1
Solutions Chapter to 1.1
1. d2y/dt2 = g dy/dt = gt + c1 y(t) = gt2/2 + c1t + c2. Now impose the initial conditions. y(0) = 0 c2 = 0. dy/dt(0) = 0 c1 = 0. Hence the solution to the initialvalue problem is: y(t) = gt2/2. The object hits the ground at the time, t0, when y(t0) = 100. Hence 100 = gt02/2, so that t0 = (200/g)1/2 4.52 s, where we have taken g = 9.8 ms2. 3.
If y(t) denotes the displacement of the object from its initial position at time t, the motion of the object can be described by the initialvalue problem d2y/dt2 = g, y(0) = 0, dy/dt(0) = v0.
We first integrate the differential equation: d2y/dt2 = g dy/dt = gt + c1 y(t) = gt2/2 + c1t + c2. Now impose the initial conditions. y(0) = 0 c2 = 0. dy/dt(0) = v0 c1 = v0. Hence the solution to the initialvalue problem is y(t) = gt2/2 + v0t. We are given that y(t0) = h. Hence h = gt02 + v0t0. Solving for v0 yields v0 = (2h gt02)/(2t0). 5. y(t) = c1cos t + c2sin t dy/dt = c1sin t + c2cos t d2y/dt2 = c1 2cos t c2 2sin t = 2(c1cos t + c2sin t) = - 2y. Consequently, d2y/dt2 + 2y = 0. To determine the amplitude of the motion we write the solution to the differential equation in the equivalent form c1 c2 y(t) = c12 + c22 cos t + sin t . c12 + c22 c12 + c22 We can now define an angle by c1 c2 cos = , sin = . c12 + c22 c12 + c22
Then the expression for the solution to the differential equation is
7.
c12 + c2 2 (cos t cos + sin t sin ) = c22 cos(t ). c12 + Consequently the motion corresponds to an oscillation with amplitude A = 2. c12 + c2 2 2
y(t) = Given family of curves satisfies: x + 4y = c1 2x + 8y dy/dx = 0 dy/dx = x/(4y) . Orthogonal trajectories satisfy: y1dy/dx = 4/x d(ln |y|)/dx = 4/x ln |y| = 4ln |x| + c2 y = kx4, where k = ec 2.
1
2
Ch. 1
First-Order Differential Equations
Given family of curves satisfies: y = cx2 c = yx2. Hence dy/dx = 2cx = 2(yx2)x = 2y/x. Orthogonal trajectories satisfy: dy/dx = x/(2y) 2ydy/dx = x d(y2)/dx = x x2 + 2y2 = k. 9.
11.
Given family of curves satisfies: y = cex dy/dx = cex = y. Orthogonal trajectories satisfy: dy/dx = 1/y ydy/dx = 1 y2/2 = x + c1 y2 = 2x + c2.
13.
y = cxm dy/dx = cmxm1 , but c = y/xm so dy/dx = my/x. Orthogonal trajectories 1 satisfy: dy/dx = x/(my) ydy/dx = x/m d(y2/2)/dx = x/m y2 = m x2 + c1. y2 = mx + c dy/dx = m/(2y) . Orthogonal trajectories satisfy: dy/dx = 2y/m
15.
Sec. 1.1
How Differential Equations Arise
3
y1dy/dx = 2/m ln|y| = 2x/m + c1 y = c2e2x/m. 17. tan(a2) tan(a) m 2 tan(a) m 1 = tan(a1) = tan(a2 a) = 1 + tan(a )tan(a) = 1 + m tan(a) . 2 2
Solutions to 1.2
1. 3. 5. 7. 2, nonlinear. 2, nonlinear. 4, linear. y(x) = (x + 4) 1 y = (x + 4) 2 so y = y 2. Therefore, y(x) = (x + 4)1 is a solution of y = y2 when x is in (, 4) or (4, ). y(x) = c1exsin 2x, y = 2c1excos 2x c1exsin 2x, and y = 3c1exsin 2x 4c1excos 2x y + 2y + 5y = 0. This solution is valid for all x on (, ). 11. y(x) = c1x3 + c2x1, y = 3c1x4 c2x2, and y = 12c1x5 + 2c2x3 x2y + 5xy + 3y = 0. Thus y(x) = c 1x3 + c2x1 is a solution which is valid for all x in (, 0) or (0, ). 13. y(x) = c1x2 + c2x3 x2sin x, y = 2c1x + 3c2x2 x2cos x 2xsin x, and y = 2c1 + 6c2x + x2sin x 2xcos x 2xcos x 2sin x. Substituting these results into the given differential equation yields x2y 4xy + 6y = x2(2c1 + 6c2x + x2sin x 4xcos x 2sin x) 4x(2c1x + 3c2x2 x2cos x 2xsin x) + 6(c1x2 + c2x3 x2sin ) = 2c1x2 + 6c2x3 + x4sin x 4x3cos x 2x2sin x 8c1x2 12c2x3 + 4x3cos x + 8x2sin x + 6c1x2 + 6c2x3 6x2sin x = x4 sin x. Hence, y(x) = c1x2 + c2x3 x2sin x is a solution of the differential equation valid for all x on (, ).
9.
4
Ch. 1
First-Order Differential Equations
15.
y(x) = eax (c1 + c2x), y = eax (c2) + aeax (c1 + c2x) = eax (c2 + ac1 + ac2x), and y = eax (ac2) + aeax (c2 + ac1 + ac2x) = aeax (2c2 + ac1 + ac2x). Substituting these results into the given differential equation yields y 2ay + a2y = aeax (2c2 + ac1 + ac2x) 2aeax (c2 + ac1 + ac2x) + a2eax (c1 + c2x) = aeax (2c2 + ac1 + ac2x 2c2 2ac1 2ac2x + ac1 + ac2x) = 0. Thus, y = e (c1 + c2x) is a solution to the given differential equation for all x in (, ).
ax
17.
y(x) = erx y(x) = rerx y(x) = r2erx. Substitution into the given differential equation yields erx (r2 + 2r 3) = 0, so that r must satisfy r2 + 2r 3 = 0, or equivalently (r + 3)(r 1) = 0. Consequently r = 3 and r = 1 are the only values of r for which y(x) = erx is a solution to the given DE. The corresponding solutions are y(x) = e3x and y(x) = ex. y(x) = xr y(x) = rxr1 y(x) = r(r 1)xr2. Substitution into the given differential equation yields xr [r(r 1) + r 1] = 0, so that r must satisfy r2 1 = 0. Consequently r = 1 and r = 1 are the only values of r for which y(x) = xr is a solution to the given DE. The corresponding solutions are y(x) = x1 and y(x) = x. 1 1 1 y(x) = 2 x(5x2 3) = 2 (5x3 3x) y(x) = 2 (15x2 3) y(x) = 15x. Substitution into the Legendre equation with N = 3 yields (1 x2)y 2xy+ 12y = (1 x 2)(15x) x(15x2 3) + 6x(5x2 3) = 0. Consequently the given function is a solution to the Legendre equation with N = 3. dy dy ex sin y xsin y ex = c xcos y dx + sin y ex = 0 dx = x cos y . dy dy dy 1 yexy exy x = c exy[x dx + y] 1 = 0 xexy dx + yexy = 1 dx = . xexy ln x y(1) = 0 e0(1) 1 = c c = 0. Therefore, exy x = 0, so that y = x . dy dy cos x 2xy2 x2y2 sin x = c 2x2y dx + 2xy2 cos x = 0 dx = . 2x2y 1 1 Since y() = , In x2y2 sin x = c 2( )2 sin() = c c = 1. Hence,
19.
21.
23.
25.
27.
Sec. 1.2
Basic Ideas and Terminology
5
x2y2 sin x = 1 so that y2 = so y(x) =
1 + sin x 1 . Since y() = , take the branch of y where x > 0 2 x
1 + sin x .
x
29.
dy 1/2 y(x) = 2x1/2 + c for all x > 0. dx = x d2y n 2 = x , where n is an integer. dx dy xn+1 xn+2 If n 1 and n 2 then dx = n + 1 + c1 y(x) = (n + 1)(n + 2) + c1x + c2. dy If n = 1 then dx = ln|x| + c1 y(x) = xln|x| + c1x + c2. dy If n = 2 then dx = c1 x1 y(x) = c1x + c2 ln|x|. The solution is valid on (, ) if n 0 and on (, 0) or (0, ) if n < 0. d2y dy 2 = cos x dx = sin x + c1. dx y(0) = 1 1 = c1 so y(x) = cos x + x + c 2. y(0) = 2 2 = 1 + 0 + c 2 c2 = 3 so y(x) = 3 + x cos x. d2y dy x dx = ex + c1 y(x) = ex + c1x + c2. y(0) = 1 c2 = 0, and 2 = e dx 1 1 y(1) = 0 c1 = e . Hence, y(x) = ex e x. a) y(x) y(0) y() y(x) y(0) y() = = = = = = c1cos x + c2sin x 0 0 = c1(1) + c2(0) c1 = 0 1 1 = c2(0), which is impossible. No solutions. c1cos x + c2sin x 0 0 = c1(1) + c2(0) c1 = 0 0 0 = c2(0), so c2 can be anything. Infinitely many solutions.
31.
33.
35.
37.
b)
38 43: Use some form of technology to define each of the given functions. Then use the technology to simplify the expression given on the left-hand side of each DE and verify that the result corresponds to the expression on the right-hand side.
6
Ch. 1
First-Order Differential Equations
45.
(a) J0(x) =
(1) (x/2) k=0 (k!)
k 2
2k
1 1 = 1 4 x2 + 64 x4 + ...
(b) A Maple plot of J(0, x, 4) is given below.
From this graph, an approximation to the first positive zero of J0(x) is 2.4. Using the Maple internal function BesselJZeros gives the approximation 2.404825558. (c) A Maple plot of the functions J0(x) and J(0, x, 4) on the interval [0, 2] is given below. We see that to the printer resolution, these graphs are indistinguishable. On a larger interval, for example, [0, 3], the two graphs would begin to differ dramatically from one another.
(d) By trial and error, we find the smallest value of m to be m = 11. A plot of the functions J(0, x), and J(0, x, 11) is given below.
