
2c3 + 6c4 L = 0.
y(x) = c1 + c2 x + c3 x2 + c4 x3 + Solving, we obtain c3 = 3w0 L2 /2EIπ 3 and c4 = −w0 L/2EIπ 3 . The deﬂection is
y(x) = w0 L
2EIπ 3 −2L2 x + 3Lx2 − x3 + 2L3
π
sin x .
π
L (b) 0.2 0.4 0.6 0.8 1 x 1 y
(c) Using a CAS we ﬁnd the maximum deﬂection to be 0.270806 when x = 0.572536.
5. (a) The general solution is w0
x5 .
120EI
The boundary conditions are y(0) = 0, y (0) = 0, y(L) = 0, y (L) = 0. The ﬁrst two conditions give c1 = 0
and c3 = 0. The conditions at x = L give the system
w0
c2 L + c4 L3 +
L5 = 0
120EI
w0 3
6c4 L +
L = 0.
6EI
Solving, we obtain c2 = 7w0 L4 /360EI and c4 = −w0 L2 /36EI. The deﬂection is
w0
y(x) =
(7L4 x − 10L2 x3 + 3x5 ).
360EI
y(x) = c1 + c2 x + c3 x2 + c4 x3 + 159 3.9 Linear Models: Boundary-Value Problems (b)
1 x 0.2 0.4 0.6 0.8 1 y (c) Using a CAS we ﬁnd the maximum deﬂection to be 0.234799 when x = 0.51933.
6. (a) ymax = y(L) = w0 L4 /8EI
(b) Replacing both L and x by L/2 in y(x) we obtain w0 L4 /128EI, which is 1/16 of the maximum deﬂection
when the length of the beam is L.
(c) ymax = y(L/2) = 5w0 L4 /384EI
(d) The maximum deﬂection in Example 1 is y(L/2) = (w0 /24EI)L4 /16 = w0 L4 /384EI, which is 1/5 of the
maximum displacement of the beam in part c.
7. The general solution of the diﬀerential equation is
y = c1 cosh P
x + c2 sinh
EI P
w0 2 w0 EI
.
x+
x +
EI
2P
P2 Setting y(0) = 0 we obtain c1 = −w0 EI/P 2 , so that
y=− w0 EI
cosh
P2 P
x + c2 sinh
EI w0 2 w0 EI
P
x+
x +
.
EI
2P
P2 Setting y (L) = 0 we ﬁnd
P w0 EI
sinh
EI P 2 c2 = P
w0 L
L−
EI
P P
cosh
EI P
L.
EI 8. The general solution of the diﬀerential equation is
y = c1 cos P
x + c2 sin
EI w0 2 w0 EI
P
x+
x +
.
EI
2P
P2 Setting y(0) = 0 we obtain c1 = −w0 EI/P 2 , so that
y=− w0 EI
cos
P2 P
x + c2 sin
EI P
w0 2 w0 EI
.
x+
x +
EI
2P
P2 Setting y (L) = 0 we ﬁnd
c2 = − P w0 EI
sin
EI P 2 w0 L
P
L−
EI
P P
cos
EI P
L.
EI 9. This is Example 2 in the text with L = π. The eigenvalues are λn = n2 π 2 /π 2 = n2 , n = 1, 2, 3, . . . and the
corresponding eigenfunctions are yn = sin(nπx/π) = sin nx, n = 1, 2, 3, . . . .
10. This is Example 2 in the text with L = π/4. The eigenvalues are λn = n2 π 2 /(π/4)2 = 16n2 , n = 1, 2, 3, . . .
and the eigenfunctions are yn = sin(nπx/(π/4)) = sin 4nx, n = 1, 2, 3, . . . . 160 3.9 Linear Models: Boundary-Value Problems 11. For λ ≤ 0 the only solution of the boundary-value problem is y = 0. For λ = α2 > 0 we have
y = c1 cos αx + c2 sin αx.
Now
y (x) = −c1 α sin αx + c2 α cos αx
and y (0) = 0 implies c2 = 0, so
y(L) = c1 cos αL = 0
gives
(2n − 1)π
2 (2n − 1)2 π 2
, n = 1, 2, 3, . . . .
4L2
(2n − 1)π
The eigenvalues (2n − 1)2 π 2 /4L2 correspond to the eigenfunctions cos
x for n = 1, 2, 3, . . . .
2L
12. For λ ≤ 0 the only solution of the boundary-value problem is y = 0. For λ = α2 > 0 we have
αL = or λ = α2 = y = c1 cos αx + c2 sin αx.
Since y(0) = 0 implies c1 = 0, y = c2 sin x dx. Now
y π
2 = c2 α cos α π
=0
2 gives π
(2n − 1)π
=
or λ = α2 = (2n − 1)2 , n = 1, 2, 3, . . . .
2
2
The eigenvalues λn = (2n − 1)2 correspond to the eigenfunctions yn = sin(2n − 1)x.
α 13. For λ = −α2 < 0 the only solution of the boundary-value problem is y = 0. For λ = 0 we have y = c1 x + c2 .
Now y = c1 and y (0) = 0 implies c1 = 0. Then y = c2 and y (π) = 0. Thus, λ = 0 is an eigenvalue with
corresponding eigenfunction y = 1.
For λ = α2 > 0 we have
y = c1 cos αx + c2 sin αx.
Now
y (x) = −c1 α sin αx + c2 α cos αx
and y (0) = 0 implies c2 = 0, so
y (π) = −c1 α sin απ = 0
gives
απ = nπ or λ = α2 = n2 , n = 1, 2, 3, . . . . The eigenvalues n2 correspond to the eigenfunctions cos nx for n = 0, 1, 2, . . . .
14. For λ ≤ 0 the only solution of the boundary-value problem is y = 0. For λ = α2 > 0 we have
y = c1 cos αx + c2 sin αx.
Now y(−π) = y(π) = 0 implies
c1 cos απ − c2 sin απ = 0
c1 cos απ + c2 sin απ = 0.
This homogeneous system will have a nontrivial solution when
cos απ
cos απ − sin απ
= 2 sin απ cos απ = sin 2απ = 0.
sin απ 161 (1) 3.9 Linear Models: Boundary-Value Problems Then n2
; n = 1, 2, 3, . . . .
4
When n = 2k − 1 is odd, the eigenvalues are (2k − 1)2 /4. Since cos(2k − 1)π/2 = 0 and sin(2k − 1)π/2 = 0,
we see from either equation in (1) that c2 = 0. Thus, the eigenfunctions corresponding to the eigenvalues
2απ = nπ λ = α2 = or (2k − 1)2 /4 are y = cos(2k − 1)x/2 for k = 1, 2, 3, . . . . Similarly, when n = 2k is even, the eigenvalues are k 2
with corresponding eigenfunctions y = sin kx for k = 1, 2, 3, . . . .
15. The auxiliary equation has solutions
1
−2 ±
2 m= 4 − 4(λ + 1) = −1 ± α. For λ = −α2 < 0 we have
y = e−x (c1 cosh αx + c2 sinh αx) .
The boundary conditions imply
y(0) = c1 = 0
y(5) = c2 e−5 sinh 5α = 0
so c1 = c2 = 0 and the only solution of the boundary-value problem is y = 0.
For λ = 0 we have
y = c1 e−x + c2 xe−x
and the only solution of the boundary-value problem is y = 0.
For λ = α2 > 0 we have
y = e−x (c1 cos αx + c2 sin αx) .
Now y(0) = 0 implies c1 = 0, so
y(5) = c2 e−5 sin 5α = 0
gives
n2 π 2
, n = 1, 2, 3, . . . .
25
n2 π 2
nπ
The eigenvalues λn =
correspond to the eigenfunctions yn = e−x sin
x for n = 1, 2, 3, . . . .
25
5
16. For λ < −1 the only solution of the boundary-value problem is y = 0. For λ = −1 we have y = c1 x + c2 .
5α = nπ or λ = α2 = Now y = c1 and y (0) = 0 implies c1 = 0. Then y = c2 and y (1) = 0. Thus, λ = −1 is an eigenvalue with
corresponding eigenfunction y = 1.
For λ > −1 or λ + 1 = α2 > 0 we have
y = c1 cos αx + c2 sin αx.
Now
y = −c1 α sin αx + c2 α cos αx
and y (0) = 0 implies c2 = 0, so
y (1) = −c1 α sin α = 0
gives
α = nπ, λ + 1 = α 2 = n2 π 2 , or λ = n2 π 2 − 1, n = 1, 2, 3, . . . . The eigenvalues n2 π 2 − 1 correspond to the eigenfunctions cos nπx for n = 0, 1, 2, . . . . 162 3.9 Linear Models: Boundary-Value Problems 17. For λ = α2 > 0 a general solution of the given diﬀerential equation is
y = c1 cos(α ln x) + c2 sin(α ln x).
Since ln 1 = 0, the boundary condition y(1) = 0 implies c1 = 0. Therefore
y = c2 sin(α ln x).
Using ln eπ = π we ﬁnd that y (eπ ) = 0 implies
c2 sin απ = 0
or απ = nπ, n = 1, 2, 3, . . . . The eigenvalues and eigenfunctions are, in turn,
λ = α 2 = n2 , n = 1, 2, 3, . . . and y = sin(n ln x). For λ ≤ 0 the only solution of the boundary-value problem is y = 0.
18. For λ = 0 the general solution is y = c1 + c2 ln x. Now y = c2 /x, so y (e−1 ) = c2 e = 0 implies c2 = 0. Then
y = c1 and y(1) = 0 gives c1 = 0. Thus y(x) = 0.
For λ = −α2 < 0, y = c1 x−α + c2 xα . The boundary conditions give c2 = c1 e2α and c1 = 0, so that c2 = 0 and
y(x) = 0.
For λ = α2 > 0, y = c1 cos(α ln x) + c2 sin(α ln x). From y(1) = 0 we obtain c1 = 0 and y = c2 sin(α ln x).
Now y = c2 (α/x) cos(α ln x), so y (e−1 ) = c2 eα cos α = 0 implies cos α = 0 or α = (2n − 1)π/2 and λ = α2 =
(2n − 1)2 π 2 /4 for n = 1, 2, 3, . . . . The corresponding eigenfunctions are
yn = sin 2n − 1
π ln x .
2 19. For λ = α4 , α > 0, the general solution of the boundary-value problem
y (4) − λy = 0, y(0) = 0, y (0) = 0, y(1) = 0, y (1) = 0 is
y = c1 cos αx + c2 sin αx + c3 cosh αx + c4 sinh αx.
The boundary conditions y(0) = 0, y (0) = 0 give c1 + c3 = 0 and −c1 α2 + c3 α2 = 0, from which we conclude
c1 = c3 = 0. Thus, y = c2 sin αx + c4 sinh αx. The boundary conditions y(1) = 0, y (1) = 0 then give
c2 sin α + c4 sinh α = 0
−c2 α2 sin α + c4 α2 sinh α = 0.
In order to have nonzero solutions of this system, we must have the determinant of the coeﬃcients equal zero,
that is,
sin α
sinh α
=0
or
2α2 sinh α sin α = 0.
2
−α sin α α2 sinh α
But since α > 0, the only way that this is satisﬁed is to have sin α = 0 or α = nπ. The system is then satisﬁed
by choosing c2 = 0, c4 = 0, and α = nπ. The eigenvalues and corresponding eigenfunctions are then
λn = α4 = (nπ)4 , n = 1, 2, 3, . . . and y = sin nπx. 20. For λ = α4 , α > 0, the general solution of the diﬀerential equation is
y = c1 cos αx + c2 sin αx + c3 cosh αx + c4 sinh αx. 163 3.9 Linear Models: Boundary-Value Problems
The boundary conditions y (0) = 0, y (0) = 0 give c2 α + c4 α = 0 and −c2 α3 + c4 α3 = 0 from which we conclude
c2 = c4 = 0. Thus, y = c1 cos αx + c3 cosh αx. The boundary conditions y(π) = 0, y (π) = 0 then give
c2 cos απ + c4 cosh απ = 0
−c2 λ cos απ + c4 λ2 cosh απ = 0.
