Unformatted text preview: Part I Ordinary Differential Equations Introduction to
Differential Equations 1 EXERCISES 1.1
Deﬁnitions and Terminology 1. Second order; linear
2. Third order; nonlinear because of (dy/dx)4
3. Fourth order; linear
4. Second order; nonlinear because of cos(r + u)
5. Second order; nonlinear because of (dy/dx)2 or 1 + (dy/dx)2 6. Second order; nonlinear because of R2
7. Third order; linear
8. Second order; nonlinear because of x2
˙
9. Writing the diﬀerential equation in the form x(dy/dx) + y 2 = 1, we see that it is nonlinear in y because of y 2 .
However, writing it in the form (y 2 − 1)(dx/dy) + x = 0, we see that it is linear in x.
10. Writing the diﬀerential equation in the form u(dv/du) + (1 + u)v = ueu we see that it is linear in v. However,
writing it in the form (v + uv − ueu )(du/dv) + u = 0, we see that it is nonlinear in u.
11. From y = e−x/2 we obtain y = − 1 e−x/2 . Then 2y + y = −e−x/2 + e−x/2 = 0.
2
12. From y = 6
5 − 6 e−20t we obtain dy/dt = 24e−20t , so that
5
dy
+ 20y = 24e−20t + 20
dt 6 6 −20t
− e
5 5 13. From y = e3x cos 2x we obtain y = 3e3x cos 2x − 2e3x sin 2x and y
y − 6y + 13y = 0. = 24.
= 5e3x cos 2x − 12e3x sin 2x, so that 14. From y = − cos x ln(sec x + tan x) we obtain y = −1 + sin x ln(sec x + tan x) and
y = tan x + cos x ln(sec x + tan x). Then y + y = tan x.
15. The domain of the function, found by solving x + 2 ≥ 0, is [−2, ∞). From y = 1 + 2(x + 2)−1/2 we have
(y − x)y = (y − x)[1 + (2(x + 2)−1/2 ]
= y − x + 2(y − x)(x + 2)−1/2
= y − x + 2[x + 4(x + 2)1/2 − x](x + 2)−1/2
= y − x + 8(x + 2)1/2 (x + 2)−1/2 = y − x + 8. 1 1.1 Deﬁnitions and Terminology
An interval of deﬁnition for the solution of the diﬀerential equation is (−2, ∞) because y is not deﬁned at
x = −2. 16. Since tan x is not deﬁned for x = π/2 + nπ, n an integer, the domain of y
{x 5x = π/2 + nπ} or {x x = π/10 + nπ/5}. From y = 25 sec2 5x we have = 5 tan 5x is y = 25(1 + tan2 5x) = 25 + 25 tan2 5x = 25 + y 2 .
An interval of deﬁnition for the solution of the diﬀerential equation is (−π/10, π/10). Another interval is
(π/10, 3π/10), and so on.
17. The domain of the function is {x 4 − x2 = 0} or {x x = −2 or x = 2}. From y = 2x/(4 − x2 )2 we have
y = 2x 1
4 − x2 2 = 2xy. An interval of deﬁnition for the solution of the diﬀerential equation is (−2, 2). Other intervals are (−∞, −2)
and (2, ∞).
√
18. The function is y = 1/ 1 − sin x , whose domain is obtained from 1 − sin x = 0 or sin x = 1. Thus, the domain
is {x x = π/2 + 2nπ}. From y = − 1 (1 − sin x)−3/2 (− cos x) we have
2
2y = (1 − sin x)−3/2 cos x = [(1 − sin x)−1/2 ]3 cos x = y 3 cos x.
An interval of deﬁnition for the solution of the diﬀerential equation is (π/2, 5π/2). Another one is (5π/2, 9π/2),
and so on.
19. Writing ln(2X −1)−ln(X −1) = t and diﬀerentiating implicitly we obtain X 2
dX
1 dX
−
=1
2X − 1 dt
X − 1 dt
2
1
−
2X − 1 X − 1 4 dX
=1
dt 2 2X − 2 − 2X + 1 dX
=1
(2X − 1)(X − 1) dt 4 dX
= −(2X − 1)(X − 1) = (X − 1)(1 − 2X).
dt
Exponentiating both sides of the implicit solution we obtain 2 2 4 t 2
4 2X − 1
= et
X −1
2X − 1 = Xet − et
(et − 1) = (et − 2)X
X= et − 1
.
et − 2 Solving et − 2 = 0 we get t = ln 2. Thus, the solution is deﬁned on (−∞, ln 2) or on (ln 2, ∞). The graph of the
solution deﬁned on (−∞, ln 2) is dashed, and the graph of the solution deﬁned on (ln 2, ∞) is solid. 2 1.1 Deﬁnitions and Terminology 20. Implicitly diﬀerentiating the solution, we obtain y dy
dy
−2x2
− 4xy + 2y
=0
dx
dx
−x2 dy − 2xy dx + y dy = 0 4
2 2xy dx + (x2 − y)dy = 0.
Using the quadratic formula to solve y 2 − 2x2 y − 1 = 0 for y, we get
√
√
y = 2x2 ± 4x4 + 4 /2 = x2 ± x4 + 1 . Thus, two explicit solutions
√
√
are y1 = x2 + x4 + 1 and y2 = x2 − x4 + 1 . Both solutions are deﬁned 4 2 2 4 x 2 on (−∞, ∞). The graph of y1 (x) is solid and the graph of y2 is dashed. 4 21. Diﬀerentiating P = c1 et / (1 + c1 et ) we obtain
dP
(1 + c1 et ) c1 et − c1 et · c1 et
c1 et [(1 + c1 et ) − c1 et ]
=
=
2
dt
1 + c1 et
1 + c1 et
(1 + c1 et )
=
x 22. Diﬀerentiating y = e−x 2 c1 et
1 + c1 et 1− c1 et
= P (1 − P ).
1 + c1 et et dt + c1 e−x we obtain
2 2 0
x y = e−x ex − 2xe−x
2 2 2 x et dt − 2c1 xe−x = 1 − 2xe−x
2 2 2 0 et dt − 2c1 xe−x .
2 2 0 Substituting into the diﬀerential equation, we have
x y + 2xy = 1 − 2xe−x 2 2 2 0 23. From y = c1 e2x + c2 xe2x we obtain x et dt − 2c1 xe−x + 2xe−x 2 et dt + 2c1 xe−x = 1.
2 2 0 dy
d2 y
= (2c1 + c2 )e2x + 2c2 xe2x and
= (4c1 + 4c2 )e2x + 4c2 xe2x , so that
dx
dx2 d2 y
dy
−4
+ 4y = (4c1 + 4c2 − 8c1 − 4c2 + 4c1 )e2x + (4c2 − 8c2 + 4c2 )xe2x = 0.