Sec. 1.2
Basic Ideas and Terminology
7
Solutions to 1.3
1. dy y y = cx1 c = xy. Hence, dx = cx2 = (xy)x2 = x. x2 + y2 dy x2 + y2 = c. Hence, 2x + 2ydx = 2c = 2x x , dy x2 + y2 dy y2 x2 so that y dx = 2x x. Consequently, dx = 2xy . x2 + y2 = 2cx 2cy = x2 c2 c2 + 2cy x2 = 0 2y 4y2 + 4x2 = y x2 + y2. c = 2 x2 + y2 x2 2cx + y2 2cy = 0 c = 2(x + y) . Differentiating the given equation yields dy x2 + y2 x2 + y2 dy 2(x c) + 2(y c) dx = 0, so that 2(x 2(x + y) ) + 2(y 2(x + y) ) dx = 0, dy x2 + 2xy y2 that is dx = 2 . y + 2xy x2 (x c)2 + (y c)2 = 2c2 dy y 3y y = cx3 dx = 3cx2 = 3( 3)x2 = x . The initial condition y(2) = 8 x 3 8 = c(2) c = 1. Thus the unique solution to the initial value problem is y = x3.
3.
5.
7.
9.
8
Ch. 1
First-Order Differential Equations
11.
(x c)2 + y2 = c2 x2 2cx + c2 + y2 = c2, so that x2 2cx + y2 = 0. (11.1) Differentiating with respect to x yields dy 2x 2c + 2y dx = 0. (11.2) x2 + y2 But, from (11.1), c = 2x which, when substituted into (11.2), yields x2 + y2 dy dy y2 x2 2x x + 2y dx = 0, that is, dx = 2xy . Imposing the initial condition y(2) = 2 implies, from (11.1), that c = 2, so that the unique solution to the initial value problem is y = + . x(4 x)
13.
dy x = 2 (y2 9), y(0) = 3. dx x +1 x f(x, y) = 2 (y2 9), which is continuous for all x, y R. x +1 f 2xy y = x2 + 1, which is continuous for all x, y R. So the IVP stated above has a unique solution on any interval containing (0, 3). By inspection we see that y(x) = 3 is the unique solution.
Sec. 1.3
The Geometry of First-Order DE
9
15.
(a) f(x, y) = 2xy2 f/y = 4xy. Both of these functions are continuous for all (x, y), and therefore the hypotheses of the existence and uniqueness theorem are satisfied for any (x0, y0). 1 2x (b) y(x) = 2 y = 2 = 2xy2. x +c (x + c)2 1 (c) y(x) = 2 . x +c 1 (i) y(0) = 1 1 = 1/c c = 1. Hence y(x) = 2 . The solution is valid on the x +1 interval (, ).
(ii) y(1) = 1 1 = 1/(1 + c) c = 0. Hence y(x) = 1/x2. This solution is valid on the interval (0, ).
(iii) y(0) = 1 1 = 1/c c = 1. Hence y(x) = valid on the interval (1, 1).
1 . This solution is x 1
2
10
Ch. 1
First-Order Differential Equations
(d) Since, by inspection, y(x) = 0 satisfies the given IVP, it must be the unique solution to the IVP.
17.
y = 4x. There are no equilibrium solutions. The slope of the solution curves is positive for x > 0 is negative for x < 0. The isoclines are the lines x = k/4 Slope of Solution Curve 4 2 0 2 4 Equation of Isocline x = 1 x = 1/2 x =0 x = 1/2 x=1
19.
y = x + y. There are no equilibrium solutions. The slope of the solution curves is positive for y > x, and negative for y < x. The isoclines are the lines with equation y + x = k. Slope of Solution Curve 2 1 Equation of Isocline y = x 2 y = x 1
Sec. 1.3
The Geometry of First-Order DE
11
0 1 2
y = x y = x + 1 y = x + 2
Since the slope of the solution curve along the isocline y = x 1 coincides with the slope of the isocline, it follows that y = x 1 is a solution to the DE. Differentiating the given DE yields: y = 1 + y = 1 + x + y. Hence the solution curves are concave up for y > x 1, and concave down for y < x 1. Putting this information together leads to the slope field shown in the figure.
21.
y = 4x/y. Slope is zero when x = 0 (y 0). The solution curves have a vertical tangent line at all points along the x-axis (except the origin). The isoclines are the lines with equation 4x/y = k y = 4x/k. Some values are given in the table below. Slope of Solution Curve 1 2 3 Equation of Isocline y = 4x y = 2x y = 4x/3
Differentiating the given DE yields y = 4/y + 4xy/y2 = 4/y 16x2/y3 = 4(y2 + 4x2)/y. Consequently the solution curves are concave up for y < 0, and concave down for y > 0. Putting this information together leads to the slope field shown in the figure.
12
Ch. 1
First-Order Differential Equations
23.
y = x2cos y. Slope is zero when x = 0. There are equilibrium solutions when y = (2k + 1)/2. The slope field is best sketched using technology. The picture below gives the slope field for /2 < y < 3/2.
25.
dT 1 dt = 80 (T 70). Equilibrium solution: T(t) = 70. Slope of solution curves is positive for T d2T 1 dT 1 > 70, and negative for T < 70. 2 = 80 dt = 6400 (T 70). Hence the solution curves are dt concave up for T > 70, and concave down for T < 70. The isoclines are the horizontal lines 1 80 (T 70) = k T = 70 80k
Slope of Solution Curve 1/4 1/5 0 1/5 1/4
Equation of Isocline T = 90 T = 86 T = 70 T = 54 T = 50
Sec. 1.3
The Geometry of First-Order DE
13
27.
y =
x sin x . 1 + y2
29.
y = 2x2sin y.
31.
y =
1 y2 . 2 + 0.5x2
14
Ch. 1
First-Order Differential Equations
33.
dy y (a) Differentiating the given equation gives dx = 2kx = 2 x. Hence the DE of the orthogonal dy x trajectories is dx = 2y. (b)
The orthogonal trajectories appear to be ellipses. This can be verified by integrating the DE derived in (a).
Solutions to 1.4
1. dy 1 dx = 2xy y dy = 2xdx
dy = 2 xdx ln|y| = x + c y
2
1
y(x) = cex .
2
3.
ex+ydy dx = 0 eydy = exdx y(x) = ln(c ex).
e dy e
y
x
dx = 0 ey + ex = c
Sec. 1.4
Separable DE
15
5.
ydx (x 2)dy = 0 y(x) = c(x 2).
xdx2 = dy ln|x 2| ln|y| = c y
1
7.
dy dy y x dx = 3 2x2 dx x(2x 1)dy = (3 y)dx ln|y 3| =
2dx dx + 2x 1 ln|y 3| = ln|x| + ln|2x 1| + c x
dx ydy3 = x(2x 1)
1
x cx 3 (y 3)(2x 1) = c2 y(x) = 2x 1 . dy x(y2 1) dy 1 xdx dx = 2(x 2)(x 1) (y + 1)(y 1) = 2 (x 2)(x 1) , y 1. 1 dy 1 dy 1 dx dx Thus, 2 y + 1 + 2 y 1 = 2 2 x 2 x 1 ln|y + 1| + ln|y 1| = 2ln|x 2| ln|x 1| + c1 y1 (x 2)2 (x 1) + c(x 2)2 y + 1 = c (x 1) y(x) = . (x 1) c(x 2)2 By inspection we see that y(x) = +1, and y(x) = 1 are also solutions of the given differential equation. The former is included in the above solution when c = 0.
9.
11.
dy (x a)(x b)dx (y c) = 0
1/(ab) 1 1 dx ln|y c| = ln c x a x a x b 1 x b 1/(ab) 1/(ab) x b = c yc = cxa (y c) x a 1 2 x b x a 1/(ab) y(x) = c + c2 x b .
dx ydy c = (x a)(x b) .
dy 1 = ab yc
13.
dy (1 x2)dx + xy = ax
adyy = 1 2xdx ln|a y| = 1 ln|1 x2| + c 2 1x 2
2 2
1
y(x) = a + c 2 , but y(0) = 2a so c = a and therefore, y(x) = a(1 + 2 ). 1 x 1 x 15. dy 3 dx = y sin x 1 dy = sin x dx for y 0. Thus 2y y
3
= cos x + c. However, we
cannot impose the initial condition y(0) = 0 on this last equation since it is not defined at y = 0. But, by inspection, y(x) = 0 is a solution to the given differential equation and, further, y(0) = 0; thus, the unique solution to the initial value problem is y(x) = 0.
16
Ch. 1
First-Order Differential Equations
17.
a)
dv dv m dt = mg kv 2 = dt 1 (k/mg) v2
dv 1 (k/mg)v
2 dv
=
dt
1 g (mg/k)1/2 tanh1[(k/mg)1/2 v] + c = t but since v(0) = 0, c = 0 so if we let a gt a = (k/mg)1/2 then we can write g tanh1(v/a) = t v(t) = atanh( a ). b) No. As t , v a and as t 0+, v 0. gt dx gt gt c) v(t) = atanh( a ) dt = atanh( a ) x(t) = a tanh( a )dt a2 gt a2 gt x(t) = g ln(cosh( a )) + c1 and if x(0) = 0 then x(t) = g ln cosh( a ) . 19. 1 dv = dt. Integrating we obtain 1 + v2 tan1(v) = t + c. The initial condition v(0) = v0 implies that c = tan1(v0), so that tan1(v) = t + tan1(v0). The object will come to rest if there is a time tr at which the velocity is zero. To determine tr we set v = 0 in the previous equation which yields tan1(0) = tr + tan1(v0). Consequently, tr = tan1(v0). The object does not remain at rest dv since we see from the given differential equation that dt < 0 at t = tr , and so v is decreasing with time. Consequently v passes through zero and becomes negative for t < tr . dv dx dv dv b) From the chain rule we have dt = dt dx = v dx. Substituting this result into the DE dv v (19.1) yields v dx = (1 + v 2). We now separate the variables: dv = dx. 1 + v2 Integrating we obtain ln(1 + v2) = 2x + c. v(0) = v 0, x(0) = 0 implies that a) Separating the variables in the given DE yields c = ln(1 + v2), so that ln(1 + v2) = 2x + ln(1 + v 2). When the object comes to rest (v = 0 0 1 0) the distance traveled by the object is x = 2 ln(1 + v2). 0 21. Solving p = p0(/ 0) for yields = 0(p/p0)1/ . Consequently the given DE can be written as dp = g 0(p/p0)1/ dy, or equivalently, p1/ dp = g 0/p01/ dy. This can be integrated directly to obtain p( 1)/ /( 1) = g 0y/p01/ + c. At the surface of the earth (y = 0) we have p = p0. Imposing this condition on the preceding solution gives c = p0(
1)/ /(
1). Substituting this value of c into the general solution of the DE we find, after some simplification, ( 1) 0gy p( 1)/ = p0( 1)/ 1 p0 , so that
Sec. 1.4
Separable DE
17
( 1) 0gy p = p0 1 p0 23.