2 The determinant of the coeﬃcients is 2α2 cosh α cos α = 0. But since α > 0, the only way that this is satisﬁed
is to have cos απ = 0 or α = (2n − 1)/2, n = 1, 2, 3, . . . . The eigenvalues and corresponding eigenfunctions are
λn = α 4 = 2n − 1
2 4 , n = 1, 2, 3, . . . and y = cos 2n − 1
2 x.
y 21. If restraints are put on the column at x = L/4, x = L/2, and x = 3L/4, then the critical load will
be P4 . L
x 22. (a) The general solution of the diﬀerential equation is
y = c1 cos P
x + c2 sin
EI P
x + δ.
EI Since the column is embedded at x = 0, the boundary conditions are y(0) = y (0) = 0. If δ = 0 this implies
that c1 = c2 = 0 and y(x) = 0. That is, there is no deﬂection.
(b) If δ = 0, the boundary conditions give, in turn, c1 = −δ and c2 = 0. Then
P
x .
EI y = δ 1 − cos In order to satisfy the boundary condition y(L) = δ we must have
δ = δ 1 − cos
This gives P
L
EI or cos P
L = 0.
EI P/EI L = nπ/2 for n = 1, 2, 3, . . . . The smallest value of Pn , the Euler load, is then
π
P1
L=
EI
2 or P1 = 1
4 π 2 EI
L2 . 23. If λ = α2 = P/EI, then the solution of the diﬀerential equation is
y = c1 cos αx + c2 sin αx + c3 x + c4 .
The conditions y(0) = 0, y (0) = 0 yield, in turn, c1 + c4 = 0 and c1 = 0. With c1 = 0 and c4 = 0 the solution
is y = c2 sin αx + c3 x. The conditions y(L) = 0, y (L) = 0, then yield
c2 sin αL + c3 L = 0 and c2 sin αL = 0. Hence, nontrivial solutions of the problem exist only if sin αL = 0. From this point on, the analysis is the same
as in Example 3 in the text. 164 3.9 Linear Models: Boundary-Value Problems 24. (a) The boundary-value problem is
d4 y
d2 y
+ λ 2 = 0,
4
dx
dx y(0) = 0, y (0) = 0, y(L) = 0, y (L) = 0, where λ = α2 = P/EI. The solution of the diﬀerential equation is y = c1 cos αx + c2 sin αx + c3 x + c4 and
the conditions y(0) = 0, y (0) = 0 yield c1 = 0 and c4 = 0. Next, by applying y(L) = 0, y (L) = 0 to
y = c2 sin αx + c3 x we get the system of equations
c2 sin αL + c3 L = 0
αc2 cos αL + c3 = 0. To obtain nontrivial solutions c2 , c3 , we must have the determinant of the coeﬃcients equal to zero:
sin αL L
=0
α cos αL 1 or tan β = β, where β = αL. If βn denotes the positive roots of the last equation, then the eigenvalues are found from
√
2
βn = αn L = λn L or λn = (βn /L)2 . From λ = P/EI we see that the critical loads are Pn = βn EI/L2 .
With the aid of a CAS we ﬁnd that the ﬁrst positive root of tan β = β is (approximately) β1 = 4.4934, and
so the Euler load is (approximately) P1 = 20.1907EI/L2 . Finally, if we use c3 = −c2 α cos αL, then the
deﬂection curves are
yn (x) = c2 sin αn x + c3 x = c2 sin βn
x −
L βn
cos βn x .
L (b) With L = 1 and c2 appropriately chosen, the general shape of the ﬁrst buckling mode,
4.4934
x −
L y1 (x) = c2 sin 4.4934
cos(4.4934) x ,
L is shown below.
y1 0.2 0.4 0.6 0.8 1 x 25. The general solution is
y = c1 cos ρ
ωx + c2 sin
T ρ
ωx.
T From y(0) = 0 we obtain c1 = 0. Setting y(L) = 0 we ﬁnd ρ/T ωL = nπ, n = 1, 2, 3, . . . . Thus, critical
√
√
speeds are ωn = nπ T /L ρ , n = 1, 2, 3, . . . . The corresponding deﬂection curves are
y(x) = c2 sin nπ
x,
L n = 1, 2, 3, . . . , where c2 = 0.
26. (a) When T (x) = x2 the given diﬀerential equation is the Cauchy-Euler equation
x2 y + 2xy + ρω 2 y = 0.
The solutions of the auxiliary equation
m(m − 1) + 2m + ρω 2 = m2 + m + ρω 2 = 0 165 3.9 Linear Models: Boundary-Value Problems are 1 1
1 1
m1 = − −
4ρω 2 − 1 i, m2 = − +
4ρω 2 − 1 i
2 2
2 2
when ρω 2 > 0.25. Thus
y = c1 x−1/2 cos(λ ln x) + c2 x−1/2 sin(λ ln x) where λ = 1
2 4ρω 2 − 1. Applying y(1) = 0 gives c1 = 0 and consequently
y = c2 x−1/2 sin(λ ln x). The condition y(e) = 0 requires c2 e−1/2 sin λ = 0. We obtain a nontrivial solution when λn = nπ,
n = 1, 2, 3, . . . . But
1
2
λn =
4ρωn − 1 = nπ.
2
Solving for ωn gives
1
ωn =
(4n2 π 2 + 1)/ρ .
2
The corresponding solutions are
yn (x) = c2 x−1/2 sin(nπ ln x).
(b) y y 1 y 1
n=1 1
n=3 n=2
ex 1 ex 1 -1 -1 1 ex -1 27. The auxiliary equation is m2 +m = m(m+1) = 0 so that u(r) = c1 r−1 +c2 . The boundary conditions u(a) = u0
and u(b) = u1 yield the system c1 a−1 + c2 = u0 , c1 b−1 + c2 = u1 . Solving gives
u0 − u1
b−a c1 =
Thus u(r) = ab and c2 = u0 − u1
b−a u1 b − u0 a
.
b−a ab u1 b − u0 a
+
.
r
b−a 28. The auxiliary equation is m2 = 0 so that u(r) = c1 + c2 ln r. The boundary conditions u(a) = u0 and u(b) = u1
yield the system c1 + c2 ln a = u0 , c1 + c2 ln b = u1 . Solving gives
c1 =
Thus
u(r) = u1 ln a − u0 ln b
ln(a/b) and c2 = u0 − u1
.
ln(a/b) u1 ln a − u0 ln b u0 − u1
u0 ln(r/b) − u1 ln(r/a)
+
ln r =
.
ln(a/b)
ln(a/b)
ln(a/b) 29. The solution of the initial-value problem
x + ω 2 x = 0, x(0) = 0, x (0) = v0 , ω 2 = 10/m is x(t) = (v0 /ω) sin ωt. To satisfy the additional boundary condition x(1) = 0 we require that ω = nπ,
n = 1, 2, 3, . . . . The eigenvalues λ = ω 2 = n2 π 2 and eigenfunctions of the problem are then x(t) =
(v0 /nπ) sin nπt. Using ω 2 = 10/m we ﬁnd that the only masses that can pass through the equilibrium position at t = 1 are mn = 10/n2 π 2 . Note for n = 1, the heaviest mass m1 = 10/π 2 will not pass through the 166 3.9 Linear Models: Boundary-Value Problems equilibrium position on the interval 0 < t < 1 (the period of x(t) = (v0 /π) sin πt is T = 2, so on 0 ≤ t ≤ 1 its
graph passes through x = 0 only at t = 0 and t = 1). Whereas for n > 1, masses of lighter weight will pass
through the equilibrium position n − 1 times prior to passing through at t = 1. For example, if n = 2, the period
of x(t) = (v0 /2π) sin 2πt is 2π/2π = 1, the mass will pass through x = 0 only once (t = 1 ) prior to t = 1; if
2
n = 3, the period of x(t) = (v0 /3π) sin 3πt is
to t = 1; and so on. 2
3 , the mass will pass through x = 0 twice (t = 1
3 and t = 2 ) prior
3 30. The initial-value problem is
x + 2
k
x + x = 0,
m
m x(0) = 0, x (0) = v0 .
√
With k = 10, the auxiliary equation has roots γ = −1/m ± 1 − 10m/m. Consider the three cases:
. The roots are γ1 = γ2 = 10 and the solution of the diﬀerential equation is x(t) = c1 e−10t +c2 te−10t .
The initial conditions imply c1 = 0 and c2 = v0 and so x(t) = v0 te−10t . The condition x(1) = 0 implies
v0 e−10 = 0 which is impossible because v0 = 0. (i) m = 1
10 (ii) 1 − 10m > 0 or 0 < m < 1
10 γ1 = − . The roots are 1
1√
1 − 10m
−
m m and γ2 = − 1
1√
1 − 10m
+
m m and the solution of the diﬀerential equation is x(t) = c1 eγ1 t + c2 eγ2 t . The initial conditions imply
c1 + c2 = 0
γ1 c1 + γ2 c2 = v0
so c1 = v0 /(γ1 − γ2 ), c2 = −v0 /(γ1 − γ2 ), and
x(t) = v0
(eγ1 t − eγ2 t ).
γ1 − γ2 Again, x(1) = 0 is impossible because v0 = 0.
(iii) 1 − 10m < 0 or m > 1
10 . The roots of the auxiliary equation are γ1 = − 1
1√
10m − 1 i
−
m m and γ2 = − 1
1√
10m − 1 i
+
m m and the solution of the diﬀerential equation is
1√
1√
10m − 1 t + c2 e−t/m sin
10m − 1 t.
m
m
√
The initial conditions imply c1 = 0 and c2 = mv0 / 10m − 1, so that
x(t) = c1 e−t/m cos x(t) = √ mv0
e−t/m sin
10m − 1 1 √
10m − 1 t ,
m The condition x(1) = 0 implies
√ mv0
1√
e−1/m sin
10m − 1 = 0
m
10m − 1
1√
10m − 1 = 0
sin
m
1√
10m − 1 = nπ
m
10m − 1
= n2 π 2 , n = 1, 2, 3, . . .
m2
(n2 π 2 )m2 − 10m + 1 = 0 167 3.9 Linear Models: Boundary-Value Problems
√
√
10 100 − 4n2 π 2
5 ± 25 − n2 π 2
m=
=
.
2n2 π 2
n2 π 2
Since m is real, 25 − n2 π 2 ≥ 0. If 25 − n2 π 2 = 0, then n2 = 25/π 2 , and n is not an integer. Thus, 25 − n2 π 2 =
(5 − nπ)(5 + nπ) > 0 and since n > 0, 5 + nπ > 0, so 5 − nπ > 0 also. Then n < 5/π, and so n = 1. Therefore,
the mass m will pass through the equilibrium position when t = 1 for
√
√
5 + 25 − π 2
5 − 25 − π 2
m1 =
and
m2 =
.
π2
π2 31. (a) The general solution of the diﬀerential equation is y = c1 cos 4x + c2 sin 4x. From y0 = y(0) = c1 we see
that y = y0 cos 4x + c2 sin 4x. From y1 = y(π/2) = y0 we see that any solution must satisfy y0 = y1 . We
also see that when y0 = y1 , y = y0 cos 4x + c2 sin 4x is a solution of the boundary-value problem for any
choice of c2 . Thus, the boundary-value problem does not have a unique solution for any choice of y0 and
y1 .
(b) Whenever y0 = y1 there are inﬁnitely many solutions.
(c) When y0 = y1 there will be no solutions.
(d) The boundary-value problem will have the trivial solution when y0 = y1 = 0. This solution will not be
unique.
32. (a) The general solution of the diﬀerential equation is y = c1 cos 4x + c2 sin 4x. From 1 = y(0) = c1 we see that
y = cos 4x + c2 sin 4x. From 1 = y(L) = cos 4L + c2 sin 4L we see that c2 = (1 − cos 4L)/ sin 4L. Thus,
y = cos 4x + 1 − cos 4L
sin 4L sin 4x will be a unique solution when sin 4L = 0; that is, when L = kπ/4 where k = 1, 2, 3, . . . .