2
dx
dx
24. From y = c1 x−1 + c2 x + c3 x ln x + 4x2 we obtain
dy
= −c1 x−2 + c2 + c3 + c3 ln x + 8x,
dx
d2 y
= 2c1 x−3 + c3 x−1 + 8,
dx2 and d3 y
= −6c1 x−4 − c3 x−2 ,
dx3 so that
x3 d3 y
d2 y
dy
+ y = (−6c1 + 4c1 + c1 + c1 )x−1 + (−c3 + 2c3 − c2 − c3 + c2 )x
+ 2x2 2 − x
3
dx
dx
dx
+ (−c3 + c3 )x ln x + (16 − 8 + 4)x2
= 12x2 . 25. From y = −x2 ,
2 x , x<0
x≥0 we obtain y = −2x,
2x, x<0
so that xy − 2y = 0.
x≥0 3 1.1 Deﬁnitions and Terminology 26. The function y(x) is not continuous at x = 0 since lim y(x) = 5 and lim y(x) = −5. Thus, y (x) does not
x→0− x→0+ exist at x = 0.
27. (a) From y = emx we obtain y = memx . Then y + 2y = 0 implies
memx + 2emx = (m + 2)emx = 0.
Since emx > 0 for all x, m = −2. Thus y = e−2x is a solution.
(b) From y = emx we obtain y = memx and y = m2 emx . Then y − 5y + 6y = 0 implies
m2 emx − 5memx + 6emx = (m − 2)(m − 3)emx = 0.
Since emx > 0 for all x, m = 2 and m = 3. Thus y = e2x and y = e3x are solutions.
28. (a) From y = xm we obtain y = mxm−1 and y = m(m − 1)xm−2 . Then xy + 2y = 0 implies
xm(m − 1)xm−2 + 2mxm−1 = [m(m − 1) + 2m]xm−1 = (m2 + m)xm−1
= m(m + 1)xm−1 = 0.
Since xm−1 > 0 for x > 0, m = 0 and m = −1. Thus y = 1 and y = x−1 are solutions.
(b) From y = xm we obtain y = mxm−1 and y = m(m − 1)xm−2 . Then x2 y − 7xy + 15y = 0 implies
x2 m(m − 1)xm−2 − 7xmxm−1 + 15xm = [m(m − 1) − 7m + 15]xm
= (m2 − 8m + 15)xm = (m − 3)(m − 5)xm = 0.
Since xm > 0 for x > 0, m = 3 and m = 5. Thus y = x3 and y = x5 are solutions.
In Problems 29–32, we substitute y = c into the diﬀerential equations and use y = 0 and y = 0
29. Solving 5c = 10 we see that y = 2 is a constant solution.
30. Solving c2 + 2c − 3 = (c + 3)(c − 1) = 0 we see that y = −3 and y = 1 are constant solutions.
31. Since 1/(c − 1) = 0 has no solutions, the diﬀerential equation has no constant solutions.
32. Solving 6c = 10 we see that y = 5/3 is a constant solution.
33. From x = e−2t + 3e6t and y = −e−2t + 5e6t we obtain
dx
= −2e−2t + 18e6t
dt and dy
= 2e−2t + 30e6t .
dt Then
x + 3y = (e−2t + 3e6t ) + 3(−e−2t + 5e6t ) = −2e−2t + 18e6t = dx
dt 5x + 3y = 5(e−2t + 3e6t ) + 3(−e−2t + 5e6t ) = 2e−2t + 30e6t = dy
.
dt and 34. From x = cos 2t + sin 2t + 1 et and y = − cos 2t − sin 2t − 1 et we obtain
5
5 and dx
1
= −2 sin 2t + 2 cos 2t + et
dt
5 and dy
1
= 2 sin 2t − 2 cos 2t − et
dt
5 1
d2 x
= −4 cos 2t − 4 sin 2t + et
2
dt
5 and d2 y
1
= 4 cos 2t + 4 sin 2t − et .
2
dt
5 Then and 1
1
d2 x
4y + et = 4(− cos 2t − sin 2t − et ) + et = −4 cos 2t − 4 sin 2t + et = 2
5
5
dt 4 1.1 Deﬁnitions and Terminology 1
1
d2 y
4x − et = 4(cos 2t + sin 2t + et ) − et = 4 cos 2t + 4 sin 2t − et = 2 .
5
5
dt
35. (y )2 + 1 = 0 has no real solutions because (y )2 + 1 is positive for all functions y = φ(x).
36. The only solution of (y )2 + y 2 = 0 is y = 0, since if y = 0, y 2 > 0 and (y )2 + y 2 ≥ y 2 > 0.
37. The ﬁrst derivative of f (x) = ex is ex . The ﬁrst derivative of f (x) = ekx is kekx . The diﬀerential equations are
y = y and y = ky, respectively.
38. Any function of the form y = cex or y = ce−x is its own second derivative. The corresponding diﬀerential
equation is y − y = 0. Functions of the form y = c sin x or y = c cos x have second derivatives that are the
negatives of themselves. The diﬀerential equation is y + y = 0.
√
39. We ﬁrst note that 1 − y 2 = 1 − sin2 x = cos2 x =  cos x. This prompts us to consider values of x for
which cos x < 0, such as x = π. In this case
dy
dx =
x=π d
(sin x)
dx = cos x x=π = cos π = −1, x=π but
1 − y 2 x=π = 1 − sin2 π = √ 1 = 1. Thus, y = sin x will only be a solution of y = 1 − y 2 when cos x > 0. An interval of deﬁnition is then
(−π/2, π/2). Other intervals are (3π/2, 5π/2), (7π/2, 9π/2), and so on.
40. Since the ﬁrst and second derivatives of sin t and cos t involve sin t and cos t, it is plausible that a linear
combination of these functions, A sin t + B cos t, could be a solution of the diﬀerential equation. Using y =
A cos t − B sin t and y = −A sin t − B cos t and substituting into the diﬀerential equation we get
y + 2y + 4y = −A sin t − B cos t + 2A cos t − 2B sin t + 4A sin t + 4B cos t
= (3A − 2B) sin t + (2A + 3B) cos t = 5 sin t.
Thus 3A − 2B = 5 and 2A + 3B = 0. Solving these simultaneous equations we ﬁnd A =
particular solution is y = 15
13 sin t − 10
13 15
13 and B = − 10 . A
13 cos t. 41. One solution is given by the upper portion of the graph with domain approximately (0, 2.6). The other solution
is given by the lower portion of the graph, also with domain approximately (0, 2.6).
42. One solution, with domain approximately (−∞, 1.6) is the portion of the graph in the second quadrant together
with the lower part of the graph in the ﬁrst quadrant. A second solution, with domain approximately (0, 1.6)
is the upper part of the graph in the ﬁrst quadrant. The third solution, with domain (0, ∞), is the part of the
graph in the fourth quadrant.
43. Diﬀerentiating (x3 + y 3 )/xy = 3c we obtain
xy(3x2 + 3y 2 y ) − (x3 + y 3 )(xy + y)
=0
x2 y 2
3x3 y + 3xy 3 y − x4 y − x3 y − xy 3 y − y 4 = 0
(3xy 3 − x4 − xy 3 )y = −3x3 y + x3 y + y 4
y = y 4 − 2x3 y
y(y 3 − 2x3 )
=
.