( 1)/
.
dT kt dt = k(T 450) T(t) = 450 + Ce . T(0) = 50 C = 400 so 1 T(t) = 450 400ekt and T(20) = 150 k = 20 ln(4/3); hence, T(t) = 450 400(3/4)t/20. i) T(40) = 450 400(3/4)2 = 225 F. ii) T(t) = 350 = 450 400(3/4)t/20 (3/4)t/20 = 1/4 20ln(4) t = ln(4) ln(3) 96.4 minutes. T(t) = 75 + cekt . T(10) = 415 75 + ce10k = 415 340 = ce10k and T(20) = 347 75 + ce20k = 347 272 = ce20k. solving these two equations yields 1 k = 10 ln(5/4) and c = 425; hence, T = 75 + 425(4/5)t/10. a) Furnace temperature: T(0) = 500 F. 10ln(17) b) If T(t) = 100 then 100 = 75 + 425(4/5)t/10 t = ln(5/4) 126.96 minutes. Thus the temperature of the coal was 100 F at 6:07 p.m.
25.
Solutions to 1.5
1. dP kt kt 3k dt = kP P(t) = P 0e . P(0) = 10 so P = 10e . P(3) = 20 2 = e
ln(2) ln(2) k = 3 . Thus P(t) = 10e3 . P(24) = 10e 3 t 24 ln(2)
= 1028 = 2560 bacteria.
3.
From P(t) = P0ekt and P(0) = 2000 it follows that P(t) = 2000ekt. Since td = 4, 1 k = 4 ln(2) so P = 2000e
ln(2)t 4
. P(t) = 10 10 = 2000e
6 6
ln(2)t 4
t 35.86 hours.
5.
50C . In formulas (1.5.5) and (1.5.6) we have P0 = 500, P 1 = 800, 50 + (C 50)ert P2 = 1000, t1 = 5, t2 = 10. Hence, 1 (1000)(300) 1 800[(800)(1500) 2(500)(1000)] r = 5 ln (500)(200) = 5 ln 3, C = 1142.86, (800)2 (500)(1000) so that P(t) =
18
Ch. 1
First-Order Differential Equations
(1142.86)(500) 571430 . 0.2tln(3) 500 + 642.86 e 500 + 642.86 e0.2tln(3) Inserting t = 15 into the preceding formula yields P(15) 1091. P(t) = P (P P ) From equation (1.5.5) r > 0 requires P2(P1 P0) > 1. Rearranging the terms in this inequality
0 2 1
7.
and using the fact that P2 > P 1 yields 2P0P P1 > P +P2 . (7.1) 0 2 P (P + P ) 2P0P2 Further, C > 0 requires that 1 0 2 2 > 0. From (7.1) we see that the numerator in P1 P0P2 the preceding inequality is positive, and therefore the denominator must also be positive. Hence, in addition to (7.1), we must also have P12 > P 0P2. 9. (a) Equilibrium solutions: P(t) = 0, P(t) = T. Slope: P > T dP/dt > 0, 0 < P < T dP/dt < 0. 1 rT2 + 4k T . We see that Isoclines: r(P T) = k P TP k/r = 0 P = 2 r the slope of the solution curves satisfies k rT2/4. Concavity: d2P/dt2 = r(2P T)dP/dt = r2(2P T)(P T)P. Hence the solution curves are concave up for P > T/2, and concave down for 0 < P < T/2. (b) A maple plot of the slope field is given below.
2
(c) For 0 < P 0 < T, the population dies out with time. For P0 > T, there is population growth. The term threshold level is appropriate since T gives the minimum value of P0 above which there is population growth. 11. dP dt = r(C P)(P T)P, P(0) = P0, r > 0, 0 < T < C.
Sec. 1.5
Some Simple Population Models
19
Equilibrium solutions: P(t) = 0, P(t) = T, P(t) = C. The slope of the solution curves is negative for 0 < P < T, and for P > C. It is positive for T < P < C. Concavity: d2P/dt2 = r2[(C P)(P T) (P T)P + (C P)P](C P)(P T)P, which simplifies to d2P/dt2 = r2(3P2 + 2PT + 2CP CT)(C P)(P T). Hence changes in 1 concavity occur when P = 3 (C + T T2). A representative slope field with some C2 CT + solution curves is shown in the figure below. We see that for 0 < P0 < T the population dies out, whereas for T < P0 < C the population grows and asymptotes to the equilibrium solution P(t) = C. If P 0 > C, then the solution decays towards the equilibrium solution P(t) = C.
13.
Separating the variables in (1.5.7) yields 1 dP P(ln C ln P) dt = r which can be integrated directly to obtain ln(ln C ln P) = rt + c so that ln(C/P) = c1ert . The initial condition P(0) = P0 requires that ln(P/P0) = c1. Hence, ln(C/P) = ert ln(C/P0) so that Since lim ert
t
P(t) = C eln(P0/k)e . = 0, it follows that lim P(t) = C.
t
rt
15.
(a) Malthusian model is P(t) = 151.3ekt. P(1) = 179.4 179.4 = 151.3e10k 1 10 k = 10 ln(179.4/151.3). Hence, P(t) = 151.3e . 151.3C (b) P(t) = . Imposing the conditions P(10) = 179.4 and P(20) = 203.3 151.3 + (C 151.3)ert gives the pair of equations
tln(179.4/151.1)
20
Ch. 1
First-Order Differential Equations
151.3C 151.3C 10r , 203.3 = 151.3 + (C 151.3)e 151.3 + (C 151.3)e20r whose solution is C 263.95, r 0.046. Using these values for C and r gives 39935.6 P(t) = . 151.3 + 112.65e0.046t (c) 179.4 =
Malthusian model: P(30) 253 million; P(40) 300 million. Logistic model: P(30) 222 million; P(40) 236 million. The logistic model fits the data better than the Malthusian model, but still gives a significant underestimate of the 1990 population.
Solutions to 1.6
In this section the function I(x) = e will represent the integrating factor for a differential equation of the form y + p(x)y = q(x).
p(x)dx
1.
y y = e2x. I(x) = e y(x) = ex(ex + c).
dx
= ex
d(exy) x x x dx = e e y = e + c
3.
2 d 2 2 2 2 = ex dx (ex y) = 2ex x3 ex y = 2 ex x3dx 2 2 2 ex y = ex (x2 1) + c y(x) = x2 1 + cex .
y + 2xy = 2x 3. I(x) = e
2
xdx
Sec. 1.6
First-Order Linear DE
21
5.
2 2x 4 1+x d 4 y + y = . I(x) = e = 1 + x 2 dx [(1 + x2)y] = 2 22 1+x (1 + x ) (1 + x2)2 dx (1 + x2)y = 4 (1 + x2)y = 4tan1 x + c (1 + x2) 1 y(x) = (4tan1 x + c). 1 + x2
2xdx
7.
1 y + xln x y = 9x2. I(x) = e
d = ln x dx (y ln x) = 9 x2ln xdx 1 d(y ln x) = 3x3ln x x3 + c y(x) = ln x (3x3ln x x3 + c) but y(e) = 2e3 so x3(3ln x 1) c = 0; thus, y(x) = . ln x
2
dx xln(x)
9.
dx 2 4et t dt + 2x = 4et x + t x = t . I(t) = e t2x = 4
d = t2. dt (t2x) = 4tet 4et (t 1) + c t 2 t te dt t x = 4e (t 1) + c x(t) = . t2
dt t
11.
(1 ysin x)dx cos x dy = 0 y + (sin x sec x)y = sec x. I(x) = e
sin xsec xdx
d = sec x dx (y sec x) = sec2 x y sec x =
sec x dx
2
y sec x = tan x + c y(x) = cos x(tan x + c) y(x) = sin x + ccos x. y + y = ex. I(x) = e
dx
13.
d = ex dx (exy) = e(+)x exy =
e
(+)x
dx.
If + = 0, then exy = x + c y(x) = ex(x + c). e( + )x ex x If + 0, then e y = + + c y(x) = + + cex.
2
15.
2 y + x y = 4x. I(x) = e
dx x
d = e2ln x = x2 dx (x2y) = 4x3 x2y = 4
x dx
3
x2y = x4 + c, but y(1) = 2 so c = 1; thus, y(x) = x2(x4 + 1).
22
Ch. 1
First-Order Differential Equations
17.
d = (4 t)2 dx ((4 t)2x) = 5(4 t)2 (4 t)2x = 5 (4 t)2dt (4 t)2x = 5(4 t)1 + c, but x(0) = 4 so c = 1; thus, x(t) = (4 t)2(5(4 t)1 1) or x(t) = (4 t)(1 + t). 1, if x 1, dx d y + y = f(x), y(0) = 3, f(x) = . I(x) = e = ex dx(exy) = exf(x) 0, if x > 1.
2 x + 4 t
x = 5. I(t) = e
2
dt 4t
19.
[exy]x =
0
exf(x) dx exy y(0) = 0 exf(x) dx exy 3 = 0 exf(x) dx 0 exf(x) dx). 0
x 1 x
x
x
x
y(x) = ex(3 +
x
If x 1,
exf(x) dx = 0 ex dx = ex 1 y(x) = ex(2 + ex). 0 exf(x) dx = 0 exdx = e 1 y(x) = ex(2 + e). 0
x
If x > 1,
21.
d On (, 1), y y = 1 I(x) = ex dx(exy) = ex y(x) = c1ex 1. Imposing the initial condition y(0) = 0 requires c1 = 1, so that y(x) = ex 1, for x < 1. d On [1, ), y y = 2 x I(x) = ex dx(exy) = (2 x)ex y(x) = x 1 + c2ex. Continuity at x = 1 requires that lim y(x) = y(1). Consequently we must choose c2 to satisfy
x1
c2e = e 1, so that c2 = 1 e1. Hence, for x 1, y(x) = x 1 + (1 e1)ex . dTm dT dT = Tm = t + c1 so dt = k(T Tm) dt = k(T t c1) dt k dt dT dt + kT = kt + kc1. e = ekt is an integrating factor for the differential equation so d kt kt kt kt dt (e T) = e (kt + kc1) e T = e (t k + c1) + c2 1 T = t k + c1 + c2ekt (t) = (t k ) + + T0ekt where = c1 and T0 = c2.
23.