(b) There will be inﬁnitely many solutions when sin 4L = 0 and 1 − cos 4L = 0; that is, when L = kπ/2 where
k = 1, 2, 3, . . . .
(c) There will be no solution when sin 4L = 0 and 1 − cos 4L = 0; that is, when L = kπ/4 where
k = 1, 3, 5, . . . .
(d) There can be no trivial solution since it would fail to satisfy the boundary conditions.
33. (a) A solution curve has the same y-coordinate at both ends of the interval [−π, π] and the tangent lines at the
endpoints of the interval are parallel.
(b) For λ = 0 the solution of y = 0 is y = c1 x + c2 . From the ﬁrst boundary condition we have
y(−π) = −c1 π + c2 = y(π) = c1 π + c2
or 2c1 π = 0. Thus, c1 = 0 and y = c2 . This constant solution is seen to satisfy the boundary-value problem.
For λ = −α2 < 0 we have y = c1 cosh αx + c2 sinh αx. In this case the ﬁrst boundary condition gives
y(−π) = c1 cosh(−απ) + c2 sinh(−απ)
= c1 cosh απ − c2 sinh απ
= y(π) = c1 cosh απ + c2 sinh απ
or 2c2 sinh απ = 0. Thus c2 = 0 and y = c1 cosh αx. The second boundary condition implies in a similar
fashion that c1 = 0. Thus, for λ < 0, the only solution of the boundary-value problem is y = 0. 168 3.9 Linear Models: Boundary-Value Problems For λ = α2 > 0 we have y = c1 cos αx + c2 sin αx. The ﬁrst boundary condition implies
y(−π) = c1 cos(−απ) + c2 sin(−απ)
= c1 cos απ − c2 sin απ
= y(π) = c1 cos απ + c2 sin απ
or 2c2 sin απ = 0. Similarly, the second boundary condition implies 2c1 α sin απ = 0. If c1 = c2 = 0 the
solution is y = 0. However, if c1 = 0 or c2 = 0, then sin απ = 0, which implies that α must be an integer, n.
Therefore, for c1 and c2 not both 0, y = c1 cos nx + c2 sin nx is a nontrivial solution of the boundary-value
problem. Since cos(−nx) = cos nx and sin(−nx) = − sin nx, we may assume without loss of generality
that the eigenvalues are λn = α2 = n2 , for n a positive integer. The corresponding eigenfunctions are
yn = cos nx and yn = sin nx.
y (c) y 3 -p 3 px -p -3 y = 2 sin 3x
2 px -3 √ y = sin 4x − 2 cos 3x
√
α x + c2 sin α x. Setting y(0) = 0 we ﬁnd c1 = 0, so that 34. For λ = α > 0 the general solution is y = c1 cos
√
y = c2 sin α x. The boundary condition y(1) + y (1) = 0 implies
√
√
√
c2 sin α + c2 α cos α = 0.
√
√
2
Taking c2 = 0, this equation is equivalent to tan α = − α . Thus, the eigenvalues are λn = αn = x2 ,
n
√
√
n = 1, 2, 3, . . . , where the xn are the consecutive positive roots of tan α = − α .
35. We see from the graph that tan x = −x has inﬁnitely many roots. Since
2
λn = αn , there are no new eigenvalues when αn < 0. For λ = 0, the
diﬀerential equation y = 0 has general solution y = c1 x + c2 . The boundary tan x
5
2.5 conditions imply c1 = c2 = 0, so y = 0. 2 4 6 8 10 12 x -2.5
-5
-7.5
-10 36. Using a CAS we ﬁnd that the ﬁrst four nonnegative roots of tan x = −x are approximately 2.02876, 4.91318,
7.97867, and 11.0855. The corresponding eigenvalues are 4.11586, 24.1393, 63.6591, and 122.889, with eigenfunctions sin(2.02876x), sin(4.91318x), sin(7.97867x), and sin(11.0855x). 169 3.9 Linear Models: Boundary-Value Problems 37. In the case when λ = −α2 < 0, the solution of the diﬀerential equation
is y = c1 cosh αx + c2 sinh αx. The condition y(0) = 0 gives c1 = 0.
The condition y(1) − 1 y (1) = 0 applied to y = c2 sinh αx gives
2
c2 (sinh α − 1 α cosh α) = 0 or tanh α = 1 α. As can be seen from
2
2
the ﬁgure, the graphs of y = tanh x and y = 1 x intersect at a single
2 y
1 point with approximate x-coordinate α1 = 1.915. Thus, there is a
2
single negative eigenvalue λ1 = −α1 ≈ −3.667 and the corresponding
eigenfuntion is y1 = sinh 1.915x. 1 x 2 For λ = 0 the only solution of the boundary-value problem is y = 0.
For λ = α2 > 0 the solution of the diﬀerential equation is y = c1 cos αx + c2 sin αx. The condition y(0) = 0
gives c1 = 0, so y = c2 sin αx. The condition y(1) − 1 y (1) = 0 gives c2 (sin α − 1 α cos α) = 0, so the eigenvalues
2
2
2
are λn = αn when αn , n = 2, 3, 4, . . . , are the positive roots of tan α = 1 α. Using a CAS we ﬁnd that
2
the ﬁrst three values of α are α2 = 4.27487, α3 = 7.59655, and α4 = 10.8127. The ﬁrst three eigenvalues
2
2
2
are then λ2 = α2 = 18.2738, λ3 = α3 = 57.7075, and λ4 = α4 = 116.9139 with corresponding eigenfunctions
y2 = sin 4.27487x, y3 = sin 7.59655x, and y4 = sin 10.8127x.
38. For λ = α4 , α > 0, the solution of the diﬀerential equation is
y = c1 cos αx + c2 sin αx + c3 cosh αx + c4 sinh αx. y
1 The boundary conditions y(0) = 0, y (0) = 0, y(1) = 0, y (1) = 0 give,
in turn,
c1 + c3 = 0 2 4 6 8 10 12 x αc2 + αc4 = 0,
c1 cos α + c2 sin α + c3 cosh α + c4 sinh α = 0
−c1 α sin α + c2 α cos α + c3 α sinh α + c4 α cosh α = 0.
The ﬁrst two equations enable us to write
c1 (cos α − cosh α) + c2 (sin α − sinh α) = 0
c1 (− sin α − sinh α) + c2 (cos α − cosh α) = 0.
The determinant cos α − cosh α
− sin α − sinh α sin α − sinh α
=0
cos α − cosh α simpliﬁes to cos α cosh α = 1. From the ﬁgure showing the graphs of 1/ cosh x and cos x, we see that this
equation has an inﬁnite number of positive roots. With the aid of a CAS the ﬁrst four roots are found to
be α1 = 4.73004, α2 = 7.8532, α3 = 10.9956, and α4 = 14.1372, and the corresponding eigenvalues are
λ1 = 500.5636, λ2 = 3803.5281, λ3 = 14,617.5885, and λ4 = 39,944.1890. Using the third equation in the
system to eliminate c2 , we ﬁnd that the eigenfunctions are
yn = (− sin αn + sinh αn )(cos αn x − cosh αn x) + (cos αn − cosh αn )(sin αn x − sinh αn x). 170 3.10 Nonlinear Models EXERCISES 3.10
Nonlinear Models 1. The period corresponding to x(0) = 1, x (0) = 1 is approximately 5.6.
The period corresponding to x(0) = 1/2, x (0) = −1 is approximately
6.2. x
2
1
4 2 6 8 -1
-2
2. The solutions are not periodic. x
10
8
6
4
2
t
-2 3. The period corresponding to x(0) = 1, x (0) = 1 is approximately
5.8. The second initial-value problem does not have a periodic x
10
8 solution. 6
4
2
4 2 6 8 10 -2
4. Both solutions have periods of approximately 6.3. x
3
2
1
-1
-2
-3 171 2 4 6 8 10 t t 3.10 Nonlinear Models
5. From the graph we see that |x1 | ≈ 1.2. x 4
3
2
1 x1=1.2
x1=1.1
t −1 5 6. From the graphs we see that the interval is approximately
(−0.8, 1.1). 10
x 3 2 1 5 10 t −1
7. Since
xe0.01x = x[1 + 0.01x +
for small values of x, a linearization is
x 8. 1
(0.01x)2 + · · · ] ≈ x
2! d2 x
+ x = 0.
dt2 3 t 5 10 15 −3
For x(0) = 1 and x (0) = 1 the oscillations are symmetric about the line x = 0 with amplitude slightly greater
than 1.
For x(0) = −2 and x (0) = 0.5 the oscillations are symmetric about the line x = −2 with small amplitude.
√
For x(0) = 2 and x (0) = 1 the oscillations are symmetric about the line x = 0 with amplitude a little greater
than 2.
For x(0) = 2 and x (0) = 0.5 the oscillations are symmetric about the line x = 2 with small amplitude.
For x(0) = −2 and x (0) = 0 there is no oscillation; the solution is constant.
√
For x(0) = − 2 and x (0) = −1 the oscillations are symmetric about the line x = 0 with amplitude a little
greater than 2. 172 3.10 9. This is a damped hard spring, so x will approach 0 as t Nonlinear Models x approaches ∞. 2 4 2 6 8 t -2 10. This is a damped soft spring, so we might expect no oscillatory solutions. However, if the initial conditions are suﬃciently small the spring
can oscillate. x
5
4
3
2
1
2 4 t -1
-2 11. x
15 x k1 = 0.01 k1 = 1 3 10 2 5 1
10 20 t 30 10 -5 -2 -15 t -1 -10 20 -3 x x k1 = 20 3 3 2 2 1 k1 = 100 1
5 10 t 1 -1 t -2 -3 3 -1 -2 2 -3 When k1 is very small the eﬀect of the nonlinearity is greatly diminished, and the system is close to pure
resonance. 173 3.10 Nonlinear Models
12. (a) x x 40 40 20 20 20 40 60 80 t 100 20 -20 40 60 80 100 t -20 k -40 0.000465 k -40 0.000466 The system appears to be oscillatory for −0.000465 ≤ k1 < 0 and nonoscillatory for k1 ≤ −0.000466.
(b) x x 3 3 2 2 1 1
20 40 60 80 100 120 140 t 20 -1
-2 40 60 80 100 120 140 t -1 k -2 0.3493 -3 k 0.3494 -3 The system appears to be oscillatory for −0.3493 ≤ k1 < 0 and nonoscillatory for k1 ≤ −0.3494.
13. For λ2 − ω 2 > 0 we choose λ = 2 and ω = 1 with
x(0) = 1 and x (0) = 2. For λ2 − ω 2 < 0 we choose
λ = 1/3 and ω = 1 with x(0) = −2 and x (0) = 4. In θ 3 λ=1/3, ω=1 both cases the motion corresponds to the overdamped λ=2, ω=1 and underdamped cases for spring/mass systems. t 5 10 15 −3
14. (a) Setting dy/dt = v, the diﬀerential equation in (13) becomes dv/dt = −gR2 /y 2 . But, by the chain rule,
dv/dt = (dv/dy)(dy/dt) = v dv/dt, so v dv/dy = −gR2 /y 2 . Separating variables and integrating we obtain
v dv = −gR2 dy
y2 and 1 2
gR2
v =
+ c.
2
y 2
Setting v = v0 and y = R we ﬁnd c = −gR + 1 v0 and
2 R2
2
− 2gR + v0 .
y
√
2
(b) As y → ∞ we assume that v → 0+ . Then v0 = 2gR and v0 = 2gR .
v 2 = 2g (c) Using g = 32 ft/s and R = 4000(5280) ft we ﬁnd
v0 =
(d) v0 = 2(32)(4000)(5280) ≈ 36765.2 ft/s ≈ 25067 mi/hr. 2(0.165)(32)(1080) ≈ 7760 ft/s ≈ 5291 mi/hr 174 3.10 Nonlinear Models 15. (a) Intuitively, one might expect that only half of a 10-pound chain could be lifted by a 5-pound vertical force.
√
(b) Since x = 0 when t = 0, and v = dx/dt = 160 − 64x/3 , we have v(0) = 160 ≈ 12.65 ft/s.