2xy 3 − x4
x(2y 3 − x3 ) 44. A tangent line will be vertical where y is undeﬁned, or in this case, where x(2y 3 − x3 ) = 0. This gives x = 0
and 2y 3 = x3 . Substituting y 3 = x3 /2 into x3 + y 3 = 3xy we get 5 1.1 Deﬁnitions and Terminology 1
1
x3 + x3 = 3x
x
2
21/3
3 3
3
x = 1/3 x2
2
2
x3 = 22/3 x2
x2 (x − 22/3 ) = 0.
Thus, there are vertical tangent lines at x = 0 and x = 22/3 , or at (0, 0) and (22/3 , 21/3 ). Since 22/3 ≈ 1.59, the
estimates of the domains in Problem 42 were close.
√
√
45. The derivatives of the functions are φ1 (x) = −x/ 25 − x2 and φ2 (x) = x/ 25 − x2 , neither of which is deﬁned
at x = ±5.
46. To determine if a solution curve passes through (0, 3) we let t = 0 and P = 3 in the equation P = c1 et /(1+c1 et ).
This gives 3 = c1 /(1 + c1 ) or c1 = − 3 . Thus, the solution curve
2
P = (−3/2)et
−3et
=
1 − (3/2)et
2 − 3et passes through the point (0, 3). Similarly, letting t = 0 and P = 1 in the equation for the oneparameter family
of solutions gives 1 = c1 /(1 + c1 ) or c1 = 1 + c1 . Since this equation has no solution, no solution curve passes
through (0, 1).
47. For the ﬁrstorder diﬀerential equation integrate f (x). For the secondorder diﬀerential equation integrate twice.
In the latter case we get y = ( f (x)dx)dx + c1 x + c2 .
48. Solving for y using the quadratic formula we obtain the two diﬀerential equations
y = 1
2 + 2 1 + 3x6
x and y = 1
2−2
x 1 + 3x6 , so the diﬀerential equation cannot be put in the form dy/dx = f (x, y).
49. The diﬀerential equation yy − xy = 0 has normal form dy/dx = x. These are not equivalent because y = 0 is a
solution of the ﬁrst diﬀerential equation but not a solution of the second.
50. Diﬀerentiating we get y = c1 + 3c2 x2 and y = 6c2 x. Then c2 = y /6x and c1 = y − xy /2, so
y= y − xy
2 y
6x x+ 1
x3 = xy − x2 y
3 and the diﬀerential equation is x2 y − 3xy + 3y = 0.
51. (a) Since e−x is positive for all values of x, dy/dx > 0 for all x, and a solution, y(x), of the diﬀerential equation
must be increasing on any interval.
2
2
dy
dy
(b) lim
= lim e−x = 0 and lim
= lim e−x = 0. Since dy/dx approaches 0 as x approaches −∞
x→−∞ dx
x→−∞
x→∞ dx
x→∞
and ∞, the solution curve has horizontal asymptotes to the left and to the right.
2 (c) To test concavity we consider the second derivative
d2 y
d
=
dx2
dx dy
dx = 2
2
d
e−x = −2xe−x .
dx Since the second derivative is positive for x < 0 and negative for x > 0, the solution curve is concave up on
(−∞, 0) and concave down on (0, ∞). 6 1.1
(d) Deﬁnitions and Terminology y x 52. (a) The derivative of a constant solution y = c is 0, so solving 5 − c = 0 we see that c = 5 and so y = 5 is a
constant solution.
(b) A solution is increasing where dy/dx = 5−y > 0 or y < 5. A solution is decreasing where dy/dx = 5−y < 0
or y > 5.
53. (a) The derivative of a constant solution is 0, so solving y(a − by) = 0 we see that y = 0 and y = a/b are
constant solutions.
(b) A solution is increasing where dy/dx = y(a − by) = by(a/b − y) > 0 or 0 < y < a/b. A solution is decreasing
where dy/dx = by(a/b − y) < 0 or y < 0 or y > a/b.
(c) Using implicit diﬀerentiation we compute
d2 y
= y(−by ) + y (a − by) = y (a − 2by).
dx2
Solving d2 y/dx2 = 0 we obtain y = a/2b. Since d2 y/dx2 > 0 for 0 < y < a/2b and d2 y/dx2 < 0 for
a/2b < y < a/b, the graph of y = φ(x) has a point of inﬂection at y = a/2b.
(d) y y=aêb y=0
x 54. (a) If y = c is a constant solution then y = 0, but c2 + 4 is never 0 for any real value of c.
(b) Since y = y 2 + 4 > 0 for all x where a solution y = φ(x) is deﬁned, any solution must be increasing on any
interval on which it is deﬁned. Thus it cannot have any relative extrema.
(c) Using implicit diﬀerentiation we compute d2 y/dx2 = 2yy = 2y(y 2 + 4). Setting d2 y/dx2 = 0 we see that
y = 0 corresponds to the only possible point of inﬂection. Since d2 y/dx2 < 0 for y < 0 and d2 y/dx2 > 0
for y > 0, there is a point of inﬂection where y = 0. 7 1.1 Deﬁnitions and Terminology (d) y x 55. In Mathematica use
Clear[y]
y[x ]:= x Exp[5x] Cos[2x]
y[x]
y''''[x] − 20y'''[x] + 158y''[x] − 580y'[x] +841y[x]//Simplify
The output will show y(x) = e5x x cos 2x, which veriﬁes that the correct function was entered, and 0, which
veriﬁes that this function is a solution of the diﬀerential equation.
56. In Mathematica use
Clear[y]
y[x ]:= 20Cos[5Log[x]]/x − 3Sin[5Log[x]]/x
y[x]
xˆ3 y'''[x] + 2xˆ2 y''[x] + 20x y'[x] − 78y[x]//Simplify
The output will show y(x) = 20 cos(5 ln x)/x − 3 sin(5 ln x)/x, which veriﬁes that the correct function was
entered, and 0, which veriﬁes that this function is a solution of the diﬀerential equation. EXERCISES 1.2
InitialValue Problems
1. Solving −1/3 = 1/(1 + c1 ) we get c1 = −4. The solution is y = 1/(1 − 4e−x ).
2. Solving 2 = 1/(1 + c1 e) we get c1 = −(1/2)e−1 . The solution is y = 2/(2 − e−(x+1) ) .
3. Letting x = 2 and solving 1/3 = 1/(4 + c) we get c = −1. The solution is y = 1/(x2 − 1). This solution is
deﬁned on the interval (1, ∞).
4. Letting x = −2 and solving 1/2 = 1/(4 + c) we get c = −2. The solution is y = 1/(x2 − 2). This solution is
√
deﬁned on the interval (−∞, − 2 ).
5. Letting x = 0 and solving 1 = 1/c we get c = 1. The solution is y = 1/(x2 + 1). This solution is deﬁned on the
interval (−∞, ∞). 8 1.2 InitialValue Problems
6. Letting x = 1/2 and solving −4 = 1/(1/4 + c) we get c = −1/2. The solution is y = 1/(x2 − 1/2) = 2/(2x2 − 1).
√
√
This solution is deﬁned on the interval (−1/ 2 , 1/ 2 ).