Sec. 1.6
First-Order Linear DE
23
25.
dT 1 Newton's law of cooling is dt = - 40 (T - 80e-t/20). This linear DE has standard form dT 1 -t/20 t/40 dt + 40 T = 2e , with integrating factor I(t) = e . Consequently the DE can be written d in the integrable form dt(et/40T) = 2e-t/40, so that T(t) = -80e-t/20 + ce-t/40. T(0) = 0 c = 80, so that T(t) = 80(e-t/40 - e-t/20). b) We see that lim T(t) = 0. This is a reasonable result since the temperature of the a)
t
surrounding medium also approaches zero as t . We would expect the temperature of the object to approach the temperature of the surrounding medium at late times. dT 1 1 b) T(t) = 80(e-t/40 - e-t/20) dt = 80(40 et/40 + 20 et/20 ). So T(t) has only one critical 1 1 point when 80(40 et/40 + 20 et/20 ) = 0 t = 40 ln 2. Since T(0) = 0, and lim T(t) = 0 t the function assumes a maximum value at tmax = 40 ln 2. T(tmax ) = 80(eln 2 e2ln 2 ) = 20 F, d) Tm(tmax ) = 80 e2ln 2 = 20 F.
27.
a)
dy dy + p(x)y = 0 y = p(x)dx dx p(x)dx yH = c1e .
dy = p(x)dx ln|y| = p(x)dx y
24
Ch. 1
First-Order Differential Equations
b)
p(x)dx Replace c1 in part a) by u(x) and let v = e . dy dv du y = uv dx = u dx + v dx . Substituting this last result into the dy dv du original equation, dx + p(x) = q(x), we obtain u dx + v dx + p(x)y = q(x), but since dv du dx = vp, the last equation reduces to v dx = q(x)
du = v1(x)q(x)dx u = into y = uv, we obtain y = e
p(x)dx
v
1
(x)q(x)dx + c. Substituting the values for u and v p(x)dx e q(x)dx + c .
29.
dy 1 1 The associated homogeneous equation is dx + x y = 0, with solution yH = c x. We 1 determine the function u(x) such that y(x) = x u(x) is a solution of the given differential dy 1 du 1 equation. We have dx = x dx 2 u. Substituting into x dy 1 du dx + x y = cos x and simplifying yields dx = xcos x, so that 1 c u = x sin x + cos x + c. Hence y(x) = sin x + x cos x + x . dy The associated homogeneous equation is dx + cot x y = 0, with solution yH = c csc x. We determine the function u(x) such that y(x) = csc x u(x) is a solution of the given differential equation. We have dy du dx = csc x dx u csc x cot x. Substituting into dy du cot x y = 2cos x and simplifying yields dx = 2cos x sin x = sin 2x, so that dx 1 1 u(x) = 2 cos 2x + c. Consequently, y(x) = 2 cos 2x csc x + ccsc x.
31.
Solutions to 1.7
1. Given V(0) = 10, A(0) = 20, c 1 = 4, r1 = 2 and r2 = 1. V = r 1t r2t dV dt = 1 V = t + 10 since V(0) = 10. c1r1t c2r2t dA A A dA 1 dt = 8 c2 = 8 V = 8 t + 10 dt + t + 10 A = 8
Sec. 1.7
Two Modeling Problems Governed by First-Order Linear DE
25
(t + 10)A = 4(t + 10) 2 + c1. A(0) = 20 c1 = 200 so 4 A(t) = t + 10 [((t + 10)2 50]. Therefore A(40) = 196 grams. 3. Given V(0) = 20, A(0) = 0, c 1 = 10, r1 = 4 and r2 = 2. V = r 1t r2t dV dt = 2 V = 2(t + 10) since V(0) = 20. Thus, V(t) = 40 for t = 10, so we must dA 2A A find A(10). c1r1t c2r2t dt = 40 2c2 = 40 V = 40 t + 10 dA 1 d dt + t + 10 A = 40 dt [(t + 10)A] = 40(t + 10)dt (t + 10)A = 20(t + 10) 2 + c. A(0) = 0 c = 2000 so 20 A(t) = t + 10 [(t + 10)2 100] and A(10) = 300 grams. Given V(0) = 10, c1 = 0, r1 = 3, r2 = 2 and A(5)/V(5) = 0.2. dV a) V = r 1t r2t dt = 1 V(t) = t + 10 since V(0) = 10. dA A 2A dA 2dt c1r1t c2r2t dt = 2c2 = 2 V = t + 10 A = t + 10 ln|A| = 2ln|t + 10| + c A = k(t + 10) 2. V(5) = 15, so A(5) = 3 since 675 A(5)/V(5) = 0.2. Thus, k = 675 and A(t) = . In particular, A(0) = 6.75 (t + 10)2 grams. 675 b) Find V(t) when A(t)/V(t) = 0.1. A(t) = and V(t) = t + 10 (t + 10)2 3 A(t) 675 A(t) V(t) = . V(t) = 0.1 (t + 10)3 = 6750 t + 10 = 15 2 so (t + 10)3 V(t) = t + 10 = 15 2 liters. 7. a) Given V(0) = w, c1 = k, r1 = r, r2 = r and A(0) = A 0. V = r 1t r2t dV dt = 0 V(t) = V(0) = w for all t. = c1r1t c2r2t dA A A r dA r dt = kr r V = kr r V = kr w A dt + w A = kr d dt (ert/w A) = krert/w A = kw + ce rt/w . A(0) = A 0 so c = A 0 kw A(t) = ert/w [kw(ert/w 1) + A 0].
3
5.
26
Ch. 1
First-Order Differential Equations
b)
A A(t) ert/w lim V(t) = lim w [kw(ert/w 1) + A 0] = lim [k + ( w0 k)ert/w ] = k. This is t t t reasonable since the volume remains constant, and the solution in the tank is gradually mixed with and replaced by the solution of concentration k flowing in.
9.
di R 1 di Let E(t) = 20, R = 4 and L = 1/10. Then dt + L i = L E(t) dt + 40i = 200 d dt (e40t i) = 200e40t i = 5 + ce40t but i(0) = 0 c = 5, consequently i(t) = 5(1 e40t). di R 1 Let R = 2, L = 2/3 and E(t) = 10sin 4t. Then dt + L i = L E(t) di d dt + 3i = 15sin 4t dt (e3ti) = 15e3tsin 4t 3e3t 3 4 e3ti = 5 (3sin 4t 4cos 4t) + c i = 3(5 sin 4t 5 cos 4t) + ce3t , 3 but i(0) = 0 c = 12/5 so i(t) = 5 (3sin 4t 4cos 4t + 4e3t ). dq 1 E dq 1 In an RC circuit for t > 0, dt + RC q = R . If E(t) = 0 then dt + RC q = 0 d dt (et/RC q) = 0 q = cet/RC and if q(0) = 5 then q(t) = 5et/RC . lim q(t) = 0. Yes, this is reasonable. As the time increases and E(t) = 0, the charge will
t
11.
13.
dissipate to zero. di R E(t) In an RL circuit, dt + L i = L and since E(t) = E0sin(t), E E di R d + L i = L0 sin(t) dt (eRt/L i) = L0 eRt/L sin(t) dt E i(t) = 2 0 2 2 [Rsin(t) Lcos(t)] + AeRt/L . R +L We can write this as E0 R L i(t) = [ 2 sin(t) cos(t)] + AeRt/L . Defining the phase 2 2 2 2 2 2 2 2 R + L R + L R + L R L by cos( ) = , sin( ) = , we have 2 2 2 2 R + L R + L2 2 E0 i(t) = [cos( )sin(t) sin( )cos(t)] + AeRt/L . That is, 2 2 2 R + L 15.
Sec. 1.7
Two Modeling Problems Governed by First-Order Linear DE
27
i(t) =
E0
2 2 2
R + L Transient part of the solution: iT(t) = AeRt/L .
Steady state part of the solution: iS(t) = E0
sin(t ) + AeRt/L .
R2 + L2 2
sin(t ).
E E dq 1 E(t) dq 1 d + RC q = R dt + RC q = R0 eat dt (et/RC q) = R0 e(1/RC a)t dt E C at E C (1/RCa)t t/RC q(t) = et/RC 1 0 e + k q(t) = 1 0 . aRC aRC e + ke Imposing the initial condition q(0) = 0 (capacitor initially uncharged) requires EC k = 1 0 , aRC so that EC at t/RC q(t) = 1 0 ). aRC (e e Thus EC dq 1 t/RC i(t) = dt = 1 0 aeat ). aRC (RC e 17. E d2q 1 dq d2q di dq di + LC q = L0. Since i = dt , 2 = dq dt = i dq . Hence the original equation can be 2 dt dt E0 E di 1 1 written i dq + LC q = L i di + LC q dq = L0 dq or E 0q i2 q2 + 2LC = L + A. (19.1) 2 q02 E 0q0 i(0) = 0 and q(0) = q0 A = 2LC L . From equation (19.1), 2 1/2 2E q (E 0C)2 (q E 0C)2 1/2 2A + 0 q i = i = 2A + LC and if we let L LC LC 2 2 1/2 (E 0C) dq q E 0C D2 = 2A + LC then i = dt = D 1 D LC q E 0C q E 0C sin1( ) = t + B. q(0) = 0 so B = Sin1( ) LC LC D D LC LC and therefore q E 0C t+B t+B = sin( ) q(t) = D sin( ) + E 0c LC D LC LC LC
19.
28
Ch. 1
First-Order Differential Equations
2A + (E0C)2 q02 E 0q0 dq t+B 2 i = dt = Dcos( ). Since D = and A = 2LC L we can LC LC |q E 0C| substitute to eliminate A and obtain D = 0 . Thus LC t+B q(t) = |q0 E 0C|sin( ) + E 0c. LC
Solutions to 1.8
y dy dv Unless otherwise indicated in this section v = x , dx = v + x dx and t > 0. tx (tx)2 (ty)2 x2 y2 ty = (tx)(ty) = xy = f(x,y). f is homogeneous of degree zero. x2 y2 1 (y/x)2 1 v2 f(x,y) = xy = = v = F(v). y/x tx ty x y (tx)sin(ty) (ty)cos(tx) xsin(y) ycos(x) f(tx, ty) = = = f(x, y). f is homogeneous of ty y x y y 1 sin(y) x cos(x) sin(v) v cos(v) degree zero. f(x,y) = = = F(v). y v x ty f(tx, ty) = tx 1 . Not homogeneous.
1.
3.
5.