5/2 ≈ 2.3717. Since the
5/2 . (This can also be obtained by (c) Since x should always be positive, we solve x(t) = 0, getting t = 0 and t =
3
4 3
2 graph of x(t) is a parabola, the maximum value occurs at tm =
solving x (t) = 0.) At this time the height of the chain is x(tm ) ≈ 7.5 ft. This is higher than predicted
because of the momentum generated by the force. When the chain is 5 feet high it still has a positive
velocity of about 7.3 ft/s, which keeps it going higher for a while.
16. (a) Setting dx/dt = v, the diﬀerential equation becomes (L − x)dv/dt − v 2 = Lg. But, by the Chain Rule,
dv/dt = (dv/dx)(dx/dt) = v dv/dx, so (L − x)v dv/dx − v 2 = Lg. Separating variables and integrating we
obtain so v
1
dv =
dx
v 2 + Lg
L−x 1
ln(v 2 + Lg) = − ln(L − x) + ln c,
2
√
v 2 + Lg = c/(L − x). When x = 0, v = 0, and c = L Lg . Solving for v and simplifying we get
and Lg(2Lx − x2 )
.
L−x dx
= v(x) =
dt Again, separating variables and integrating we obtain
L−x dx = dt √ and 2Lx − x2
√
= t + c1 .
Lg Lg(2Lx − x2 )
√
√
Since x(0) = 0, we have c1 = 0 and 2Lx − x2 / Lg = t. Solving for x we get
√
Lgt
dx
2 − Lgt2
x(t) = L − L
and v(t) =
.
=
dt
L − gt2
(b) The chain will be completely on the ground when x(t) = L or t = L/g . (c) The predicted velocity of the upper end of the chain when it hits the ground is inﬁnity.
17. (a) The weight of x feet of the chain is 2x, so the corresponding mass is m = 2x/32 = x/16. The only force
acting on the chain is the weight of the portion of the chain hanging over the edge of the platform. Thus,
by Newton’s second law,
d
d x
(mv) =
v
dt
dt 16
and x dv/dt + v 2 = 32x.
xv dv/dx + v 2 = 32x. = 1
dv
dx
x
+v
16
dt
dt = 1
dv
x
+ v2
16
dt = 2x Now, by the Chain Rule, dv/dt = (dv/dx)(dx/dt) = v dv/dx, so (b) We separate variables and write the diﬀerential equation as (v 2 − 32x) dx + xv dv = 0. This is not an exact
form, but µ(x) = x is an integrating factor. Multiplying by x we get (xv 2 − 32x2 ) dx + x2 v dv = 0. This
form is the total diﬀerential of u = 1 x2 v 2 − 32 x3 , so an implicit solution is 1 x2 v 2 − 32 x3 = c. Letting x = 3
2
3
2
3
and v = 0 we ﬁnd c = −288. Solving for v we get
√
dx
8 x3 − 27
√
, 3 ≤ x ≤ 8.
=v=
dt
3x
(c) Separating variables and integrating we obtain
√ x
8
dx = √ dt
3
x3 − 27 x and
3 175 √ s
8
ds = √ t + c.
3
s3 − 27 3.10 Nonlinear Models
Since x = 3 when t = 0, we see that c = 0 and
√
3
t=
8 x s
ds.
− 27
3
We want to ﬁnd t when x = 7. Using a CAS we ﬁnd t(7) = 0.576 seconds.
√ s3 18. (a) There are two forces acting on the chain as it falls from the platform. One is the force due to gravity on
the portion of the chain hanging over the edge of the platform. This is F1 = 2x. The second is due to the
motion of the portion of the chain stretched out on the platform. By Newton’s second law this is
F2 =
=
From d
d (8 − x)2
d 8−x
[mv] =
v =
v
dt
dt
32
dt 16
8 − x dv
1 dx
1
dv
− v
=
(8 − x)
− v2 .
16 dt
16 dt
16
dt d
[mv] = F1 − F2 we have
dt
d 2x
1
dv
v = 2x −
(8 − x)
− v2
dt 32
16
dt
x dv
1 dx
1
dv
+
v
= 2x −
(8 − x)
− v2
16 dt
16 dt
16
dt
x dv
dv
+ v 2 = 32x − (8 − x)
+ v2
dt
dt
x dv
dv
dv
= 32x − 8
+x
dt
dt
dt dv
= 32x.
dt
By the Chain Rule, dv/dt = (dv/dx)(dx/dt) = v dv/dx, so
8 8 dv
dv
dv
= 8v
= 32x and v
= 4x.
dt
dx
dx (b) Integrating v dv = 4x dx we get 1 v 2 = 2x2 + c. Since v = 0 when x = 3, we have c = −18. Then
2
√
v 2 = 4x2 − 36 and v = 4x2 − 36 . Using v = dx/dt, separating variables, and integrating we obtain
dx
x
= 2 dt and cosh−1 = 2t + c1 .
2−9
3
x
Solving for x we get x(t) = 3 cosh(2t + c1 ). Since x = 3 when t = 0, we have cosh c1 = 1 and c1 = 0. Thus,
√ x(t) = 3 cosh 2t. Diﬀerentiating, we ﬁnd v(t) = dx/dt = 6 sinh 2t.
(c) To ﬁnd the time when the back end of the chain leaves the platform we solve x(t) = 3 cosh 2t = 8. This
gives t1 = 1 cosh−1 8 ≈ 0.8184 seconds. The velocity at this instant is
2
3
v(t1 ) = 6 sinh cosh−1 8
3 √
= 2 55 ≈ 14.83 ft/s. (d) Replacing 8 with L and 32 with g in part (a) we have L dv/dt = gx. Then
dv
dv
dv
g
= Lv
= gx and v
= x.
dt
dx
dx
L
Integrating we get 1 v 2 = (g/2L)x2 + c. Setting x = x0 and v = 0, we ﬁnd c = −(g/2L)x2 . Solving for v
0
2
we ﬁnd
g 2 g 2
v(x) =
x − x0 .
L
L
L 176 3.10 Nonlinear Models Then the velocity at which the end of the chain leaves the edge of the platform is
g 2
(L − x2 ) .
0
L v(L) = 19. Let (x, y) be the coordinates of S2 on the curve C. The slope at (x, y) is then
dy/dx = (v1 t − y)/(0 − x) = (y − v1 t)/x or xy − y = −v1 t.
Diﬀerentiating with respect to x and using r = v1 /v2 gives
dt
dx
dt ds
xy = −v1
ds dx
1
xy = −v1 (− 1 + (y )2 )
v2 xy + y − y = −v1 xy = r 1 + (y )2 . Letting u = y and separating variables, we obtain
du
= r 1 + u2
dx
du
r
√
= dx
2
x
1+u
x sinh−1 u = r ln x + ln c = ln(cxr )
u = sinh(ln cxr )
dy
1
=
dx
2 cxr − 1
cxr . At t = 0, dy/dx = 0 and x = a, so 0 = car − 1/car . Thus c = 1/ar and
dy
1
=
dx
2 x
a r a
x − r = 1
2 r x
a x
a − −r . If r > 1 or r < 1, integrating gives
y= a
1
2 1+r x
a 1+r − 1
1−r x
a 1−r + c1 . When t = 0, y = 0 and x = a, so 0 = (a/2)[1/(1 + r) − 1/(1 − r)] + c1 . Thus c1 = ar/(1 − r2 ) and
y= a
1
2 1+r x
a 1+r − 1
1−r x
a 1−r + ar
.
1 − r2 To see if the paths ever intersect we ﬁrst note that if r > 1, then v1 > v2 and y → ∞ as x → 0+ . In other
words, S2 always lags behind S1 . Next, if r < 1, then v1 < v2 and y = ar/(1 − r2 ) when x = 0. In other words,
when the submarine’s speed is greater than the ship’s, their paths will intersect at the point (0, ar/(1 − r2 )).
Finally, if r = 1, then integration gives
y= 1 x2
1
− ln x + c2 .
2 2a a When t = 0, y = 0 and x = a, so 0 = (1/2)[a/2 − (1/a) ln a] + c2 . Thus c2 = −(1/2)[a/2 − (1/a) ln a] and
y= 1 x2
1
1 a 1
1 1 2
1 a
− ln x −
− ln a =
(x − a2 ) + ln
.
2 2a a
2 2 a
2 2a
a x Since y → ∞ as x → 0+ , S2 will never catch up with S1 . 177 3.10 Nonlinear Models
20. (a) Let (r, θ) denote the polar coordinates of the destroyer S1 . When S1 travels the 6 miles from (9, 0) to (3, 0)
it stands to reason, since S2 travels half as fast as S1 , that the polar coordinates of S2 are (3, θ2 ), where θ2
is unknown. In other words, the distances of the ships from (0, 0) are the same and r(t) = 15t then gives
the radial distance of both ships. This is necessary if S1 is to intercept S2 .
(b) The diﬀerential of arc length in polar coordinates is (ds)2 = (r dθ)2 + (dr)2 , so that
ds
dt 2 = r2 2 dθ
dt dr
dt + 2 . Using ds/dt = 30 and dr/dt = 15 then gives
dθ
dt 2 900 = 225t2 dθ
dt 2 675 = 225t2 + 225 √
dθ
3
=
dt
t
√
√
r
θ(t) = 3 ln t + c = 3 ln
+ c.
15
√
When r = 3, θ = 0, so c = − 3 ln 1 and
5
θ(t) =
Thus r = 3eθ/ √ 3 √ 3 ln r
1
− ln
15
5 = √ 3 ln r
.
3 , whose graph is a logarithmic spiral. (c) The time for S1 to go from (9, 0) to (3, 0) = 1
5 hour. Now S1 must intercept the path of S2 for some angle β, where 0 < β < 2π. At the time of interception t2 we have 15t2 = 3eβ/
is then
t= √ 3 √ or t = 1 eβ/
5 3 . The total time √
1 1 β/√3
1
< (1 + e2π/ 3 ).
+ e
5 5
5 21. Since (dx/dt)2 is always positive, it is necessary to use |dx/dt|(dx/dt) in order to account for the fact that the
motion is oscillatory and the velocity (or its square) should be negative when the spring is contracting. y 22. (a) From the graph we see that the approximations appears to
be quite good for 0 ≤ x ≤ 0.4. Using an equation solver to
solve sin x − x = 0.05 and sin x − x = 0.005, we ﬁnd that the 1.25 approximation is accurate to one decimal place for θ1 = 0.67 1 and to two decimal places for θ1 = 0.31. 1.5 0.75
0.5
0.25
0.25 0.5 0.75 178 1 1.25 1.5 x 3.10
(b) Θ Nonlinear Models Θ 1 1
Θ1 0.5
1 2 1, 3, 5 3 Θ1 0.5 4 5 6 t 1 -0.5 3 4 5 6 -0.5 -1 2 7, 9, 11 -1 23. (a) Write the diﬀerential equation as θ 2 d2 θ
+ ω 2 sin θ = 0,
dt2 moon
earth 2 where ω = g/l. To test for diﬀerences between the earth
and the moon we take l = 3, θ(0) = 1, and θ (0) = 2. t 5 Using g = 32 on the earth and g = 5.5 on the moon we -2 obtain the graphs shown in the ﬁgure. Comparing the
apparent periods of the graphs, we see that the pendulum oscillates faster on the earth than on the moon.
(b) The amplitude is greater on the moon than on the earth.
(c) The linear model is θ d2 θ
+ ω 2 θ = 0,
dt2 2 moon
earth where ω 2 = g/l. When g = 32, l = 3, θ(0) = 1, and
θ (0) = 2, the solution is
θ(t) = cos 3.266t + 0.612 sin 3.266t. t 5
-2 When g = 5.5 the solution is
θ(t) = cos 1.354t + 1.477 sin 1.354t.
As in the nonlinear case, the pendulum oscillates faster on the earth than on the moon and still has greater
amplitude on the moon.
24. (a) The general solution of
d2 θ
+θ =0
dt2
is θ(t) = c1 cos t + c2 sin t. From θ(0) = π/12 and θ (0) = −1/3 we ﬁnd
θ(t) = (π/12) cos t − (1/3) sin t.