In Problems 7–10, we use x = c1 cos t + c2 sin t and x = −c1 sin t + c2 cos t to obtain a system of two equations in
the two unknowns c1 and c2 .
7. From the initial conditions we obtain the system
c1 = −1
c2 = 8.
The solution of the initialvalue problem is x = − cos t + 8 sin t.
8. From the initial conditions we obtain the system
c2 = 0
−c1 = 1.
The solution of the initialvalue problem is x = − cos t.
9. From the initial conditions we obtain Solving, we ﬁnd c1 = √ √ 3
1
1
c1 + c2 =
2
2
2
√
1
3
− c1 +
c2 = 0.
2
2
3/4 and c2 = 1/4. The solution of the initialvalue problem is
√
x = ( 3/4) cos t + (1/4) sin t. 10. From the initial conditions we obtain √ √
√
2
2
c1 +
c2 = 2
2
2
√
√
√
2
2
−
c1 +
c2 = 2 2 .
2
2
Solving, we ﬁnd c1 = −1 and c2 = 3. The solution of the initialvalue problem is x = − cos t + 3 sin t. In Problems 11–14, we use y = c1 ex + c2 e−x and y = c1 ex − c2 e−x to obtain a system of two equations in the two
unknowns c1 and c2 .
11. From the initial conditions we obtain
c1 + c2 = 1
c1 − c2 = 2.
Solving, we ﬁnd c1 = 3
2 and c2 = − 1 . The solution of the initialvalue problem is y = 3 ex − 1 e−x .
2
2
2 12. From the initial conditions we obtain ec1 + e−1 c2 = 0
ec1 − e−1 c2 = e. Solving, we ﬁnd c1 = 1
2 and c2 = − 1 e2 . The solution of the initialvalue problem is
2
y= 13. From the initial conditions we obtain 1 x 1 2 −x
1
1
e − e e = ex − e2−x .
2
2
2
2
e−1 c1 + ec2 = 5
e−1 c1 − ec2 = −5. 9 1.2 InitialValue Problems
Solving, we ﬁnd c1 = 0 and c2 = 5e−1 . The solution of the initialvalue problem is y = 5e−1 e−x = 5e−1−x . 14. From the initial conditions we obtain
c1 + c2 = 0
c1 − c2 = 0.
Solving, we ﬁnd c1 = c2 = 0. The solution of the initialvalue problem is y = 0.
15. Two solutions are y = 0 and y = x3 .
16. Two solutions are y = 0 and y = x2 . (Also, any constant multiple of x2 is a solution.)
∂f
2
= y −1/3 . Thus, the diﬀerential equation will have a unique solution in any
∂y
3
rectangular region of the plane where y = 0.
√
18. For f (x, y) = xy we have ∂f /∂y = 1 x/y . Thus, the diﬀerential equation will have a unique solution in any
2
17. For f (x, y) = y 2/3 we have region where x > 0 and y > 0 or where x < 0 and y < 0.
19. For f (x, y) = y
∂f
1
we have
= . Thus, the diﬀerential equation will have a unique solution in any region
x
∂y
x where x = 0.
20. For f (x, y) = x + y we have ∂f
= 1. Thus, the diﬀerential equation will have a unique solution in the entire
∂y plane.
21. For f (x, y) = x2 /(4 − y 2 ) we have ∂f /∂y = 2x2 y/(4 − y 2 )2 . Thus the diﬀerential equation will have a unique
solution in any region where y < −2, −2 < y < 2, or y > 2.
x2
∂f
−3x2 y 2
we have
=
2 . Thus, the diﬀerential equation will have a unique solution in
1 + y3
∂y
(1 + y 3 )
any region where y = −1. 22. For f (x, y) = y2
2x2 y
∂f
=
we have
2 . Thus, the diﬀerential equation will have a unique solution in
x2 + y 2
∂y
(x2 + y 2 )
any region not containing (0, 0). 23. For f (x, y) = 24. For f (x, y) = (y + x)/(y − x) we have ∂f /∂y = −2x/(y − x)2 . Thus the diﬀerential equation will have a unique
solution in any region where y < x or where y > x.
y 2 − 9 and ∂f /∂y = y/ y 2 − 9.
We see that f and
∂f /∂y are both continuous in the regions of the plane determined by y < −3 and y > 3 with no restrictions on
x. In Problems 25–28, we identify f (x, y) = 25. Since 4 > 3, (1, 4) is in the region deﬁned by y > 3 and the diﬀerential equation has a unique solution through
(1, 4).
26. Since (5, 3) is not in either of the regions deﬁned by y < −3 or y > 3, there is no guarantee of a unique solution
through (5, 3).
27. Since (2, −3) is not in either of the regions deﬁned by y < −3 or y > 3, there is no guarantee of a unique solution
through (2, −3).
28. Since (−1, 1) is not in either of the regions deﬁned by y < −3 or y > 3, there is no guarantee of a unique solution
through (−1, 1).
29. (a) A oneparameter family of solutions is y = cx. Since y = c, xy = xc = y and y(0) = c · 0 = 0. 10 1.2 InitialValue Problems
(b) Writing the equation in the form y = y/x, we see that R cannot contain any point on the yaxis. Thus,
any rectangular region disjoint from the yaxis and containing (x0 , y0 ) will determine an interval around x0
and a unique solution through (x0 , y0 ). Since x0 = 0 in part (a), we are not guaranteed a unique solution
through (0, 0).
(c) The piecewisedeﬁned function which satisﬁes y(0) = 0 is not a solution since it is not diﬀerentiable at
x = 0.
d
30. (a) Since
tan(x + c) = sec2 (x + c) = 1 + tan2 (x + c), we see that y = tan(x + c) satisﬁes the diﬀerential
dx
equation.
(b) Solving y(0) = tan c = 0 we obtain c = 0 and y = tan x. Since tan x is discontinuous at x = ±π/2, the
solution is not deﬁned on (−2, 2) because it contains ±π/2.
(c) The largest interval on which the solution can exist is (−π/2, π/2).
d
1
1
1
31. (a) Since
= y 2 , we see that y = −
−
=
is a solution of the diﬀerential equation.
dx
x+c
(x + c)2
x+c
(b) Solving y(0) = −1/c = 1 we obtain c = −1 and y = 1/(1 − x). Solving y(0) = −1/c = −1 we obtain c = 1
and y = −1/(1 + x). Being sure to include x = 0, we see that the interval of existence of y = 1/(1 − x) is
(−∞, 1), while the interval of existence of y = −1/(1 + x) is (−1, ∞).
32. (a) Solving y(0) = −1/c = y0 we obtain c = −1/y0 and
y=− 1
y0
=
,
−1/y0 + x
1 − y0 x y0 = 0. Since we must have −1/y0 + x = 0, the largest interval of existence (which must contain 0) is either
(−∞, 1/y0 ) when y0 > 0 or (1/y0 , ∞) when y0 < 0.
(b) By inspection we see that y = 0 is a solution on (−∞, ∞).