7.
f(tx, ty) = f(x, y) =
(tx)2 + (ty) 2
tx x =
=
x2 + y2
x
= f(x, y). Homogeneous of degree zero.
x2 + y2
|x| 2 1 + (y/x) = x
y 1 + (x)2 = v2 = F(v). 1 +
9.
dy y dy y dv (3x 2y) dx = 3y (3 2 x) dx = 3 x (3 2v)(v + x dx) = 3v dv 3v 3 2v dx 3 x dx = 3 2v v 2 dv = x 2v ln|v| = ln|x| + c1 2v 3x y 3x 2y ln|x| = ln|x| + c1 ln|y| = 2y + c2 y2 = ce3x/y .
Sec. 1.8
Change of Variables
29
11.
dy dy sin(y/x)(xdx y) = xcos(y/x) sin(y/x)(dx y/x) = cos(y/x) dv dv sin v sin v(v + xdx v) = cos v sin v (xdx ) = cos v cos v dv = ln|cos v| = ln|x| + c1 |xcos(y/x)| = c2 y(x) = xcos1(c/x).
dx x
13.
We first rewrite the given differential equation in the equivalent form (9x2 + y2)1/2 + y y = . Factoring out an x2 from the square root yields x |x|[9 + (y/x)2]1/2 + y y = . Since we are told to solve the differential equation on the interval x y x > 0 we have |x|= x, so that y = [9 + (y/x)2] + x , which we recognize as being homogeneous. We therefore let y = xV, so that y = xV + V. Substitution into the preceding DE yields xV + V = (9 + V 2)1/2 + V, that is xV = (9 + V 2)1/2. Separating the 1 1 variables in this equation we obtain 2 1/2 dV = x dx. Integrating we obtain (9 + V ) 2 1/2 ln[V + (9 + V ) ] = ln c1x. Exponentiating both sides yields V + (9 + V2)1/2 = c1x. Substituting y/x = V and multiplying through by x yields the general solution y + y2 = c1x2. 9x2 +
15.
dy dy y dv x dx + yln x = yln y dx = x ln(y/x) v + x dx = vln(v) dv dx ln(y/x) 1 = c v(ln(v) 1) = x ln|ln(v) 1| = ln|x| + c1 x
y(x) = xe1 + cx. 17.
2 2 y dy 2 2 2xydy (x2ey /x + 2y2)dx = 0 2 x dx ( ey /x + 2(y/x)2) = 0 dv 2 dv 2 2 dx 2v(v + x dx ) (ev + 2v2) = 0 2vxdx = ev ev (2vdv) = x v2 y2/x2 2 2 e = ln|x| + c1 e = ln(cx) y = x ln(ln(cx)).
19.
dy dx =
dy dx = y dv 1 + v2 1 v2 x dx = v v dv = 2 1 + v 1 v2
x2 + y 2 x
1 + (y/x) 2 1
y/x
dv v + x dx =
1 + v2 1
v
dx. If we let u x
2
= 1 + v 2 so that udu = vdv, then the last
30
Ch. 1
First-Order Differential Equations
equation simplifies to
|x(1 u)| = c2 1 u = c/x u = c /x 2c/x + 1
2 2 2
1duu = dx ln|1 u| = ln|x| + c x
1
v2 = c2/x2 2c/x y2 = c2 2cx. 21. dy dv dv x dx = xtan(y/x) + y v + x dx = tan(v) + v x dx = tan(v) dx cot(v)dv = x ln|sin(v)| = ln|x| + c1 sin(v) = cx
v = sin1(cx) y(x) = xsin1(cx). 23. The given DE can be written as (x 4y) dy = (4x + y) dx. Converting to polar coordinates we have x = rcos dx = cos dr rsin d, and y = rsin dy = sin dr + rcos d. Substituting these results into the preceding DE and simplifying yields the separable equation 4r1 dr = d which can be integrated directly to yield 4 ln r = + c, so that r = c1e/4. dy 2x y dy 2 y/x dv 2v dv 2 2v 4v2 = x + 4y dx = 1 + 4y/x v + x dx = 1 + 4v x dx = dx 1 + 4v 1 1 + 4v dx 1 2 dv = x 2 ln|2v2 + v 1| = ln|x| + c 2v2 + v 1 1 1 1 2 ln|x2(2v2 + v 1)| = c 2 ln|2y2 + yx x2| = c, but y(1) = 1 so c = 2 ln(2). Thus, 1 1 ln|2y2 + yx x2| = 2 ln(2) and since y(1) = 1 it must be the case that 2y2 yx + x2 = 2. 2
25.
27.
dy dx y/x =
4 (y/x) 2
dv v + x dx = v + 2 4 v
y sin1(v/2) = ln|x| + c sin1( 2x ) = ln x + c since x > 0. 29. Given family of curves satisfies: x2 + y2 = 2cy c =
dv v 4
2
=
dx x
x2 + y2 2y . Hence
dy dy dy x 2xy 2x + 2y dx = 2c dx dx = c y = 2 2 . Orthogonal trajectories satisfy: x y dy y2 x2 dy dv = 2xy . Let y = vx so that dx = v + x dx. Substituting these results into the last dx c dv v2 + 1 y2 equation yields x dx = 2v ln|v2 + 1| = ln|x| + c1 2 + 1 = x2 x2 + y2 = x 2kx.
Sec. 1.8
Change of Variables
31
31.
a)
Let r represent the radius of one of the circles with center at (a, ma) and passing through (0, 0). r = 2 = |a| 2 . Thus, the circle's equation can be (a 0)2 + (ma 0) 1 + m written as (x a)2 + (y ma) 2 = (|a| 2 )2 or (x a)2 + (y ma) 2 = a2(1 + m 2). 1 + m x2 + y2 (x a)2 + (y ma) 2 = a2(1 + m 2) a = 2[x + my] . Differentiating the first dy ax equation with respect to x and solving we obtain dx = y ma . Substituting for a and dy y2 x2 2mxy simplifying yields dx = . Orthogonal trajectories satisfy: my 2 mx 2 + 2xy dy mx 2 my 2 2xy dy m m (y/x)2 2(y/x) = 2 2 dx = . Let v = y/x so that dx y x 2mxy (y/x)2 1 2m(y/x) dy dv = v + x dx . Substituting these results into the last equation yields dx dv m m v2 2v xdv (m v)(1 + v 2) v + x dx = 2 dx = v 1 2mv v2 2mv 1 v2 2mv 1 dx dv 2vdv dx 2 dv = 2 = x vm x (m v)(1 + v ) 1+v 2 2 ln|v m| ln(1 + v ) = ln|x| + c1 v m = c 2x(1 + v ) y mx = c 2x2 + c2y2 x2 + y2 + cmx cy = 0. Completing the square we c2 obtain (x + cm/2)2 + (y c/2)2 = 4 (m 2 + 1) . Now letting b = c/2, the last equation
b)
32
Ch. 1
First-Order Differential Equations
becomes (x + bm) 2 + (y b)2 = b2(m 2 + 1) which is a family of circles lying on the line y = my and passing through the origin. c)
33.
m 2 tan(/4) dy 6y/x 1 6y x y = cx6 dx = 6y/x = m 2. m 1 = 1 + m tan(/4) = 1 + 6y/x = 6y + x . 2 dy dv Let v = y/x so that dx = v + x dx . Substituting these results into the last equation yields v dv 6v 1 dv (3v 1)(1 2v) + x dx = 6v + 1 x dx = 6v + 1 dx 9 8 dv = 3v 1 2v 1 x 3ln|3v 1| 4ln|2v 1| = ln|x| + c1 3ln|3y x| = 4ln|2y x| + c1 Oblique trajectories: (3y x)3 = k(2y x)4.
35.
a)
dy y = cx1 dx = cx2
m 2 tan( 0) y = x . m 1 = 1 + m tan( ) = 2 0
. y 1 + ( x)tan( 0)
y x tan( 0)
dy dv Let v = y/x so that dx = v + x dx . Substituting these results into the last equation yields tan( 0) + v 2vtan( 0) 2 dv 2dx v + x dx = vtan( ) 1 2 dv = x v tan( 0) 2v tan( 0) 0 ln|v2tan( 0) 2v tan( 0)| = 2ln|x| + c1 (y2 x2)tan( 0) 2xy = k. c)
Sec. 1.8
Change of Variables
33
37.
dy 1 2 1 dx x y = 4x y cos x. This is a Bernoulli equation. Multiplying both sides by y results in dy 1 y dx x y2 = 4x2cos x. (37.1) du dy dy 1 du Let u = y2 so dx = 2y dx or y dx = 2 dx . Substituting these results into equation (37.1) du 2 yields dx x u = 8x2cos x which has an integrating factor I(x) = x2 d(x2u) = 8cos x x2u = 8 u = x2(8sin x + c) y2 = x2(8sin x + c).
cos x dx x
2
u = 8sin x + c
39.
dy 3 1/3 2 dx 2x y = 6y x ln x or 1 dy 3 2/3 = 6x2ln x. (39.1) 1/3 dx 2x y y du 2 dy du 1 Let u = y2/3 dx = 3 y1/3 dx. Substituting these results into equation (39.1) yields dx x 1 u = 4x2ln x. An integrating factor for this equation is I(x) = x so d 1 1 1 2 2 dx (x u) = 4xln x x u = 4 xln x dx + c x u = 2x ln x x + c
u(x) = x(2x2ln x x2 + c) y2/3 = x(2x2ln x x2 + c). 41. dy 2 2 4 dx + x y = 6 y x or dy 2 y2 dx + x y1 = 6x4. du dy Let u = y1 dx = y2 dx. Substituting these results into equation (41.1) du 2 yields dx x u = 6x4. An integrating factor for this equation is I(x) = x2 so (41.1)
34
Ch. 1
First-Order Differential Equations
d 2 2 2 3 5 2 1 5 2 dx (x u) = 6x x u = 2x + c u = 2x + cx y = 2x + cx 1 y(x) = 2 . x (c 2x3) 43. dy (x a)(x b) (dx y1/2 ) = 2(b a) y or dy 2(b a) y1/2 dx (x a)(x b) y1/2 = 1. du dy Let u = y1/2 2 dx = y1/2 dx. Substituting these results into (43.1) yields du (b a) 1 (x a) dx (x a)(x b) u = 2. An integrating factor is I(x) = (x b) so d (x a) (x a) (x a) 1 dx (x b) u = 2(x b) (x b) u = 2 {x + (b a) ln|x b|+ c} (x b) y1/2 = 2(x a) {x + (b a) ln |x b| + c} 1 xb 2 y(x) = 4 x a [x + (b a)ln|x b| + c]2. 45. dy 3 1/2 dx + 4xy = 4x y or dy y1/2 dx + 4xy1/2 = 4x3. du dy Let u = y1/2 2 dx = y1/2 dx. Substituting these results into (45.1) gives du 2 + 2xu = 2x 3. An integrating factor is ex so dx d x2 2 x2 3 x2 x2 2 1/2 = x2 1 + cex dx (e u) = 2e x e u = e (x 1) + c y 2 y(x) = [(x2 1) + cex ]2. 47. dy 1 3 dx ( 1)x y = (1 ) xy or dy 1 3x y dx ( 1)x y1 = 1 . 1 du dy Let u = y1 1 dx = y dx. Substituting these results into (47.1) gives du 1 d 2 dx + x u = 3x. An integrating factor is I(x) = x so dx (xu) = 3x 1/(1 ) x3 + c x3 + c xu = x3 + c y1 = x y(x) = x . (45.1) (43.1)
(47.1)
Sec. 1.8
Change of Variables
35
49.