Setting θ(t) = 0 we have tan t = π/4 which implies t1 = tan−1 (π/4) ≈ 0.66577.
(b) We set θ(t) = θ(0) + θ (0)t + 1 θ (0)t2 + 1 θ (0)t3 + · · · and use θ (t) = − sin θ(t) together with
2
6
θ(0) = π/12 and θ (0) = −1/3. Then
√ √
θ (0) = − sin(π/12) = − 2 ( 3 − 1)/4
and
√ √
θ (0) = − cos θ(0) · θ (0) = − cos(π/12)(−1/3) = 2 ( 3 + 1)/12. 179 3.10 Nonlinear Models
Thus
θ(t) = π
1
− t−
12 3 √ √
√ √
2 ( 3 − 1) 2
2 ( 3 + 1) 3
t +
t + ···.
8
72 (c) Setting π/12 − t/3 = 0 we obtain t1 = π/4 ≈ 0.785398.
(d) Setting √ √
π
2 ( 3 − 1) 2
1
− t−
t =0
12 3
8
and using the positive root we obtain t1 ≈ 0.63088. (e) Setting √ √
√ √
π
2 ( 3 − 1) 2
2 ( 3 + 1) 3
1
− t−
t +
t =0
12 3
8
72
we ﬁnd with the help of a CAS that t1 ≈ 0.661973 is the ﬁrst positive root. (f ) From the output we see that y(t) is an interpolating function on the
interval 0 ≤ t ≤ 5, whose graph is shown. The positive root of y(t) = 0
near t = 1 is t1 = 0.666404. 0.4
0.2
1 2 3 4 5 2 4 6 8 10 -0.2
-0.4
(g) To ﬁnd the next two positive roots we change the interval used in
NDSolve and Plot from {t,0,5} to {t,0,10}. We see from the graph
that the second and third positive roots are near 4 and 7, respectively.
Replacing {t,1} in FindRoot with {t,4} and then {t,7} we obtain
t2 = 3.84411 and t3 = 7.0218. 0.4
0.2
-0.2
-0.4 25. From the table below we see that the pendulum ﬁrst passes the vertical position between 1.7 and 1.8 seconds.
To reﬁne our estimate of t1 we estimate the solution of the diﬀerential equation on [1.7, 1.8] using a step size of
h = 0.01. From the resulting table we see that t1 is between 1.76 and 1.77 seconds. Repeating the process with
h = 0.001 we conclude that t1 ≈ 1.767. Then the period of the pendulum is approximately 4t1 = 7.068. The
error when using t1 = 2π is 7.068 − 6.283 = 0.785 and the percentage relative error is (0.785/7.068)100 = 11.1.
h=0.1 tn
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
2.00 h=0.01
θn
0.78540
0.78523
0.78407
0.78092
0.77482
0.76482
0.75004
0.72962
0.70275
0.66872
0.62687
0.57660
0.51744
0.44895
0.37085
0.28289
0.18497
0.07706
-0.04076
-0.16831
-0.30531 tn
1.70
1.71
1.72
1.73
1.74
1.75
1.76
1.77
1.78
1.79
1.80 θn
0.07706
0.06572
0.05428
0.04275
0.03111
0.01938
0.00755
-0.00438
-0.01641
-0.02854
-0.04076 h=0.001
1.763
1.764
1.765
1.766
1.767
1.768
1.769
1.770 0.00398
0.00279
0.00160
0.00040
-0.00079
-0.00199
-0.00318
-0.00438 180 3.11 Solving Systems of Linear Equations EXERCISES 3.11
Solving Systems of Linear Equations 1. From Dx = 2x − y and Dy = x we obtain y = 2x − Dx, Dy = 2Dx − D2 x, and (D2 − 2D + 1)x = 0. The
solution is
x = c1 et + c2 tet
y = (c1 − c2 )et + c2 tet .
2. From Dx = 4x + 7y and Dy = x − 2y we obtain y = 1 Dx − 4 x, Dy = 1 D2 x − 4 Dx, and (D2 − 2D − 15)x = 0.
7
7
7
7
The solution is
x = c1 e5t + c2 e−3t
1
y = c1 e5t − c2 e−3t .
7
3. From Dx = −y + t and Dy = x − t we obtain y = t − Dx, Dy = 1 − D2 x, and (D2 + 1)x = 1 + t. The solution is
x = c1 cos t + c2 sin t + 1 + t
y = c1 sin t − c2 cos t + t − 1.
4. From Dx − 4y = 1 and x + Dy = 2 we obtain y = 1 Dx −
4 1
4 , Dy = 1 D2 x, and (D2 + 1)x = 2. The solution is
4 x = c1 cos t + c2 sin t + 2
1
1
1
y = c2 cos t − c1 sin t − .
4
4
4
5. From (D2 + 5)x − 2y = 0 and −2x + (D2 + 2)y = 0 we obtain y = 1 (D2 + 5)x, D2 y =
2
(D2 + 1)(D2 + 6)x = 0. The solution is
√
√
x = c1 cos t + c2 sin t + c3 cos 6 t + c4 sin 6 t
√
√
1
1
y = 2c1 cos t + 2c2 sin t − c3 cos 6 t − c4 sin 6 t.
2
2 1
4
2 (D + 5D2 )x, and 6. From (D + 1)x + (D − 1)y = 2 and 3x + (D + 2)y = −1 we obtain x = − 1 − 1 (D + 2)y, Dx = − 1 (D2 + 2D)y,
3
3
3
and (D2 + 5)y = −7. The solution is
√
7
5 t + c2 sin 5 t −
5
√
√
2
5
− c1 −
c2 cos 5 t +
3
3 y = c1 cos
x= √ √ 5
2
c1 − c2
3
3 √
3
sin 5 t + .
5 7. From D2 x = 4y + et and D2 y = 4x − et we obtain y = 1 D2 x − 1 et , D2 y = 1 D4 x − 1 et , and
4
4
4
4
(D2 + 4)(D − 2)(D + 2)x = −3et . The solution is
1
x = c1 cos 2t + c2 sin 2t + c3 e2t + c4 e−2t + et
5
1
2t
−2t
y = −c1 cos 2t − c2 sin 2t + c3 e + c4 e
− et .
5 181 3.11 Solving Systems of Linear Equations
8. From (D2 +5)x+Dy = 0 and (D +1)x+(D −4)y = 0 we obtain (D −5)(D2 +4)x = 0 and (D −5)(D2 +4)y = 0.
The solution is
x = c1 e5t + c2 cos 2t + c3 sin 2t
y = c4 e5t + c5 cos 2t + c6 sin 2t.
Substituting into (D + 1)x + (D − 4)y = 0 gives
(6c1 + c4 )e5t + (c2 + 2c3 − 4c5 + 2c6 ) cos 2t + (−2c2 + c3 − 2c5 − 4c6 ) sin 2t = 0
so that c4 = −6c1 , c5 = 1 c3 , c6 = − 1 c2 , and
2
2
1
1
y = −6c1 e5t + c3 cos 2t − c2 sin 2t.
2
2
9. From Dx + D2 y = e3t and (D + 1)x + (D − 1)y = 4e3t we obtain D(D2 + 1)x = 34e3t and D(D2 + 1)y = −8e3t .
The solution is
4
y = c1 + c2 sin t + c3 cos t − e3t
15
17
x = c4 + c5 sin t + c6 cos t + e3t .
15
3t
Substituting into (D + 1)x + (D − 1)y = 4e gives
(c4 − c1 ) + (c5 − c6 − c3 − c2 ) sin t + (c6 + c5 + c2 − c3 ) cos t = 0
so that c4 = c1 , c5 = c3 , c6 = −c2 , and
x = c1 − c2 cos t + c3 sin t + 17 3t
e .
15 10. From D2 x − Dy = t and (D + 3)x + (D + 3)y = 2 we obtain D(D + 1)(D + 3)x = 1 + 3t and
D(D + 1)(D + 3)y = −1 − 3t. The solution is
1
x = c1 + c2 e−t + c3 e−3t − t + t2
2
1 2
y = c4 + c5 e−t + c6 e−3t + t − t .
2
Substituting into (D + 3)x + (D + 3)y = 2 and D2 x − Dy = t gives
3(c1 + c4 ) + 2(c2 + c5 )e−t = 2
and
(c2 + c5 )e−t + 3(3c3 + c6 )e−3t = 0
so that c4 = −c1 , c5 = −c2 , c6 = −3c3 , and
1
y = −c1 − c2 e−t − 3c3 e−3t + t − t2 .
2
11. From (D2 − 1)x − y = 0 and (D − 1)x + Dy = 0 we obtain y = (D2 − 1)x, Dy = (D3 − D)x, and
(D − 1)(D2 + D + 1)x = 0. The solution is
√ t −t/2 x = c1 e + e y= √
3
3
c2 cos
t + c3 sin
t
2
2 √
3
3
− c2 −
c3
2
2 √ −t/2 e 3
t+
cos
2 182 √ 3
3
c2 − c3
2
2 √
−t/2 e sin 3
t.
2 3.11 Solving Systems of Linear Equations
12. From (2D2 − D − 1)x − (2D + 1)y = 1 and (D − 1)x + Dy = −1 we obtain (2D + 1)(D − 1)(D + 1)x = −1 and
(2D + 1)(D + 1)y = −2. The solution is
x = c1 e−t/2 + c2 e−t + c3 et + 1
y = c4 e−t/2 + c5 e−t − 2.
Substituting into (D − 1)x + Dy = −1 gives
3
1
− c1 − c4 e−t/2 + (−2c2 − c5 )e−t = 0
2
2
so that c4 = −3c1 , c5 = −2c2 , and
y = −3c1 e−t/2 − 2c2 e−t − 2.
13. From (2D − 5)x + Dy = et and (D − 1)x + Dy = 5et we obtain Dy = (5 − 2D)x + et and (4 − D)x = 4et . Then
4
x = c1 e4t + et
3
and Dy = −3c1 e4t + 5et so that
3
y = − c1 e4t + c2 + 5et .
4
14. From Dx + Dy = et and (−D2 + D + 1)x + y = 0 we obtain y = (D2 − D − 1)x, Dy = (D3 − D2 − D)x, and
D2 (D − 1)x = et . The solution is
x = c1 + c2 t + c3 et + tet
y = −c1 − c2 − c2 t − c3 et − tet + et .
15. Multiplying the ﬁrst equation by D + 1 and the second equation by D2 + 1 and subtracting we obtain
(D4 − D2 )x = 1. Then
1
x = c1 + c2 t + c3 et + c4 e−t − t2 .
2
Multiplying the ﬁrst equation by D + 1 and subtracting we obtain D2 (D + 1)y = 1. Then
1
y = c5 + c6 t + c7 e−t − t2 .
2
Substituting into (D − 1)x + (D2 + 1)y = 1 gives
(−c1 + c2 + c5 − 1) + (−2c4 + 2c7 )e−t + (−1 − c2 + c6 )t = 1
so that c5 = c1 − c2 + 2, c6 = c2 + 1, and c7 = c4 . The solution of the system is
1
x = c1 + c2 t + c3 et + c4 e−t − t2
2
1
y = (c1 − c2 + 2) + (c2 + 1)t + c4 e−t − t2 .
2
16. From D2 x−2(D2 +D)y = sin t and x+Dy = 0 we obtain x = −Dy, D2 x = −D3 y, and D(D2 +2D+2)y = − sin t.
The solution is
1
2
y = c1 + c2 e−t cos t + c3 e−t sin t + cos t + sin t
5
5
1
2
x = (c2 + c3 )e−t sin t + (c2 − c3 )e−t cos t + sin t − cos t.
5
5 183 3.11 Solving Systems of Linear Equations
17. From Dx = y, Dy = z. and Dz = x we obtain x = D2 y = D3 x so that (D − 1)(D2 + D + 1)x = 0,
√
√
3
3
t
−t/2
x = c1 e + e
c2 sin
t + c3 cos
t ,
2
2
y = c1 et + √
3
1
− c2 −
c3
2
2 e−t/2 sin z = c1 et + √
3
1
− c2 +
c3
2
2 e−t/2 sin and √ 3
t+
2 3
1
c2 − c3
2
2 √ 3
t+
2 √ √ e−t/2 cos 3
t,
2
√ √
− 3
1
c2 − c3
2
2 e−t/2 cos 3
t.