33. (a) Diﬀerentiating 3x2 − y 2 = c we get 6x − 2yy = 0 or yy = 3x.
y (b) Solving 3x2 − y 2 = 3 for y we get
y = φ1 (x) = 3(x2 − 1) , y = φ2 (x) = − 3(x2 − 1) ,
y = φ3 (x) = 3(x2 − 1) , y = φ4 (x) = − 3(x2 − 1) , 4 1 < x < ∞, 2 1 < x < ∞,
−∞ < x < −1, 4 2 −∞ < x < −1. 4 x 2 2 4 x 2
4 (c) Only y = φ3 (x) satisﬁes y(−2) = 3.
y 34. (a) Setting x = 2 and y = −4 in 3x2 − y 2 = c we get 12 − 16 = −4 = c, so the
explicit solution is 4 y = − 3x2 + 4 , −∞ < x < ∞.
(b) Setting c = 0 we have y = √ √
3x and y = − 3x, both deﬁned on 2
4 2
2 (−∞, ∞). 4 11 1.2 InitialValue Problems In Problems 35–38, we consider the points on the graphs with xcoordinates x0 = −1, x0 = 0, and x0 = 1. The
slopes of the tangent lines at these points are compared with the slopes given by y (x0 ) in (a) through (f).
35. The graph satisﬁes the conditions in (b) and (f).
36. The graph satisﬁes the conditions in (e).
37. The graph satisﬁes the conditions in (c) and (d).
38. The graph satisﬁes the conditions in (a).
39. Integrating y = 8e2x + 6x we obtain
y= (8e2x + 6x)dx = 4e2x + 3x2 + c. Setting x = 0 and y = 9 we have 9 = 4 + c so c = 5 and y = 4e2x + 3x2 + 5.
40. Integrating y = 12x − 2 we obtain
y = (12x − 2)dx = 6x2 − 2x + c1 . Then, integrating y we obtain
y= (6x2 − 2x + c1 )dx = 2x3 − x2 + c1 x + c2 . At x = 1 the ycoordinate of the point of tangency is y = −1 + 5 = 4. This gives the initial condition y(1) = 4.
The slope of the tangent line at x = 1 is y (1) = −1. From the initial conditions we obtain
2 − 1 + c1 + c2 = 4 or c1 + c2 = 3 6 − 2 + c1 = −1 or c1 = −5. and Thus, c1 = −5 and c2 = 8, so y = 2x3 − x2 − 5x + 8.
41. When x = 0 and y = 1
2 , y = −1, so the only plausible solution curve is the one with negative slope at (0, 1 ),
2 or the black curve.
42. If the solution is tangent to the xaxis at (x0 , 0), then y = 0 when x = x0 and y = 0. Substituting these values
into y + 2y = 3x − 6 we get 0 + 0 = 3x0 − 6 or x0 = 2.
43. The theorem guarantees a unique (meaning single) solution through any point. Thus, there cannot be two
distinct solutions through any point.
44. When y = 1 4
16 x , y = 1 x3 = x( 1 x2 ) = xy 1/2 , and y(2) =
4
4
y= 1
16 (16) = 1. When 0, x<0 1 4
16 x , x≥0 we have
y =
and y(2) = 1
16 (16) 0, x<0 1 3
4x , x≥0 =x 0, x<0 1 2
4x , x≥0 = xy 1/2 , = 1. The two diﬀerent solutions are the same on the interval (0, ∞), which is all that is required by Theorem 1.1.
45. At t = 0, dP/dt = 0.15P (0) + 20 = 0.15(100) + 20 = 35. Thus, the population is increasing at a rate of 3,500
individuals per year. 12 1.3 Diﬀerential Equations as Mathematical Models
If the population is 500 at time t = T then
dP
dt = 0.15P (T ) + 20 = 0.15(500) + 20 = 95.
t=T Thus, at this time, the population is increasing at a rate of 9,500 individuals per year. EXERCISES 1.3
Differential Equations as Mathematical Models
dP
dP
= kP + r;
= kP − r
dt
dt
2. Let b be the rate of births and d the rate of deaths. Then b = k1 P and d = k2 P . Since dP/dt = b − d, the
diﬀerential equation is dP/dt = k1 P − k2 P . 1. 3. Let b be the rate of births and d the rate of deaths. Then b = k1 P and d = k2 P 2 . Since dP/dt = b − d, the
diﬀerential equation is dP/dt = k1 P − k2 P 2 .
dP
4.
= k1 P − k2 P 2 − h, h > 0
dt
5. From the graph in the text we estimate T0 = 180◦ and Tm = 75◦ . We observe that when T = 85, dT /dt ≈ −1.
From the diﬀerential equation we then have
k= dT /dt
−1
= −0.1.
=
T − Tm
85 − 75 6. By inspecting the graph in the text we take Tm to be Tm (t) = 80 − 30 cos πt/12. Then the temperature of the
body at time t is determined by the diﬀerential equation
dT
π
= k T − 80 − 30 cos t
dt
12 , t > 0. 7. The number of students with the ﬂu is x and the number not infected is 1000 − x, so dx/dt = kx(1000 − x).
8. By analogy, with the diﬀerential equation modeling the spread of a disease, we assume that the rate at which the
technological innovation is adopted is proportional to the number of people who have adopted the innovation
and also to the number of people, y(t), who have not yet adopted it. If one person who has adopted the
innovation is introduced into the population, then x + y = n + 1 and
dx
= kx(n + 1 − x),
dt x(0) = 1. 9. The rate at which salt is leaving the tank is
Rout (3 gal/min) · A
lb/gal
300 = A
lb/min.
100 Thus dA/dt = A/100. The initial amount is A(0) = 50.
10. The rate at which salt is entering the tank is
Rin = (3 gal/min) · (2 lb/gal) = 6 lb/min. 13 1.3 Diﬀerential Equations as Mathematical Models
Since the solution is pumped out at a slower rate, it is accumulating at the rate of (3 − 2)gal/min = 1 gal/min.
After t minutes there are 300 + t gallons of brine in the tank. The rate at which salt is leaving is
Rout = (2 gal/min) · A
lb/gal
300 + t = 2A
lb/min.
300 + t The diﬀerential equation is
dA
2A
=6−
.
dt
300 + t
11. The rate at which salt is entering the tank is
Rin = (3 gal/min) · (2 lb/gal) = 6 lb/min.
Since the tank loses liquid at the net rate of
3 gal/min − 3.5 gal/min = −0.5 gal/min,
after t minutes the number of gallons of brine in the tank is 300 − 1 t gallons. Thus the rate at which salt is
2
leaving is
A
3.5A
7A
Rout =
lb/gal · (3.5 gal/min) =
lb/min =
lb/min.
300 − t/2
300 − t/2
600 − t
The diﬀerential equation is
dA
7A
=6−
dt
600 − t dA
7
+
A = 6.
dt
600 − t or 12. The rate at which salt is entering the tank is
Rin = (cin lb/gal) · (rin gal/min) = cin rin lb/min.
Now let A(t) denote the number of pounds of salt and N (t) the number of gallons of brine in the tank at time
t. The concentration of salt in the tank as well as in the outﬂow is c(t) = x(t)/N (t). But the number of gallons
of brine in the tank remains steady, is increased, or is decreased depending on whether rin = rout , rin > rout ,
or rin < rout . In any case, the number of gallons of brine in the tank at time t is N (t) = N0 + (rin − rout )t.