dy 3 (1 3) dx + y sec x = y sec x or
1 3 = secx. (49.1) dx + (sec x) y du 1 3 3 dy Let u = y dx = (1 3) y dx. Substituting these results into du (49.1) gives dx + (sec x) u = sec x. An integrating factor is I(x) = sec x + tan x so d dx ((sec x + tan x)u) = sec x (sec x + tan x) (sec x + tan x) u = tan x + sec x + c 1/(1 3) c c 1 3 y = 1 + sec x + tan x y(x) = 1 + sec x + tan x .
(1 3) y
3 dy
51.
dy 3 3 dx + y cot x = y sin x or dy y3 dx + (cot x) y2 = sin3x. 1 du dy Let u = y2 2 dx = y3 dx. Substituting these results into (51.1) gives du 3 2 dx (2 cot x)u = 2sin x. An integrating factor is I(x) = csc x so d 2 2 dx (u csc x) = 2sin x u csc x = 2cos x + c. y(/2) = 1 c = 1. Thus, 1 y2 = . 2 sin x (2 cos x + 1) (51.1)
53.
dy dy dv dv dv = (9x y)2. Let v = 9x y so that dx = 9 dx dx = 9 v2 = dx 9 v2 1 3 tanh1(v/3) = x + c1 but y(0) = 0 so c = 0. Thus, tanh1(3x y/3) = 3x or y(x) = 3(3x tanh 3x). dy dy 1 dv = sin2(3x 3y + 1). Let v = 3x 3y + 1 so that dx = 1 3 dx dx 1 dv dv 1 3 dx = sin2v dx = 3cos2v sec2v dv = 3 dx
dx
55.
tan v = 3x + c tan(3x 3y + 1) = 3x + c 1 y(x) = 3 [3x tan1(3x + c) + 1]. 57. 1 1 Substituting into (56.1) for F(V) = ln V 1 yields V ln V dV = x dx ln(ln V) = ln cx V 1 = ecx y(x) = x ecx .
36
Ch. 1
First-Order Differential Equations
59.
a) y = Y(x) + v 1(x) y = Y(x) v 2(x)v(x). Now substitute into the given DE and simplify algebraically to obtain Y(x) + p(x)Y(x) + q(x)Y2(x) v2(x)v(x) + v1p(x) + q(x)[2Y(x)v 1(x) + v2(x)] = r(x). We are told that Y(x) is a particular solution to the given DE, and therefore Y(x) + p(x)Y(x) + q(x)Y 2(x) = r(x). Consequently the transformed DE reduces to v2(x)v(x) + v1p(x) + q(x)[2Y(x)v1(x) + v2(x)] = 0, or equivalently v [p(x) + 2Y(x)q(x)]v = q(x). b) The given DE can be written as y x1y y2 = x2, which is a Riccati DE with p(x) = x1, q(x) = 1, and r(x) = x2. Since y(x) = x1 is a solution to the given DE, we make the substitution y(x) = x1 + v1(x). According to the result from a), the given DE then reduces to v (x1 + 2x1)v = 1, or equivalently v x1v = 1. This linear DE has d integrating actor I(x) = x1, so that v must satisfy dx(x1v) = x1 v(x) = x(c ln x). 1 1 1 1 Hence the solution to the original equation is y(x) = x + x(c ln x) = x c ln x 1 . y = x1 + w(x) y = x2 + w. Substituting into the given DE yields (x2 + w) + 7x 1(x1 + w) 3(x 2 + 2x1w + w 2) = 3x2, which simplifies to w + x1w 3w 2 = 0. b) The preceding equation can be written in the equivalent form w2w + x 1w1 = 3. We let u = w1, so that u = w2w. Substitution into the DE gives, after simplification, u x1u = 3. An integrating factor for this linear DE is I = x1, so that d the DE can be written in the integrable form dx(x1u) = 3x1. Integrating we obtain 1 u(x) = x(3ln x + c), so that w(x) = x(c 3ln x). Consequently the solution to the original Riccati equation is 1 1 y(x) = x 1 + c 3ln x . a) dy 2 1 2ln x y1 dx x ln y = . Let u = ln y so using the technique of the preceding problem: x du 2 1 2ln x xu = and dx x dx 2 2 dx 1 2ln x 1 2ln x x x 2 u = e dx + c1 = ln x + cx2, and e dx + c1 = x x3 x 2 2 since u = ln y, ln y = ln x + cx . Now y(1) = e so c = 1 y(x) = xex .
61.
63.
Sec. 1.8
Change of Variables
37
65.
dy sec2 y dx +
1
2x 2 x 1 + 1 + du 1 1 equation becomes dx + u = which is first order linear. An integrating 2x 2 x 1 + 1 +
tan y =
1
du dy . Let u = tan y so that dx = sec2 y dx and the given
1 + x factor for this equation is I(x) = e 1 + x 1 + x d x e e 1 + 1 + x 1 + x 1 + x dx (e u) = e u = e u = e +c 2x 2 x 1 + 1 + x 1 + u = 1+ce . But u = tan y so tan y = 1 + ce x or 1 +
y(x) = tan1(1 + c e x ). 1 +
Solutions to 1.9
1. (y + 3x2)dx + xdy = 0. M = y + 3x2 and N = x My = 1 and Nx = 1 My = N x the differential equation is exact. yexydx + (2y xexy)dy = 0. M = ye xy and N = 2y xexy My = yxexy + exy and Nx = xyexy exy My Nx the differential equation is not exact. (y2 + cos x)dx + (2xy + sin y)dy = 0. M = y2 + cos x and N = 2xy + sin y My = 2y and Nx = 2y My = N x the differential equation is exact so there exists a potential function such that a) x = y2 + cos x and dh(y) b) y = 2xy + sin y. From a) (x, y) = xy2 + sin x + h(y) y = 2xy + dy so from dh(y) dh b), 2xy + dy = 2xy + sin y dy = sin y h(y) = cos y where the constant of integration has been set to zero since we just need one potential function. (x, y) = xy2 + sin x cos y xy2 + sin x cos y = c. (4e2x + 2xy y2)dx + (x y)2dy = 0. M = 4e 2x + 2xy y2 and N = (x y)2 My = 2(x y) and Nx = 2(x y) My = N x the differential equation is exact so there exists a potential function such that a) x = 4e2x + 2xy y2 and b) y = (x y)2. 3 dh(x) 2 2 y From b) (x, y) = x y xy + 3 + h(x) x = 2xy y2 + dx so from a), dh(x) dh(x) 2xy y2 + dx = 4e2x + 2xy y2 dx = 4e2x h(x) = 2e2x where the constant of integration has been set to zero since we just need one potential function.
3.
5.
7.
38
Ch. 1
First-Order Differential Equations
3 y3 2x 2 2 y (x, y) = x y xy + 3 + 2e x y xy + 3 + 2e2x = c1 6e2x + 3x2y 3xy2 + y3 = c. 2 2
9.
x 1 y dx + x dy = 0. M = 1 y and N = x x2 + y2 2 2 2 2 2 x x +y x +y x + y2 y2 x2 y2 x2 My = 2 and Nx = 2 My = N x the differential equation is exact (x + y2)2 (x + y2)2 so there exists a potential function such that 1 y x a) x = x 2 . From b) (x, y) = Tan1(y/x) + h(x) 2 and b) y = 2 x +y x + y2 y dh(x) y dh(x) 1 y dh 1 x = 2 2 + dx so from a), 2 2 + dx = x 2 2 dx = x x +y x +y x +y h(x) = ln|x| where the constant of integration has been set to zero since we just need one potential function. (x, y) = tan1(y/x) + ln|x| tan1(y/x) + ln|x| = c. [ycos(xy) sin x]dx + xcos(xy)dy = 0. M = ycos(xy) sin x and N = xcos(xy) My = xysin(xy) + cos(xy) and N x = xysin(xy) + cos(xy) My = N x the differential equation is exact so there exists a potential function such that a) x = ycos(xy) sin x and b) y = xcos(xy). dh(x) From b) (x, y) = sin(xy) + h(x) x = ycos(xy) + dx so from a), dh(x) dh ycos(xy) + dx = ycos(xy) sin x dx = sin x h(x) = cos x where the constant of integration has been set to zero since we just need one potential function. (x, y) = sin(xy) + cos x sin(xy) + cos x = c.
11.
13.
(3x2ln x + x2 y)dx xdy = 0. M = 3x 2ln x + x2 y and N = x My = 1 and Nx = 1 My = N x the differential equation is exact so there exists a potential function such that a) x = 3x2ln x + x2 y and b) y = x. From b) (x, y) = xy + h(x) dh(x) x = y + dx so from a), dh(x) dh(x) y + dx = 3x2ln x + x2 y dx = 3x2ln x + x2 h(x) = x3ln x where the constant of integration has been set to zero since we just need one potential function.