2 18. From Dx + z = et , (D − 1)x + Dy + Dz = 0, and x + 2y + Dz = et we obtain z = −Dx + et , Dz = −D2 x + et , and
the system (−D2 + D − 1)x + Dy = −et and (−D2 + 1)x + 2y = 0. Then y = 1 (D2 − 1)x, Dy = 1 D(D2 − 1)x,
2
2
and (D − 2)(D2 + 1)x = −2et so that the solution is
x = c1 e2t + c2 cos t + c3 sin t + et
3 2t
c1 e − c2 cos t − c3 sin t
2
z = −2c1 e2t − c3 cos t + c2 sin t. y= 19. Write the system in the form
Dx − 6y = 0
x − Dy + z = 0
x + y − Dz = 0.
Multiplying the second equation by D and adding to the third equation we obtain (D + 1)x − (D2 − 1)y = 0.
Eliminating y between this equation and Dx − 6y = 0 we ﬁnd
(D3 − D − 6D − 6)x = (D + 1)(D + 2)(D − 3)x = 0.
Thus
x = c1 e−t + c2 e−2t + c3 e3t ,
and, successively substituting into the ﬁrst and second equations, we get
1
1
1
y = − c1 e−t − c2 e−2t + c3 e3t
6
3
2
5 −t 1 −2t 1 3t
z = − c1 e − c2 e
+ c3 e .
6
3
2
20. Write the system in the form
(D + 1)x − z = 0
(D + 1)y − z = 0
x − y + Dz = 0.
Multiplying the third equation by D + 1 and adding to the second equation we obtain
(D +1)x+(D2 +D −1)z = 0. Eliminating z between this equation and (D +1)x−z = 0 we ﬁnd D(D +1)2 x = 0.
Thus
x = c1 + c2 e−t + c3 te−t ,
and, successively substituting into the ﬁrst and third equations, we get
y = c1 + (c2 − c3 )e−t + c3 te−t
z = c1 + c3 e−t . 184 3.11 Solving Systems of Linear Equations
21. From (D + 5)x + y = 0 and 4x − (D + 1)y = 0 we obtain y = −(D + 5)x so that Dy = −(D2 + 5D)x. Then
4x + (D2 + 5D)x + (D + 5)x = 0 and (D + 3)2 x = 0. Thus
x = c1 e−3t + c2 te−3t
y = −(2c1 + c2 )e−3t − 2c2 te−3t .
Using x(1) = 0 and y(1) = 1 we obtain
c1 e−3 + c2 e−3 = 0
−(2c1 + c2 )e−3 − 2c2 e−3 = 1
or
c1 + c2 = 0
2c1 + 3c2 = −e3 .
Thus c1 = e3 and c2 = −e3 . The solution of the initial value problem is
x = e−3t+3 − te−3t+3
y = −e−3t+3 + 2te−3t+3 .
22. From Dx − y = −1 and 3x + (D − 2)y = 0 we obtain x = − 1 (D − 2)y so that Dx = − 1 (D2 − 2D)y. Then
3
3
− 1 (D2 − 2D)y = y − 1 and (D2 − 2D + 3)y = 3. Thus
3
√
√
y = et c1 cos 2 t + c2 sin 2 t + 1
and 1 t
e
3
Using x(0) = y(0) = 0 we obtain
x= Thus c1 = −1 and c2 = √ c1 − √ 2 c2 cos √ 2t + √ √
2
2 c1 + c2 sin 2 t + .
3 c1 + 1 = 0
√
1
2
c1 − 2 c2 + = 0.
3
3
2/2. The solution of the initial value problem is
√
√
√
2
2
2
t
x = e − cos 2 t −
sin 2 t +
3
6
3
y = et − cos √ √
2t + √
2
sin 2 t
2 + 1. 23. Equating Newton’s law with the net forces in the x- and y-directions gives m d2 x/dt2 = 0 and m d2 y/dt2 = −mg,
respectively. From mD2 x = 0 we obtain x(t) = c1 t + c2 , and from mD2 y = −mg or D2 y = −g we obtain
y(t) = − 1 gt2 + c3 t + c4 .
2
24. From Newton’s second law in the x-direction we have
d2 x
dx
1 dx
m 2 = −k cos θ = −k
= −|c| .
dt
v dt
dt
In the y-direction we have
d2 y
1 dy
dy
= −mg − k sin θ = −mg − k
= −mg − |c| .
2
dt
v dt
dt
From mD2 x + |c|Dx = 0 we have D(mD + |c|)x = 0 so that (mD + |c|)x = c1 or (D + |c|/m)x = c2 . This is a
m linear ﬁrst-order diﬀerential equation. An integrating factor is e |c|dt/m d |c|t/m
x] = c2 e|c|t/m
[e
dt 185 = e|c|t/m so that 3.11 Solving Systems of Linear Equations
and e|c|t/m x = (c2 m/|c|)e|c|t/m + c3 . The general solution of this equation is x(t) = c4 + c3 e−|c|t/m . From
(mD2 +|c|D)y = −mg we have D(mD+|c|)y = −mg so that (mD+|c|)y = −mgt+c1 or (D+|c|/m)y = −gt+c2 .
|c|dt/m This is a linear ﬁrst-order diﬀerential equation with integrating factor e = e|c|t/m . Thus d |c|t/m
y] = (−gt + c2 )e|c|t/m
[e
dt
mg |c|t/m m2 g |c|t/m
e|c|t/m y = −
+ 2 e
+ c3 e|c|t/m + c4
te
|c|
c
and
y(t) = − mg
m2 g
t + 2 + c3 + c4 e−|c|t/m .
|c|
c 25. The FindRoot application of Mathematica gives a solution of x1 (t) = x2 (t) as approximately t = 13.73 minutes.
So tank B contains more salt than tank A for t > 13.73 minutes.
26. (a) Separating variables in the ﬁrst equation, we have dx1 /x1 = −dt/50, so x1 = c1 e−t/50 . From x1 (0) = 15
we get c1 = 15. The second diﬀerential equation then becomes
dx2
15 −t/50
2
=
e
− x2
dt
50
75 or 2
dx2
3 −t/50
+ x2 =
e
.
dt
75
10 This diﬀerential equation is linear and has the integrating factor e 2 dt/75 = e2t/75 . Then d 2t/75
3 −t/50+2t/75
3 t/150
x2 ] =
=
[e
e
e
dt
10
10
so
e2t/75 x2 = 45et/150 + c2
and
x2 = 45e−t/50 + c2 e−2t/75 .
From x2 (0) = 10 we get c2 = −35. The third diﬀerential equation then becomes or dx3
1
90 −t/50 70 −2t/75
− e
− x3
=
e
dt
75
75
25
dx3
1
6
14
+ x3 = e−t/50 − e−2t/75 .
dt
25
5
15 This diﬀerential equation is linear and has the integrating factor e dt/25 = et/25 . Then d t/25
6
14
6
14
x3 ] = e−t/50+t/25 − e−2t/75+t/25 = et/50 − et/75 ,
[e
dt
5
15
5
15
so
et/25 x3 = 60et/50 − 70et/75 + c3
and
x3 = 60e−t/50 − 70e−2t/75 + c3 e−t/25 .
From x3 (0) = 5 we get c3 = 15. The solution of the initial-value problem is
x1 (t) = 15e−t/50
x2 (t) = 45e−t/50 − 35e−2t/75
x3 (t) = 60e−t/50 − 70e−2t/75 + 15e−t/25 . 186 3.11 Solving Systems of Linear Equations
(b) pounds salt
14
12
10
8
6
4
2 x1 x2
x3 50 100 150 200 time (c) Solving x1 (t) = 1 , x2 (t) = 1 , and x3 (t) = 1 , FindRoot gives, respectively, t1 = 170.06 min, t2 =
2
2
2
214.7 min, and t3 = 224.4 min. Thus, all three tanks will contain less than or equal to 0.5 pounds of salt
after 224.4 minutes.
27. (a) Write the system as
(D2 + 3)x1 − x2 = 0 −2x1 + (D2 + 2)x2 = 0.
Then
(D2 + 2)(D2 + 3)x1 − 2x1 = (D2 + 1)(D2 + 4)x1 = 0,
and
x1 (t) = c1 cos t + c2 sin t + c3 cos 2t + c4 sin 2t.
Since x2 = (D2 + 3)x1 , we have
x2 (t) = 2c1 cos t + 2c2 sin t − c3 cos 2t − c4 sin 2t.
The initial conditions imply
x1 (0) = c1 + c3 = 2
x1 (0) = c1 + 2c4 = 1
x2 (0) = 2c1 − c3 = −1
x2 (0) = 2c2 − 2c4 = 1,
so c1 = 1
3 , c2 = 2
3 , c3 = 5
3 , and c4 = 1
6 . Thus 1
cos t +
3
2
x2 (t) = cos t +
3 x1 (t) = (b) x1
3
2
1
-1
-2
-3 2
sin t +
3
4
sin t −
3 5
cos 2t +
3
5
cos 2t −
3 1
sin 2t
6
1
sin 2t.
6 x2
3
2
1
5 10 15 20 t
-1
-2
-3 5 10 15 20 t In this problem the motion appears to be periodic with period 2π. In Figure 3.59 of Example 4 in the text
the motion does not appear to be periodic. 187 3.11 Solving Systems of Linear Equations
(c)
x2
3
2
1
-2 -1 1 2 x1 -1
-2 CHAPTER 3 REVIEW EXERCISES 1. y = 0
2. Since yc = c1 ex + c2 e−x , a particular solution for y − y = 1 + ex is yp = A + Bxex .
3. It is not true unless the diﬀerential equation is homogeneous. For example, y1 = x is a solution of y + y = x,
but y2 = 5x is not.
4. True
5. 8 ft, since k = 4
6. 2π/5, since 1 x + 6.25x = 0
4
7. 5/4 m, since x = − cos 4t + 3 sin 4t
4
√
√
8. From x(0) = ( 2/2) sin φ = −1/2 we see that sin φ = −1/ 2 , so φ is an angle in the third or fourth quadrant.
√
√
Since x (t) = 2 cos(2t + φ), x (0) = 2 cos φ = 1 and cos φ > 0. Thus φ is in the fourth quadrant and
φ = −π/4.
9. The set is linearly independent over (−∞, ∞) and linearly dependent over (0, ∞).
10. (a) Since f2 (x) = 2 ln x = 2f1 (x), the set of functions is linearly dependent.
(b) Since xn+1 is not a constant multiple of xn , the set of functions is linearly independent.
(c) Since x + 1 is not a constant multiple of x, the set of functions is linearly independent.
(d) Since f1 (x) = cos x cos(π/2) − sin x sin(π/2) = − sin x = −f2 (x), the set of functions is linearly dependent.
(e) Since f1 (x) = 0 · f2 (x), the set of functions is linearly dependent.
(f ) Since 2x is not a constant multiple of 2, the set of functions is linearly independent.
(g) Since 3(x2 ) + 2(1 − x2 ) − (2 + x2 ) = 0, the set of functions is linearly dependent.
(h) Since xex+1 + 0(4x − 5)ex − exex = 0, the set of functions is linearly dependent. 188 CHAPTER 3 REVIEW EXERCISES
11. (a) The auxiliary equation is (m − 3)(m + 5)(m − 1) = m3 + m2 − 17m + 15 = 0, so the diﬀerential equation is
y + y − 17y + 15y = 0.
(b) The form of the auxiliary equation is
m(m − 1)(m − 2) + bm(m − 1) + cm + d = m3 + (b − 3)m2 + (c − b + 2)m + d = 0.
Since (m − 3)(m + 5)(m − 1) = m3 + m2 − 17m + 15 = 0, we have b − 3 = 1, c − b + 2 = −17, and d = 15.
Thus, b = 4 and c = −15, so the diﬀerential equation is y + 4y − 15y + 15y = 0.