The output rate of salt is then
A
A
lb/gal · (rout gal/min) = rout
lb/min.
N0 + (rin − rout )t
N0 + (rin − rout )t Rout = The diﬀerential equation for the amount of salt, dA/dt = Rin − Rout , is
dA
A
= cin rin − rout
dt
N0 + (rin − rout )t or dA
rout
+
A = cin rin .
dt
N0 + (rin − rout )t 13. The volume of water in the tank at time t is V = Aw h. The diﬀerential equation is then
dh
1 dV
1
−cAh
=
=
dt
Aw dt
Aw
Using Ah = π 2
12 2 = 2gh =− cAh
Aw 2gh . π
, Aw = 102 = 100, and g = 32, this becomes
36
dh
cπ √
cπ/36 √
64h = −
h.
=−
dt
100
450 14. The volume of water in the tank at time t is V = 1 πr2 h where r is the radius of the tank at height h. From
3
the ﬁgure in the text we see that r/h = 8/20 so that r = 2 h and V = 1 π
5
3
4
respect to t we have dV /dt = 25 πh2 dh/dt or
dh
25 dV
=
.
dt
4πh2 dt 14 2
2
5h h= 4
3
75 πh . Diﬀerentiating with 1.3 Diﬀerential Equations as Mathematical Models
√
From Problem 13 we have dV /dt = −cAh 2gh where c = 0.6, Ah = π
√
−2π h/15 and
√
2π h
dh
25
5
−
=
= − 3/2 .
dt
4πh2
15
6h 2 2
12 , and g = 32. Thus dV /dt = 15. Since i = dq/dt and L d2 q/dt2 + R dq/dt = E(t), we obtain L di/dt + Ri = E(t).
dq
1
16. By Kirchhoﬀ’s second law we obtain R + q = E(t).
dt
C
dv
17. From Newton’s second law we obtain m
= −kv 2 + mg.
dt
18. Since the barrel in Figure 1.35(b) in the text is submerged an additional y feet below its equilibrium position the number of cubic feet in the additional submerged portion is the volume of the circular cylinder:
π×(radius)2 ×height or π(s/2)2 y. Then we have from Archimedes’ principle
upward force of water on barrel = weight of water displaced
= (62.4) × (volume of water displaced)
= (62.4)π(s/2)2 y = 15.6πs2 y.
It then follows from Newton’s second law that
w d2 y
= −15.6πs2 y
g dt2 d2 y 15.6πs2 g
+
y = 0,
dt2
w or where g = 32 and w is the weight of the barrel in pounds.
19. The net force acting on the mass is
F = ma = m d2 x
= −k(s + x) + mg = −kx + mg − ks.
dt2 Since the condition of equilibrium is mg = ks, the diﬀerential equation is
m d2 x
= −kx.
dt2 20. From Problem 19, without a damping force, the diﬀerential equation is m d2 x/dt2 = −kx. With a damping
force proportional to velocity, the diﬀerential equation becomes
m d2 x
dx
= −kx − β
2
dt
dt or m d2 x
dx
+ kx = 0.
+β
2
dt
dt 21. Let x(t) denote the height of the top of the chain at time t with the positive direction upward. The weight of
the portion of chain oﬀ the ground is W = (x ft) · (1 lb/ft) = x. The mass of the chain is m = W/g = x/32.
The net force is F = 5 − W = 5 − x. By Newton’s second law,
d x
v =5−x
dt 32 or x dv
dx
+v
= 160 − 32x.
dt
dt Thus, the diﬀerential equation is
x d2 x
dx
+
dt2
dt 2 + 32x = 160. 22. The force is the weight of the chain, 2L, so by Newton’s second law,
of chain oﬀ the ground is m = 2(L − x)/g, we have
d 2(L − x)
v = 2L
dt
g (L − x) or 15 d
[mv] = 2L. Since the mass of the portion
dt dv
dx
+v −
= Lg.
dt
dt 1.3 Diﬀerential Equations as Mathematical Models Thus, the diﬀerential equation is
(L − x) d2 x
dx
−
2
dt
dt 2 = Lg. 23. From g = k/R2 we ﬁnd k = gR2 . Using a = d2 r/dt2 and the fact that the positive direction is upward we get
d2 r
k
gR2
= −a = − 2 = − 2
dt2
r
r or d2 r gR2
+ 2 = 0.
dt2
r 24. The gravitational force on m is F = −kMr m/r2 . Since Mr = 4πδr3 /3 and M = 4πδR3 /3 we have Mr = r3 M/R3
and
F = −k Mr m
r3 M m/R3
mM
= −k
= −k 3 r.
2
r
r2
R Now from F = ma = d2 r/dt2 we have
m d2 r
mM
= −k 3 r
2
dt
R d2 r
kM
= − 3 r.
2
dt
R or dA
= k(M − A).
dt
dA
26. The diﬀerential equation is
= k1 (M − A) − k2 A.
dt
27. The diﬀerential equation is x (t) = r − kx(t) where k > 0.
25. The diﬀerential equation is 28. By the Pythagorean Theorem the slope of the tangent line is y = −y
s2 − y 2 . 29. We see from the ﬁgure that 2θ + α = π. Thus y y
2 tan θ
.
= tan α = tan(π − 2θ) = − tan 2θ = −
−x
1 − tan2 θ
Since the slope of the tangent line is y = tan θ we have y/x = 2y [1 − (y )2 ] or
y − y(y )2 = 2xy , which is the quadratic equation y(y )2 + 2xy − y = 0 in y .
Using the quadratic formula, we get
y = −2x ± 4x2 + 4y 2
−x ± x2 + y 2
=
.
2y
y (x,y) θ
θα
θ Since dy/dx > 0, the diﬀerential equation is
dy
−x + x2 + y 2
=
dx
y or y dy
−
dx x α y
φ
x x2 + y 2 + x = 0. 30. The diﬀerential equation is dP/dt = kP , so from Problem 37 in Exercises 1.1, P = ekt , and a oneparameter
family of solutions is P = cekt .
31. The diﬀerential equation in (3) is dT /dt = k(T − Tm ). When the body is cooling, T > Tm , so T − Tm > 0.
Since T is decreasing, dT /dt < 0 and k < 0. When the body is warming, T < Tm , so T − Tm < 0. Since T is
increasing, dT /dt > 0 and k < 0.
32. The diﬀerential equation in (8) is dA/dt = 6 − A/100. If A(t) attains a maximum, then dA/dt = 0 at this time
and A = 600. If A(t) continues to increase without reaching a maximum, then A (t) > 0 for t > 0 and A cannot
exceed 600. In this case, if A (t) approaches 0 as t increases to inﬁnity, we see that A(t) approaches 600 as t
increases to inﬁnity.