Sec. 1.9
Exact DE
39
(x, y) = xy + x 3ln x xy + x3ln x = c. Now since y(1) = 5, x3ln x + 5 c = 5; thus, x3ln x xy = 5 or y(x) = . x 15. (yexy + cos x)dx + xexydy = 0. M = ye xy + cos x and N = xexy My = xyexy + exy and Nx = xyexy + exy My = N x the differential equation is exact so there exists a potential function such that a) x = yexy + cos x and b) y = xexy. From b), (x, y) = exy + h(x) dh(x) dh(x) dh(x) x = yexy + dx so from a), yexy + dx = yexy + cos x dx = cos x h(x) = sin x where the constant of integration has been set to zero since we just need one potential function. (x, y) = exy + sin x exy + sin x = c. Now since y(/2) = 0, c = 2; ln(2 sin x) thus, exy + sin x = 2 or y(x) = . x 17. M = cos(xy)[tan(xy) + xy] and N = x 2cos(xy) My = 2xcos(xy) x2ysin(xy) = N x My = N x Mdx + Ndy = 0 is exact I(x, y) = cos(xy) is an integrating factor for [tan(xy) + xy]dx + x2dy = 0. 19. M = ex/y(x2y1 2x) and N = ex/yx3y2 My = ex/y(x3y3 3x2y2) = N x Mdx + Ndy = 0 is exact I(x, y) = y2ex/y is an integrating factor for y[x2 2xy]dx x3dy = 0. ydx (2x + y4)dy = 0. M = y and N = (2x + y4) My = 1 and Nx = 2
21.
My Nx g(y)dy = 3y1 = g(y) is a function of y alone so I(y) = e = 1/y3 is an M integrating factor for the given equation. Multiplying the given equation by I(y) results in the 2 2 3 2 y exact equation y dx (2xy + y)dy = 0. We find that (x, y) = xy 2 and hence, the y2 general solution of our differential equation is xy2 2 = c1 2x y4 = cy2. 23. (y x2)dx + 2xdy = 0. M = y x2 and N = 2x My = 1 and Nx = 2 M N 1 f(x)dx = 1 yN x = 2x f(x) is a function of x alone so I(x) = e x is an integrating factor for the given equation. Multiplying the given equation by I(x) results in 2x5/2 the exact equation (x1/2y x3/2 )dx + 2x1/2 dy = 0. We find that (x, y) = 2x1/2 y 5
40
Ch. 1
First-Order Differential Equations
25.
and hence, the general solution of our differential equation is 2x1/2 y \f(2x5/2 ,5) = c or y(x) c + 2x5/2 = . 10 x dy 2xy 1 + (2xy + 2x3y 1)dx + (1 + x 2)2dy = 0. 2 = dx 1 + x (1 + x2)2 M = 2xy + 2x 3y 1 and N = (1 + x 2)2 My = 2x + 2x 3 and Nx = 4x(1 + x 2)
My Nx 2x f(x)dx = 1 = 2 f(x) is a function of x alone so I(x) = e N 1+x 1 + x2 is an integrating factor for the given equation. Multiplying the given equation by I(x) results in 1 the exact equation (2xy )dx + (1 + x 2)dy = 0. We find that (x, y) = (1 + x 2)y 1 + x2 tan1 x and hence, the general solution of our differential equation is tan1 x + c 2 1 (1 + x )y tan x = c or y(x) = . 1 + x2 27. (y1 x1)dx + (xy2 2y1)dy = 0 xr ys (y1 x1)dx + xr ys (xy2 2y1)dy = 0 (xr ys 1 xr 1ys )dx + (xr + 1 ys 2 2xr ys 1 )dy = 0. M = x r ys 1 xr 1ys and N = x r + 1 ys 2 2xr ys 1 My = xr (s 1)ys 2 xr 1sys 1 and Nx = (r + 1)x r ys 2 2rxr 1ys 1 . The equation is exact if and only if My = N x s1 s r + 1 2r xr (s 1)ys 2 xr 1sys 1 = (r + 1)x r ys 2 2rxr 1ys 1 2 xy = xy y y2 sr2 s 2r = xy . From the last equation we require that s r 2 = 0 and 2 y s 2r = 0. Solving this system yields r = 2 and s = 4. 2y(y + 2x2)dx + x(4y + 3x 2)dy = 0 xr ys 2y(y + 2x2)dx + xr ys x(4y + 3x2)dy = 0 M = 2xr ys + 2 + 4xr + 2 ys + 1 and N = 4xr + 1 ys + 1 + 3xr + 3 ys My = 2xr (s + 2)y s + 1 + 4xr + 2 (s + 1)y s and Nx = 4(r + 1)x r ys + 1 + 3(r + 3)x r + 2 ys . The equation is exact if and only if My = N x 2xr (s + 2)y s + 1 + 4xr + 2 (s + 1)y s = 4(r + 1)x r ys + 1 + 3(r + 3)x r + 2 ys 2(s + 2)y + 4x 2(s + 1) = 4(r + 1)y + 3(r + 3)x2. From this last equation we require that 2(s + 2) = 4(r + 1) and 4(s + 1) = 3(r + 3). Solving this system yields r = 1 and s = 2. 31. a) dy dx + py = q can be written in differential form as (py q)dx + dy = 0.
29.
(31.1)
Sec. 1.9
Exact DE
41
This has M = (py q), N = 1, so that factor for (31.1) is I(x) = e b)
My Nx = p(x). Consequently, an integrating N
p(t)dt. p(t)dt yields the exact equation p(t)dt
x x
x
Multiplying (31.1) by I(x) = e
e (py q)dx + e dy = 0. Hence, there exists a potential function such that a) x = e
x
x
p(t)dt
p(t)dt
(py q) and b) y = e
x
p(t)dt
. From b),
x
(x, y) = ye p(x) ye
x
p(t)dt
+ h(x) x = p(x) ye
x
p(t)dt
dh(x) + dx so from a),
x
x
p(t)dt
dh(x) + dx = e
x
p(t)dt
(py q)
dh(x) dx = q(x)e
p(t)dt
h(x) = q(x) e
p(t)dtdx where the constant of integration has been set to zero
since we just need one potential function. Consequently, (x, y) =
ye
x
p(t)dt
p(t)dtdx q(x) e
x
y(x) = I
1
x
Iq(t)dt + c ,
where I(x) = e
p(t)dt.
Solutions to 1.10
x
1.
dy 2ln x 2ln x Separable. dx = xy ydy = x dx y2 = 2(ln x)2 + c.
ydy = 2lnxxdx
3.
dy 2xy Exact. dx = 2 2xydx + (x2 + 2y)dy = 0. N = x2 + 2y and M = 2xy x + 2y My = 2x = N x There exists a potential function (x, y) such that a) x = 2xy and b) y = x2 + 2y . From a) (x, y) = yx2 + h(y) dh(y) dh y = x2 + dy so from b), x2 + h(y) = x 2 + 2y dy = 2y h(y) = y2 where the constant of integration has been set to zero since we just need one potential function. (x, y) = x2y + y2 y2 + x2y + c = 0.
42
Ch. 1
First-Order Differential Equations
5.
dy 1 dy 1 Bernoulli. dx + y(tan x + ysin x) = 0 2 dx + tan x y = sin x. y du 1 dy Let u = y1 so that dx = 2 dx . Substituting these results into the last differential equation y du d yields dx tan x u = sin x which has an integrating factor I(x) = cos x. Hence, dx (u cos x) = cos xsin x cos2 x u cos x = cos xsin x dx u cos x = 2 + c1 cos2 x + c2 1 cos2 x 2cos x 1 u(x) = cos x 2 + c1 y = y(x) = . 2cos x sin2 x + c
7.
dy 1 x2 y 2 dy = y + 1 (y/x)2 . First order homogeneous. dx x y = x dx x dy dv Let v = y/x so that dx = v + x dx . Upon substituting these results into the last differential dv dv dx equation results in x dx = 2 = 1 v 2 x 1 v
sin1v = ln x + c1 sin1(y/x) = ln(cx) y(x) = xsin[ln(cx)]. 9.
dy 1 25x2ln x dy 1 25x2ln x Bernoulli. dx + x y = y dx + x y2 = . Let u = y2 so that 2y 2 dy 1 du y dx = 2 dx . Substituting these results into the last equation gives du 2 2 dx + x u = 25x ln x. An integrating factor for this equation is
I(x) = e
2
dx x
d = x2 dx (x2u) = 25x4ln x x2u = 25
x ln x dx
4
x2u = 5x5ln x x5 + c u = x2[x5(5ln x 1) + c] 1 y2 = 2 [x5(5ln x 1) + c]. x 11. dy First order linear. dx + cot x y = sec x. An integrating factor for this equation is given by e
cot x dx =
y sin x =
tan xdx y sin x = ln(csec x) y(x) = ln(csec x) . sin x
d sin x. dx (y sin x) = sin x sec x
Sec. 1.10
Summary of Techniques
43
13.
dy y First order homogeneous. y[ln(y/x) + 1]dx xdy = 0 dx = x [ln(y/x) + 1]. Let dy dv v = y/x so that dx = v + x dx . Substituting these results into the last differential equation dv dv dv dx yields v + x dx = v(ln v + 1) x dx = vln v v ln v = x ln|ln(v)| = ln|x| + c1 ln(v) = cx ln(y/x) = cx y(x) = xecx .
15.
dy First order linear. dx + sin xy = sin x. An integrating factor is given by I(x) = e
sin xdx =
ecos xy =
e
d ecos x. Hence, dx (ecos xy) = ecos xsin x sin x dx ecos xy = ecos x + c y(x) = 1 + cecos x .
cos x
17.
dy 1 dy 1 1 9x2 Bernoulli. 2xln x dx y = 9x3y3ln x 3 dx 2xln x 2 = 2 . Let u = y2 so that y y 1 dy 1 du = 2 dx . Substituting these results into the last differential equation yields y3 dx du 1 2 dx + xln x u = 9x
which has an integrating factor I(x) = e
so that u(x) = But, u = y2, consequently, y2 =
dx xln x 2
d 2 dx (u ln x) = 9x ln x u ln x = 9
x ln xdx u ln x = 3x ln x x + c
3 3
= ln x. Hence,
3x3ln x x3 + c . ln x
ln x . c + x (3ln x 1)
3
19.
dy dy 2 First order linear. (x2 1)( dx 1) + 2y = 0 dx + 2 y = 1. An integrating factor x 1 for this equation is given by I(x) = e
2 dx 2 x 1
d x1 x1 x1 dx x + 1 y = x + 1 x + 1 y = x+1 y(x) = x 1 (x 2ln|x + 1| + c).
x1 = x+1 x1 x1 dx x + 1 y = x 2ln|x + 1| + c x+1
44
Ch. 1
First-Order Differential Equations
21.