12. (a) The auxiliary equation is am(m − 1) + bm + c = am2 + (b − a)m + c = 0. If the roots are 3 and −1, then
we want (m − 3)(m + 1) = m2 − 2m − 3 = 0. Thus, let a = 1, b = −1, and c = −3, so that the diﬀerential
equation is x2 y − xy − 3y = 0.
(b) In this case we want the auxiliary equation to be m2 + 1 = 0, so let a = 1, b = 1, and c = 1. Then the
diﬀerential equation is x2 y + xy + y = 0.
√
13. From m2 − 2m − 2 = 0 we obtain m = 1 ± 3 so that
y = c1 e(1+ √ 3 )x √ + c2 e(1− 3 )x . √
14. From 2m2 + 2m + 3 = 0 we obtain m = −1/2 ± ( 5/2)i so that
√
√
5
5
−x/2
c1 cos
y=e
x + c2 sin
x .
2
2
15. From m3 + 10m2 + 25m = 0 we obtain m = 0, m = −5, and m = −5 so that
y = c1 + c2 e−5x + c3 xe−5x .
16. From 2m3 + 9m2 + 12m + 5 = 0 we obtain m = −1, m = −1, and m = −5/2 so that
y = c1 e−5x/2 + c2 e−x + c3 xe−x .
√
17. From 3m3 + 10m2 + 15m + 4 = 0 we obtain m = −1/3 and m = −3/2 ± ( 7/2)i so that
√
√
7
7
−x/3
−3x/2
y = c1 e
+e
c2 cos
x + c3 sin
x .
2
2
√
18. From 2m4 + 3m3 + 2m2 + 6m − 4 = 0 we obtain m = 1/2, m = −2, and m = ± 2 i so that
√
√
y = c1 ex/2 + c2 e−2x + c3 cos 2 x + c4 sin 2 x.
19. Applying D4 to the diﬀerential equation we obtain D4 (D2 − 3D + 5) = 0. Then
√
√
11
11
3x/2
y=e
c1 cos
x + c2 sin
x + c3 + c4 x + c5 x2 + c6 x3
2
2
yc
2 3 and yp = A + Bx + Cx + Dx . Substituting yp into the diﬀerential equation yields
(5A − 3B + 2C) + (5B − 6C + 6D)x + (5C − 9D)x2 + 5Dx3 = −2x + 4x3 .
Equating coeﬃcients gives A = −222/625, B = 46/125, C = 36/25, and D = 4/5. The general solution is
√
√
222
46
36
11
11
4
3x/2
c1 cos
x + c2 sin
x −
+
x + x2 + x3 .
y=e
2
2
625 125
25
5 189 CHAPTER 3 REVIEW EXERCISES
20. Applying (D − 1)3 to the diﬀerential equation we obtain (D − 1)3 (D − 2D + 1) = (D − 1)5 = 0. Then
y = c1 ex + c2 xex + c3 x2 ex + c4 x3 ex + c5 x4 ex
yc
2 x 3 x 4 x and yp = Ax e + Bx e + Cx e . Substituting yp into the diﬀerential equation yields
12Cx2 ex + 6Bxex + 2Aex = x2 ex .
Equating coeﬃcients gives A = 0, B = 0, and C = 1/12. The general solution is
y = c1 ex + c2 xex + 1 4 x
x e .
12 21. Applying D(D2 + 1) to the diﬀerential equation we obtain
D(D2 + 1)(D3 − 5D2 + 6D) = D2 (D2 + 1)(D − 2)(D − 3) = 0.
Then
y = c1 + c2 e2x + c3 e3x + c4 x + c5 cos x + c6 sin x
yc and yp = Ax + B cos x + C sin x. Substituting yp into the diﬀerential equation yields
6A + (5B + 5C) cos x + (−5B + 5C) sin x = 8 + 2 sin x.
Equating coeﬃcients gives A = 4/3, B = −1/5, and C = 1/5. The general solution is
1
1
4
y = c1 + c2 e2x + c3 e3x + x − cos x + sin x.
3
5
5
22. Applying D to the diﬀerential equation we obtain D(D3 − D2 ) = D3 (D − 1) = 0. Then
y = c1 + c2 x + c3 ex + c4 x2
yc and yp = Ax2 . Substituting yp into the diﬀerential equation yields −2A = 6. Equating coeﬃcients gives A = −3.
The general solution is
y = c1 + c2 x + c3 ex − 3x2 .
23. The auxiliary equation is m2 − 2m + 2 = [m − (1 + i)][m − (1 − i)] = 0, so yc = c1 ex sin x + c2 ex cos x and
W = ex sin x ex cos x ex cos x + ex sin x −ex sin x + ex cos x = −e2x . Identifying f (x) = ex tan x we obtain
u1 = −
u2 = (ex cos x)(ex tan x)
= sin x
−e2x (ex sin x)(ex tan x)
sin2 x
=−
= cos x − sec x.
−e2x
cos x Then u1 = − cos x, u2 = sin x − ln | sec x + tan x|, and
y = c1 ex sin x + c2 ex cos x − ex sin x cos x + ex sin x cos x − ex cos x ln | sec x + tan x|
= c1 ex sin x + c2 ex cos x − ex cos x ln | sec x + tan x|.
24. The auxiliary equation is m2 − 1 = 0, so yc = c1 ex + c2 e−x and
W = ex e−x ex −e−x 190 = −2. CHAPTER 3 REVIEW EXERCISES
Identifying f (x) = 2ex /(ex + e−x ) we obtain
u1 = 1
ex
=
ex + e−x
1 + e2x u2 = − e2x
e3x
ex
=−
= −ex +
.
ex + e−x
1 + e2x
1 + e2x Then u1 = tan−1 ex , u2 = −ex + tan−1 ex , and
y = c1 ex + c2 e−x + ex tan−1 ex − 1 + e−x tan−1 ex .
25. The auxiliary equation is 6m2 − m − 1 = 0 so that
y = c1 x1/2 + c2 x−1/3 .
26. The auxiliary equation is 2m3 + 13m2 + 24m + 9 = (m + 3)2 (m + 1/2) = 0 so that
y = c1 x−3 + c2 x−3 ln x + c3 x−1/2 .
27. The auxiliary equation is m2 − 5m + 6 = (m − 2)(m − 3) = 0 and a particular solution is yp = x4 − x2 ln x so
that
y = c1 x2 + c2 x3 + x4 − x2 ln x.
28. The auxiliary equation is m2 − 2m + 1 = (m − 1)2 = 0 and a particular solution is yp = 1 x3 so that
4
1
y = c1 x + c2 x ln x + x3 .
4
29. (a) The auxiliary equation is m2 + ω 2 = 0, so yc = c1 cos ωt + c2 sin ωt. When ω = α, yp = A cos αt + B sin αt
and
y = c1 cos ωt + c2 sin ωt + A cos αt + B sin αt.
When ω = α, yp = At cos ωt + Bt sin ωt and
y = c1 cos ωt + c2 sin ωt + At cos ωt + Bt sin ωt.
(b) The auxiliary equation is m2 − ω 2 = 0, so yc = c1 eωt + c2 e−ωt . When ω = α, yp = Aeαt and
y = c1 eωt + c2 e−ωt + Aeαt .
When ω = α, yp = Ateωt and
y = c1 eωt + c2 e−ωt + Ateωt .
30. (a) If y = sin x is a solution then so is y = cos x and m2 + 1 is a factor of the auxiliary equation
m4 + 2m3 + 11m2 + 2m + 10 = 0. Dividing by m2 + 1 we get m2 + 2m + 10, which has roots −1 ± 3i. The
general solution of the diﬀerential equation is
y = c1 cos x + c2 sin x + e−x (c3 cos 3x + c4 sin 3x).
(b) The auxiliary equation is m(m + 1) = m2 + m = 0, so the associated homogeneous diﬀerential equation
is y + y = 0. Letting y = c1 + c2 e−x + 1 x2 − x and computing y + y we get x. Thus, the diﬀerential
2
equation is y + y = x.
31. (a) The auxiliary equation is m4 − 2m2 + 1 = (m2 − 1)2 = 0, so the general solution of the diﬀerential equation
is
y = c1 sinh x + c2 cosh x + c3 x sinh x + c4 x cosh x. 191 CHAPTER 3 REVIEW EXERCISES (b) Since both sinh x and x sinh x are solutions of the associated homogeneous diﬀerential equation, a particular
solution of y (4) − 2y + y = sinh x has the form yp = Ax2 sinh x + Bx2 cosh x.
32. Since y1 = 1 and y1 = 0, x2 y1 − (x2 + 2x)y1 + (x + 2)y1 = −x2 − 2x + x2 + 2x = 0, and y1 = x is a solution of the
associated homogeneous equation. Using the method of reduction of order, we let y = ux. Then y = xu + u
and y = xu + 2u , so
x2 y − (x2 + 2x)y + (x + 2)y = x3 u + 2x2 u − x3 u − 2x2 u − x2 u − 2xu + x2 u + 2xu
= x3 u − x3 u = x3 (u − u ).
To ﬁnd a second solution of the homogeneous equation we note that u = ex is a solution of u − u = 0. Thus,
yc = c1 x + c2 xex . To ﬁnd a particular solution we set x3 (u − u ) = x3 so that u − u = 1. This diﬀerential
equation has a particular solution of the form Ax. Substituting, we ﬁnd A = −1, so a particular solution of the
original diﬀerential equation is yp = −x2 and the general solution is y = c1 x + c2 xex − x2 .
33. The auxiliary equation is m2 − 2m + 2 = 0 so that m = 1 ± i and y = ex (c1 cos x + c2 sin x). Setting y(π/2) = 0
and y(π) = −1 we obtain c1 = e−π and c2 = 0. Thus, y = ex−π cos x.
34. The auxiliary equation is m2 + 2m + 1 = (m + 1)2 = 0, so that y = c1 e−x + c2 xe−x . Setting y(−1) = 0 and
y (0) = 0 we get c1 e − c2 e = 0 and −c1 + c2 = 0. Thus c1 = c2 and y = c1 (e−x + xe−x ) is a solution of the
boundary-value problem for any real number c1 .
35. The auxiliary equation is m2 − 1 = (m − 1)(m + 1) = 0 so that m = ±1 and y = c1 ex + c2 e−x . Assuming
yp = Ax + B + C sin x and substituting into the diﬀerential equation we ﬁnd A = −1, B = 0, and C = − 1 .
2
Thus yp = −x − 1
2 sin x and
y = c1 ex + c2 e−x − x − 1
sin x.
2 Setting y(0) = 2 and y (0) = 3 we obtain Solving this system we ﬁnd c1 = 13
4 c1 + c2 = 2
3
c1 − c2 − = 3.
2
5
and c2 = − 4 . The solution of the initial-value problem is
y= 13 x 5 −x
1
e − e − x − sin x.
4
4
2 36. The auxiliary equation is m2 + 1 = 0, so yc = c1 cos x + c2 sin x and
W = cos x sin x − sin x cos x = 1. Identifying f (x) = sec3 x we obtain
u1 = − sin x sec3 x = − sin x
cos3 x u2 = cos x sec3 x = sec2 x.
Then
u1 = − 1 1
1
= − sec2 x
2 cos2 x
2 u2 = tan x.
Thus
y = c1 cos x + c2 sin x − 1
cos x sec2 x + sin x tan x
2 192 CHAPTER 3 REVIEW EXERCISES = c1 cos x + c2 sin x − 1 − cos2 x
1
sec x +
2
cos x = c3 cos x + c2 sin x + 1
sec x.
2 and y = −c3 sin x + c2 cos x + 1
sec x tan x.
2 The initial conditions imply
c3 + 1
=1
2
1
c2 = .
2 Thus c3 = c2 = 1/2 and
y= 1
1
1
cos x + sin x + sec x.
2
2
2 37. Let u = y so that u = y . The equation becomes u du/dx = 4x. Separating variables we obtain
1
u du = 4x dx =⇒ u2 = 2x2 + c1 =⇒ u2 = 4x2 + c2 .