33. This diﬀerential equation could describe a population that undergoes periodic ﬂuctuations. 16 1.3 Diﬀerential Equations as Mathematical Models
34. (a) As shown in Figure 1.43(b) in the text, the resultant of the reaction force of magnitude F and the weight
of magnitude mg of the particle is the centripetal force of magnitude mω 2 x. The centripetal force points
to the center of the circle of radius x on which the particle rotates about the yaxis. Comparing parts of
similar triangles gives
F cos θ = mg and F sin θ = mω 2 x. (b) Using the equations in part (a) we ﬁnd
tan θ = F sin θ
mω 2 x
ω2 x
=
=
F cos θ
mg
g or dy
ω2 x
=
.
dx
g 35. From Problem 23, d2 r/dt2 = −gR2 /r2 . Since R is a constant, if r = R + s, then d2 r/dt2 = d2 s/dt2 and, using
a Taylor series, we get
d2 s
R2
2gs
= −g
= −gR2 (R + s)−2 ≈ −gR2 [R−2 − 2sR−3 + · · · ] = −g + 3 + · · · .
2
dt
(R + s)2
R
Thus, for R much larger than s, the diﬀerential equation is approximated by d2 s/dt2 = −g.
36. (a) If ρ is the mass density of the raindrop, then m = ρV and
dm
dr
dV
d 4 3
=ρ
=ρ
πr = ρ 4πr2
dt
dt
dt 3
dt = ρS dr
.
dt If dr/dt is a constant, then dm/dt = kS where ρ dr/dt = k or dr/dt = k/ρ. Since the radius is decreasing,
k < 0. Solving dr/dt = k/ρ we get r = (k/ρ)t + c0 . Since r(0) = r0 , c0 = r0 and r = kt/ρ + r0 .
d
(b) From Newton’s second law, [mv] = mg, where v is the velocity of the raindrop. Then
dt
m dv
dm
+v
= mg
dt
dt or ρ 4 3 dv
4
πr
+ v(k4πr2 ) = ρ πr3 g.
3
dt
3 Dividing by 4ρπr3 /3 we get
dv 3k
+
v=g
dt
ρr dv
3k/ρ
v = g, k < 0.
+
dt
kt/ρ + r0 or 37. We assume that the plow clears snow at a constant rate of k cubic miles per hour. Let t be the time in hours
after noon, x(t) the depth in miles of the snow at time t, and y(t) the distance the plow has moved in t hours.
Then dy/dt is the velocity of the plow and the assumption gives
wx dy
= k,
dt where w is the width of the plow. Each side of this equation simply represents the volume of snow plowed in
one hour. Now let t0 be the number of hours before noon when it started snowing and let s be the constant rate
in miles per hour at which x increases. Then for t > −t0 , x = s(t + t0 ). The diﬀerential equation then becomes
dy
k
1
.
=
dt
ws t + t0
Integrating, we obtain
k
[ ln(t + t0 ) + c ]
ws
where c is a constant. Now when t = 0, y = 0 so c = − ln t0 and
y= y= k
t
ln 1 +
ws
t0 17 . 1.3 Diﬀerential Equations as Mathematical Models Finally, from the fact that when t = 1, y = 2 and when t = 2, y = 3, we obtain
1+ 2
t0 2 = 1+ 1
t0 3 . Expanding and simplifying gives t2 + t0 − 1 = 0. Since t0 > 0, we ﬁnd t0 ≈ 0.618 hours ≈ 37 minutes. Thus it
0
started snowing at about 11:23 in the morning.
dP
dA
38. (1):
= kP is linear
(2):
= kA is linear
dt
dt
dx
dT
(5):
(3):
= k(T − Tm ) is linear
= kx(n + 1 − x) is nonlinear
dt
dt
dX
dA
A
(6):
= k(α − X)(β − X) is nonlinear
(8):
=6−
is linear
dt
dt
100
dh
d2 q
dq
Ah
1
(10):
2gh is nonlinear
(11): L 2 + R + q = E(t) is linear
=−
dt
Aw
dt
dt
C
d2 s
dv
(12):
= −g is linear
(14): m
= mg − kv is linear
2
dt
dt
d2 s
ds
d2 x 64
(15): m 2 + k
− x = 0 is linear
= mg is linear
(16):
dt
dt
dt2
L
(17): linearity or nonlinearity is determined by the manner in which W and T1 involve x.
39. At time t, when the population is 2 million cells, the diﬀerential equation P (t) = 0.15P (t) gives the rate of
increase at time t. Thus, when P (t) = 2 (million cells), the rate of increase is P (t) = 0.15(2) = 0.3 million cells
per hour or 300,000 cells per hour.
40. Setting A (t) = −0.002 and solving A (t) = −0.0004332A(t) for A(t), we obtain
A(t) = A (t)
−0.002
=
≈ 4.6 grams.
−0.0004332
−0.0004332 CHAPTER 1 REVIEW EXERCISES d
dy
c1 ekx = c1 kekx ;
= ky
dx
dx
d
dy
dy
2.
(5 + c1 e−2x ) = −2c1 e−2x = −2(5 + c1 e−2x − 5);
= −2(y − 5) or
= −2y + 10
dx
dx
dx
d
3.
(c1 cos kx + c2 sin kx) = −kc1 sin kx + kc2 cos kx;
dx
d2
(c1 cos kx + c2 sin kx) = −k 2 c1 cos kx − k 2 c2 sin kx = −k 2 (c1 cos kx + c2 sin kx);
dx2
d2 y
d2 y
= −k 2 y or
+ k2 y = 0
dx2
dx2
d
(c1 cosh kx + c2 sinh kx) = kc1 sinh kx + kc2 cosh kx;
4.
dx
d2
(c1 cosh kx + c2 sinh kx) = k 2 c1 cosh kx + k 2 c2 sinh kx = k 2 (c1 cosh kx + c2 sinh kx);
dx2 1. 18 CHAPTER 1 REVIEW EXERCISES
d2 y
= k2 y
dx2 d2 y
− k2 y = 0
dx2 or 5. y = c1 ex + c2 xex ; y = c1 ex + c2 xex + c2 ex ; y = c1 ex + c2 xex + 2c2 ex ; y + y = 2(c1 ex + c2 xex ) + 2c2 ex = 2(c1 ex + c2 xex + c2 ex ) = 2y ; y − 2y + y = 0 6. y = −c1 ex sin x + c1 ex cos x + c2 ex cos x + c2 ex sin x;
y = −c1 ex cos x − c1 ex sin x − c1 ex sin x + c1 ex cos x − c2 ex sin x + c2 ex cos x + c2 ex cos x + c2 ex sin x
= −2c1 ex sin x + 2c2 ex cos x;
y − 2y = −2c1 ex cos x − 2c2 ex sin x = −2y;
7. a,d 8. c y − 2y + 2y = 0
10. a,c 9. b 11. b 12. a,b,d 13. A few solutions are y = 0, y = c, and y = ex .
14. Easy solutions to see are y = 0 and y = 3.
15. The slope of the tangent line at (x, y) is y , so the diﬀerential equation is y = x2 + y 2 .
16. The rate at which the slope changes is dy /dx = y , so the diﬀerential equation is y = −y or y + y = 0.
17. (a) The domain is all real numbers.