2
When x = 1, y = u = 2, so 4 = 4 + c2 and c2 = 0. Then
dy
dy
u2 = 4x2 =⇒
= 2x or
= −2x
dx
dx
=⇒ y = x2 + c3 or y = −x2 + c4 .
When x = 1, y = 5, so 5 = 1 + c3 and 5 = −1 + c4 . Thus c3 = 4 and c4 = 6. We have y = x2 + 4 and
y = −x2 + 6. Note however that when y = −x2 + 6, y = −2x and y (1) = −2 = 2. Thus, the solution of the
initial-value problem is y = x2 + 4.
38. Let u = y so that y = u du/dy. The equation becomes 2u du/dy = 3y 2 . Separating variables we obtain
2u du = 3y 2 dy =⇒ u2 = y 3 + c1 .
When x = 0, y = 1 and y = u = 1 so 1 = 1 + c1 and c1 = 0. Then
u2 = y 3 =⇒ dy
dx 2 = y 3 =⇒ =⇒ −2y −1/2 = x + c2 dy
= y 3/2 =⇒ y −3/2 dy = dx
dx
4
=⇒ y =
.
(x + c2 )2 When x = 0, y = 1, so 1 = 4/c2 and c2 = ±2. Thus, y = 4/(x + 2)2 and y = 4/(x − 2)2 . Note, however, that
2
when y = 4/(x + 2)2 , y = −8/(x + 2)3 and y (0) = −1 = 1. Thus, the solution of the initial-value problem is
y = 4/(x − 2)2 .
39. (a) The auxiliary equation is 12m4 + 64m3 + 59m2 − 23m − 12 = 0 and has roots −4, − 3 , − 1 , and
2
3
general solution is
y = c1 e−4x + c2 e−3x/2 + c3 e−x/3 + c4 ex/2 .
(b) The system of equations is
c1 + c2 + c3 + c4
3
1
1
−4c1 − c2 − c3 + c4
2
3
2
9
1
1
16c1 + c2 + c3 + c4
4
9
4
27
1
1
−64c1 − c2 − c3 + c4
8
27
8 193 = −1
=2
=5
= 0. 1
2. The CHAPTER 3 REVIEW EXERCISES
73
Using a CAS we ﬁnd c1 = − 495 , c2 = 109 , c3 = − 3726 , and c4 = 257 . The solution of the initial-value
35
385
45
problem is
73 −4x 109 −3x/2 3726 −x/3 257 x/2
y=−
e
e
e
e .
+
−
+
495
35
385
45 40. Consider xy + y = 0 and look for a solution of the form y = xm . Substituting
into the diﬀerential equation we have
xy + y = m(m − 1)xm−1 + mxm−1 = m2 xm−1 . −1/2 Identifying f (x) = −x 1
0 1 2 3 4 5 x -1 Thus, the general solution of xy +y = 0 is yc = c1 +c2 ln x. To ﬁnd a particular
√
solution of xy + y = − x we use variation of parameters. The Wronskian is
W = y ln x
1
= .
x
1/x -2
-3
-4
-5 we obtain
u1 = √
x−1/2 ln x √
−x−1/2
= x ln x and u2 =
= − x,
1/x
1/x so that
u1 = x3/2 2
4
ln x −
3
9 2
and u2 = − x3/2 .
3 Then 4
2
4
2
ln x −
− x3/2 ln x = − x3/2
3
9
3
9
and the general solution of the diﬀerential equation is
yp = x3/2 4
y = c1 + c2 ln x − x3/2 .
9
The initial conditions are y(1) = 0 and y (1) = 0. These imply that c1 =
initial-value problem is
4 2
4
y = + ln x − x3/2 .
9 3
9
The graph is shown above. 4
9 and c2 = 2
3 . The solution of the 41. From (D − 2)x + (D − 2)y = 1 and Dx + (2D − 1)y = 3 we obtain (D − 1)(D − 2)y = −6 and Dx = 3 − (2D − 1)y.
Then
3
y = c1 e2t + c2 et − 3 and x = −c2 et − c1 e2t + c3 .
2
Substituting into (D − 2)x + (D − 2)y = 1 gives c3 = 5 so that
2
3
5
x = −c2 et − c1 e2t + .
2
2
42. From (D − 2)x − y = t − 2 and −3x + (D − 4)y = −4t we obtain (D − 1)(D − 5)x = 9 − 8t. Then and 8
3
x = c1 et + c2 e5t − t −
5
25
y = (D − 2)x − t + 2 = −c1 et + 3c2 e5t + 16 11
+ t.
25 25 43. From (D − 2)x − y = −et and −3x + (D − 4)y = −7et we obtain (D − 1)(D − 5)x = −4et so that
x = c1 et + c2 e5t + tet .
Then 194 CHAPTER 3 REVIEW EXERCISES
y = (D − 2)x + et = −c1 et + 3c2 e5t − tet + 2et .
44. From (D + 2)x + (D + 1)y = sin 2t and 5x + (D + 3)y = cos 2t we obtain (D2 + 5)y = 2 cos 2t − 7 sin 2t. Then
y = c1 cos t + c2 sin t −
and 2
7
cos 2t + sin 2t
3
3 1
1
x = − (D + 3)y + cos 2t
5
5
1
1
1
5
3
3
=
c1 − c2 sin t + − c2 − c1 cos t − sin 2t − cos 2t.
5
5
5
5
3
3 45. The period of a spring/mass system is given by T = 2π/ω where ω 2 = k/m = kg/W , where k is the spring
constant, W is the weight of the mass attached to the spring, and g is the acceleration due to gravity. Thus,
√
√
√
√
the period of oscillation is T = (2π/ kg ) W . If the weight of the original mass is W , then (2π/ kg ) W = 3
√
√
√
√
and (2π/ kg ) W − 8 = 2. Dividing, we get W / W − 8 = 3/2 or W = 9 (W − 8). Solving for W we ﬁnd
4
that the weight of the original mass was 14.4 pounds.
46. (a) Solving 3 x + 6x = 0 subject to x(0) = 1 and x (0) = −4 we obtain
8
√
x = cos 4t − sin 4t = 2 sin (4t + 3π/4) .
(b) The amplitude is √ 2, period is π/2, and frequency is 2/π. (c) If x = 1 then t = nπ/2 and t = −π/8 + nπ/2 for n = 1, 2, 3, . . . .
(d) If x = 0 then t = π/16 + nπ/4 for n = 0, 1, 2, . . .. The motion is upward for n even and downward for n
odd.
(e) x (3π/16) = 0
(f ) If x = 0 then 4t + 3π/4 = π/2 + nπ or t = 3π/16 + nπ.
47. From mx + 4x + 2x = 0 we see that nonoscillatory motion results if 16 − 8m ≥ 0 or 0 < m ≤ 2.
48. From x + βx + 64x = 0 we see that oscillatory motion results if β 2 − 256 < 0 or 0 ≤ β < 16.
49. From q + 104 q = 100 sin 50t, q(0) = 0, and q (0) = 0 we obtain qc = c1 cos 100t + c2 sin 100t, qp =
and
1
(a) q = − 150 sin 100t + 1
75 (b) i = − 2 cos 100t +
3 cos 50t, and 2
3 sin 50t, (c) q = 0 when sin 50t(1 − cos 50t) = 0 or t = nπ/50 for n = 0, 1, 2, . . . .
50. By Kirchhoﬀ’s second law,
d2 q
dq
1
+R
+ q = E(t).
dt2
dt
C
Using q (t) = i(t) we can write the diﬀerential equation in the form
L L di
1
+ Ri + q = E(t).
dt
C Then diﬀerentiating we obtain
L d2 i
di
1
+R
+ i = E (t).
dt2
dt C 51. For λ = α2 > 0 the general solution is y = c1 cos αx + c2 sin αx. Now
y(0) = c1 and y(2π) = c1 cos 2πα + c2 sin 2πα, 195 1
75 sin 50t, CHAPTER 3 REVIEW EXERCISES so the condition y(0) = y(2π) implies
c1 = c1 cos 2πα + c2 sin 2πα
which is true when α = √ λ = n or λ = n2 for n = 1, 2, 3, . . . . Since
y = −αc1 sin αx + αc2 cos αx = −nc1 sin nx + nc2 cos nx, we see that y (0) = nc2 = y (2π) for n = 1, 2, 3, . . . . Thus, the eigenvalues are n2 for n = 1, 2, 3, . . . ,
with corresponding eigenfunctions cos nx and sin nx. When λ = 0, the general solution is y = c1 x + c2 and the
corresponding eigenfunction is y = 1.
For λ = −α2 < 0 the general solution is y = c1 cosh αx + c2 sinh αx. In this case y(0) = c1 and y(2π) =
c1 cosh 2πα + c2 sinh 2πα, so y(0) = y(2π) can only be valid for α = 0. Thus, there are no eigenvalues corresponding to λ < 0.
52. (a) The diﬀerential equation is d2 r/dt2 − ω 2 r = −g sin ωt. The auxiliary equation is m2 − ω 2 = 0, so
rc = c1 eωt + c2 e−ωt . A particular solution has the form rp = A sin ωt + B cos ωt. Substituting into
the diﬀerential equation we ﬁnd −2Aω 2 sin ωt − 2Bω 2 cos ωt = −g sin ωt. Thus, B = 0, A = g/2ω 2 , and
rp = (g/2ω 2 ) sin ωt. The general solution of the diﬀerential equation is r(t) = c1 eωt +c2 e−ωt +(g/2ω 2 ) sin ωt.
The initial conditions imply c1 + c2 = r0 and g/2ω − ωc1 + ωc2 = v0 . Solving for c1 and c2 we get
c1 = (2ω 2 r0 + 2ωv0 − g)/4ω 2 and c2 = (2ω 2 r0 − 2ωv0 + g)/4ω 2 , so that
r(t) = 2ω 2 r0 + 2ωv0 − g ωt 2ω 2 r0 − 2ωv0 + g −ωt
g
e +
e
+
sin ωt.
4ω 2
4ω 2
2ω 2 (b) The bead will exhibit simple harmonic motion when the exponential terms are missing. Solving c1 = 0,
c2 = 0 for r0 and v0 we ﬁnd r0 = 0 and v0 = g/2ω.
To ﬁnd the minimum length of rod that will accommodate simple harmonic motion we determine the
amplitude of r(t) and double it. Thus L = g/ω 2 .
(c) As t increases, eωt approaches inﬁnity and e−ωt approaches 0. Since sin ωt is bounded, the distance, r(t), of
the bead from the pivot point increases without bound and the distance of the bead from P will eventually
exceed L/2.
(d) r
17 20 16.1
16 10
2 4 6 8 10 12 14 -10
-20 0 10 15 196 t CHAPTER 3 REVIEW EXERCISES
(e) For each v0 we want to ﬁnd the smallest value of t for which r(t) = ±20. Whether we look for r(t) = −20
or r(t) = 20 is determined by looking at the graphs in part (d). The total times that the bead stays on the
rod is shown in the table below. v0 0 10 15 16.1 17 r
t -20
1.55007 -20
2.35494 -20
3.43088 20
6.11627 20
4.22339 When v0 = 16 the bead never leaves the rod.
53. Unlike the derivation given in Section 3.8 in the text, the weight mg of the mass m does not appear in the net
force since the spring is not stretched by the weight of the mass when it is in the equilibrium position (i.e. there
is no mg − ks term in the net force). The only force acting on the mass when it is in motion is the restoring
force of the spring. By Newton’s second law,
m d2 x
= −kx
dt2 or d2 x
k
+ x = 0.
dt2
m 54. The force of kinetic friction opposing the motion of the mass in µN , where µ is the coeﬃcient of sliding friction
and N is the normal component of the weight. Since friction is a force opposite to the direction of motion
and since N is pointed directly downward (it is simply the weight of the mass), Newton’s second law gives, for
motion to the right (x > 0) ,
m d2 x
= −kx − µmg,
dt2 and for motion to the left (x < 0),
d2 x
= −kx + µmg.
dt2
Traditionally, these two equations are written as one expression
m m d2 x
+ fx sgn(x ) + kx = 0,
dt2 where fk = µmg and
sgn(x ) = 1, x > 0
−1, x < 0. 197 ...

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