(b) Since y = 2/3x1/3 , the solution y = x2/3 is undeﬁned at x = 0. This function is a solution of the diﬀerential
equation on (−∞, 0) and also on (0, ∞).
18. (a) Diﬀerentiating y 2 − 2y = x2 − x + c we obtain 2yy − 2y = 2x − 1 or (2y − 2)y = 2x − 1.
(b) Setting x = 0 and y = 1 in the solution we have 1 − 2 = 0 − 0 + c or c = −1. Thus, a solution of the
initialvalue problem is y 2 − 2y = x2 − x − 1.
(c) Solving y 2 − 2y − (x2 − x − 1) = 0 by the quadratic formula we get y = (2 ± 4 + 4(x2 − x − 1) )/2
√
= 1± x2 − x = 1± x(x − 1) . Since x(x−1) ≥ 0 for x ≤ 0 or x ≥ 1, we see that neither y = 1+ x(x − 1)
nor y = 1 − x(x − 1) is diﬀerentiable at x = 0. Thus, both functions are solutions of the diﬀerential
equation, but neither is a solution of the initialvalue problem.
19. Setting x = x0 and y = 1 in y = −2/x + x, we get
1=− 2
+ x0
x0 or x2 − x0 − 2 = (x0 − 2)(x0 + 1) = 0.
0 Thus, x0 = 2 or x0 = −1. Since x = 0 in y = −2/x+x, we see that y = −2/x+x is a solution of the initialvalue
problem xy + y = 2x, y(−1) = 1, on the interval (−∞, 0) and y = −2/x + x is a solution of the initialvalue
problem xy + y = 2x, y(2) = 1, on the interval (0, ∞).
20. From the diﬀerential equation, y (1) = 12 + [y(1)]2 = 1 + (−1)2 = 2 > 0, so y(x) is increasing in some
neighborhood of x = 1. From y = 2x + 2yy we have y (1) = 2(1) + 2(−1)(2) = −2 < 0, so y(x) is concave
down in some neighborhood of x = 1.
21. (a) y
3 y
3 2 2 1 1 3 2 1
1 1 2 3 x 3 2 1
1 2 2 2 3 1 3 y = x2 + c1 y = −x2 + c2 19 3 x CHAPTER 1 REVIEW EXERCISES
(b) When y = x2 + c1 , y = 2x and (y )2 = 4x2 . When y = −x2 + c2 , y = −2x and (y )2 = 4x2 .
(c) Pasting together x2 , x ≥ 0, and −x2 , x ≤ 0, we get y =
22. The slope of the tangent line is y (−1,4) √
= 6 4 + 5(−1)3 = 7. −x2 , x ≤ 0
x2 ,
x > 0. 23. Diﬀerentiating y = x sin x + x cos x we get
y = x cos x + sin x − x sin x + cos x
and
y = −x sin x + cos x + cos x − x cos x − sin x − sin x
= −x sin x − x cos x + 2 cos x − 2 sin x.
Thus
y + y = −x sin x − x cos x + 2 cos x − 2 sin x + x sin x + x cos x = 2 cos x − 2 sin x.
An interval of deﬁnition for the solution is (−∞, ∞).
24. Diﬀerentiating y = x sin x + (cos x) ln(cos x) we get
− sin x
cos x y = x cos x + sin x + cos x − (sin x) ln(cos x) = x cos x + sin x − sin x − (sin x) ln(cos x)
= x cos x − (sin x) ln(cos x)
and
y = −x sin x + cos x − sin x − sin x
cos x − (cos x) ln(cos x) sin2 x
− (cos x) ln(cos x)
cos x
1 − cos2 x
= −x sin x + cos x +
− (cos x) ln(cos x)
cos x
= −x sin x + cos x + sec x − cos x − (cos x) ln(cos x)
= −x sin x + cos x + = −x sin x + sec x − (cos x) ln(cos x).
Thus
y + y = −x sin x + sec x − (cos x) ln(cos x) + x sin x + (cos x) ln(cos x) = sec x.
To obtain an interval of deﬁnition we note that the domain of ln x is (0, ∞), so we must have cos x > 0. Thus,
an interval of deﬁnition is (−π/2, π/2).
25. Diﬀerentiating y = sin(ln x) we obtain y = cos(ln x)/x and y = −[sin(ln x) + cos(ln x)]/x2 . Then
x2 y + xy + y = x2 − sin(ln x) + cos(ln x)
x2 +x cos(ln x)
+ sin(ln x) = 0.
x An interval of deﬁnition for the solution is (0, ∞).
26. Diﬀerentiating y = cos(ln x) ln(cos(ln x)) + (ln x) sin(ln x) we obtain
y = cos(ln x)
=− 1
cos(ln x) − sin(ln x)
x + ln(cos(ln x)) − ln(cos(ln x)) sin(ln x) (ln x) cos(ln x)
+
x
x 20 sin(ln x)
x + ln x cos(ln x) sin(ln x)
+
x
x CHAPTER 1 REVIEW EXERCISES and
y = −x ln(cos(ln x)) 1
sin(ln x)
cos(ln x)
+ sin(ln x)
−
x
cos(ln x)
x + ln(cos(ln x)) sin(ln x)
= 1
x2 1
sin(ln x)
1
cos(ln x) 1
+ x (ln x) −
− (ln x) cos(ln x) 2
+
x2
x
x
x2
x sin2 (ln x)
1
− ln(cos(ln x)) cos(ln x) +
+ ln(cos(ln x)) sin(ln x)
2
x
cos(ln x)
− (ln x) sin(ln x) + cos(ln x) − (ln x) cos(ln x) . Then
x2 y + xy + y = − ln(cos(ln x)) cos(ln x) + sin2 (ln x)
+ ln(cos(ln x)) sin(ln x) − (ln x) sin(ln x)
cos(ln x) + cos(ln x) − (ln x) cos(ln x) − ln(cos(ln x)) sin(ln x)
+ (ln x) cos(ln x) + cos(ln x) ln(cos(ln x)) + (ln x) sin(ln x)
2 = sin (ln x)
sin2 (ln x) + cos2 (ln x)
1
+ cos(ln x) =
=
= sec(ln x).
cos(ln x)
cos(ln x)
cos(ln x) To obtain an interval of deﬁnition, we note that the domain of ln x is (0, ∞), so we must have cos(ln x) > 0.
Since cos x > 0 when −π/2 < x < π/2, we require −π/2 < ln x < π/2. Since ex is an increasing function, this is
equivalent to e−π/2 < x < eπ/2 . Thus, an interval of deﬁnition is (e−π/2 , eπ/2 ). (Much of this problem is more
easily done using a computer algebra system such as Mathematica or Maple.)
27. From the graph we see that estimates for y0 and y1 are y0 = −3 and y1 = 0.
28. The diﬀerential equation is
dh
cA0
=−
dt
Aw 2gh . Using A0 = π(1/24)2 = π/576, Aw = π(2)2 = 4π, and g = 32, this becomes
dh
c √
cπ/576 √
64h =
h.
=−
dt
4π
288 21 ...
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 Spring '08
 Storm
 Differential Equations, Equations, Derivative, Quadratic equation, Elementary algebra

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