(x5 + ym)dx xn y3dy = 0. a) M = x 5 + ym and N = xn y3 My = my m1 and Nx = nxn1 y3. Now the equation will be exact My = N x my m1 = nxn1 y3 m = n = 0. b) If m = 0, then the differential equation can be written as separable for all n R. c) dy x5 + ym Rewrite the differential equation as dx = = f(x, y). Now xn y3 (tx)5 + (ty)m t5x5 + tmym for any t > 0, f(tx, ty) = = n n 3 3 . The only time that (tx)n (ty)3 tx ty f(x, y) can be recovered from the last equation is if m = 5 and n = 2 in which case f(tx, x5 + y5 ty) = 2 3 = f(x, y). xy dy Rewriting the given equation as dx xn ym 3 = x5 n y3 it is clear that there exits no values of m or n that will make this equation linear, because the right hand side can never be made a function of x alone. e) dy If we consider the form dx xn ym 3 = x5 n y3 then we see that a Bernoulli equation will result when m = 4 and n R. x5 + 1 dx y3dy = 0 which is xn
d)
Solutions to 1.11
1. Applying Euler's method with f(x, y) = 4y 1, x0 = 0, y0 = 1, and h = 0.05 we have yn+1 = yn + 0.05(4yn 1). This generates the sequence of approximants given in the table below. n 1 2 3 4 5 6 7 8 9 xn 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 yn 1.15 1.33 1.546 1.805 2.116 2.489 2.937 3.475 4.120
Sec. 1.11
Numerical Solution to First-Order DE
45
10
0.50
4.894
Consequently the Euler approximation to y(0.5) is y10 = 4.894. (Actual value: y(0.5) = 5.792 rounded to 3 decimal places.) 3. Applying Euler's method with f(x, y) = x y2, x0 = 0, y0 = 2, and h = 0.05 we have yn+1 = yn + 0.05(xn yn 2). This generates the sequence of approximants given in the table below. n 1 2 3 4 5 6 7 8 9 10 xn 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 yn 1.80 1.641 1.511 1.404 1.316 1.242 1.180 1.127 1.084 1.048
Consequently the Euler approximation to y(0.5) is y10 = 1.048. (Actual value: y(0.5) = 1.088 rounded to 3 decimal places.) 5. Applying Euler's method with f(x, y) = 2xy2, x0 = 0, y0 = 1, and h = 0.1 we have yn+1 = yn + 0.1xn yn 2. This generates the sequence of approximants given in the table below.
n 1 2 3 4 5 6 7 8 9 10
xn 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
yn 0.5 0.505 0.515 0.531 0.554 0.584 0.625 0.680 0.754 0.856
46
Ch. 1
First-Order Differential Equations
Consequently the Euler approximation to y(1) is y10 = 0.856. (Actual value: y(1) = 1.) 7. Applying the modified Euler method with f(x, y) = 2xy/(1 + x2), x0 = 0, y0 = 1, and h = 0.1 we have y*n+1 = yn 0.2xn yn /(1 + xn 2). yn+1 = yn + 0.05[ 2xn yn /(1 + xn 2) 2xn+1y*n+1/(1 + xn+12)] This generates the sequence of approximants given in the table below. n 1 2 3 4 5 6 7 8 9 10 xn 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 yn 0.9900 0.9616 0.9177 0.8625 0.8007 0.7361 0.6721 0.6108 0.5536 0.5012
Consequently the modified Euler approximation to y(1) is y10 = 0.5012. (Actual value: y(1) = 0.5.) 9. Applying the modified Euler method with f(x, y) = x2y, x0 = 0, y0 = 1, and h = 0.2 we have y*n+1 = yn 0.2xn 2yn , yn+1 = yn 0.1[xn 2yn + (xn+1)2y*n+1]. This generates the sequence of approximants given in the table below. n 1 2 3 4 5 xn 0.2 0.4 0.6 0.8 1.0 yn 0.9960 0.9762 0.9266 0.8382 0.7114
Consequently the modified Euler approximation to y(1) is y5 = 0.7114. (Actual value: y(1) = 0.7165 rounded to 4 decimal places.)
Sec. 1.11
Numerical Solution to First-Order DE
47
11.
We have f(x, y) = 4y 1, x0 = 0, y0 = 1, and h = 0.05. So, 1 k1 = 0.05(4yn 1), k2 = 0.05[4(yn + 2 k1) 1] 1 k3 = 0.05[4(yn + 2 k2) 1], k4 = 0.05[4(yn + k3) 1], 1 yn+1 = yn + 6 (k1 + 2k2 + 2k3 + k4). This generates the sequence of approximants given in the table below (computations rounded to five decimal places).
n 1 2 3 4 5 6 7 8 9 10
xn 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50
yn 1.16605 1.36886 1.61658 1.91914 2.28868 2.74005 3.29135 3.96471 4.78714 5.79167
Consequently the RungeKutta approximation to y(0.5) is y10 = 5.79167. (Actual value: y(0.5) = 5.79179 rounded to 5 decimal places.) 13. We have f(x, y) = x y2, x0 = 0, y0 = 2, and h = 0.05. Hence, k1 = 0.05(xn yn 2), k2 = 0.05[xn + 0.025 (yn + k1/2)2], k3 = 0.05[xn + 0.025 (yn + k2/2)2], k4 = 0.05[xn+1 (yn + k3)]2, 1 yn+1 = yn + 6 (k1 + 2k2 + 2k3 + k4). This generates the sequence of approximants given in the table below (computations rounded to 6 decimal places). n 1 2 3 4 xn 0.05 0.10 0.15 0.20 yn 1.181936 1.671135 1.548079 1.445025
48
Ch. 1
First-Order Differential Equations
5 6 7 8 9 10
0.25 0.30 0.35 0.40 0.45 0.50
1.358189 1.284738 1.222501 1.169789 1.125263 1.087845
Consequently the RungeKutta approximation to y(0.5) is y10 = 1.087845. (Actual value: y(0.5) = 1.087844 rounded to 6 decimal places.) 15. We have f(x, y) = 2xy2, x0 = 0, y0 = 1, and h = 0.1. Hence, k1 = 0.2xn yn 2, k2 = 0.2(xn + 0.05)(yn + k1/2)2, k3 = 0.2(xn + 0.05)(yn + k2/2)2, k4 = 0.2xn+1(yn + k3)2, 1 yn+1 = yn + 6 (k1 + 2k2 + 2k3 + k4). This generates the sequence of approximants given in the table below (computations rounded to 6 decimal places). n 1 2 3 4 5 6 7 8 9 10 xn 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 yn 0.502513 0.510204 0.523560 0.543478 0.571429 0.609756 0.662252 0.735295 0.840336 0.999996
Consequently the RK4 approximation to y(1) is y10 = 0.999996. (Actual value: y(1) = 1.)
Solutions to 1.12
1. d2y 2 dy dy du d2y = x dx + 4x2. Let u = dx so that dx = . Substituting these results into the first dx2 dx2 du 2 du 2 equation yields dx = x u + 4x2 dx x u = 4x2. An appropriate integrating factor for this
2
equation is I = e
dx x
= x2 d(x2u) = 4
Sec. 1.12
Some Higher Order DE
49
x2u = 4
y(x) = c1x3 + x4 + c2. 3. d2y 2 dy 2 dy dy du du d2y + y ( dx ) = dx . Let u = dx so that dx = udy = . Substituting these results into dx2 dx2 du 2 du 2 the first equation yields u dy + y u2 = u u = 0 or dy + y u = 1. An appropriate
dx x
2
dy u = 4x + c1 u = 4x3 + c1x2 dx = 4x3 + c1x2
=y integrating factor for the last equation is I(y) = e d dy dx (y u) = y y u = y dy y u = y /3 + c dx = y/3 + c /y
2 dy y 3 2 2 2 2 2 2 1 1
2
ln|y3 + c2| = x + c3 y(x) = (c4ex + c5)1/3 . 5. d2y dy 2 + tan x dx = dx
( dy ) . dx
2
dy du d2y Let u = dx so that dx = . Substituting these results into the dx2
du 1 first equation yields dx + tan xu = u2 which is a Bernoulli equation. Letting z = u1 gives 2 u du dz dz dx = dx . Substituting these results into the last equation yields dx tan x z = 1. I(x) = e
tan xdx
= cos x is an integrating factor for this equation
sin x + c1 d dx (z cos x) = cos x z cos x = cos x dx z = cos x cos x dy cos x u(x) = c sin x dx = c sin x y(x) = c2 ln|c1 sin x|. 1 1 7. d2y 2 dy dy du d2y x dx = 6x4. Let u = dx so that dx = . Substituting these results into the first dx2 dx2 du 2 equation yields dx x u = 6x4. An appropriate integrating factor for this equation is d = x2 dx (x2u) = 6x2 x2u = 6 x2dx dy 1 u = 2x5 + cx2 dx = 2x5 + cx2 y(x) = 3 x6 + c1x3 + c2. d2y dy dy dy du d2y ( dx )2 dx = 0. Let u = dx so that dx = . Substituting these results into the dx2 dx2 du first equation yields dx u = u2 which is a Bernoulli equation. If u = 0 then y is a
I(x) = e
2 dx x
9.
50
Ch. 1
First-Order Differential Equations
dz constant and satisfies the equation. Now suppose that u 0. Let z = u1 so that dx = u2
dx du dz . Substituting these results into the last equation yields dx + z = . I(x) = e = ex dx d is an integrating factor for the last differential equation dx (exz) = ex ex dy ex exz = exdx z = + cex u = dx = c ex c ex ex 1 x y = x dx y(x) = ln|c1 + c2e |. c e
11.
d2y 2x dy dy du d2y = . Let u = dx so that dx = . If u = 0 then y is a constant and satisfies dx2 1 + x2 dx dx2 the equation. Now suppose that u 0. Substituting these results into the first equation yields c1 c1 du 2x dy 2 = 2 u ln|u| = ln(1 +x ) + c u = 2 dx = dx 1+x 1+x 1 + x2 y(x) = c1tan1 x + c2 where c1 and c2 are constants. d2y dy dy du d2y tan x dx = 1. Let u = dx so that dx = . Substituting these results into the last dx2 dx2 du equation yields dx tan x u = 1. An appropriate integrating factor for this equation is I(x) = e
13.
tan x dx
u cos x = y(x) =
(tan x + csec x)dx y(x) = ln(sec x) + c ln(sec x + tan x) + c .
1 2
cos x dx u(x) = tan x + csec x dy = tan x + csec x dx
d = cos x dx (u cos x) = cos x
15.
d2y dy du du d2y = 2y where a > 0 and > 0. Let u = dx so that dx = udy = . Substituting these dx2 dx2 du results into the first equation yields u dy = 2y u2 = 2y2 + c2. Using the given that y(0) = a and y(0) = 0 we find that c2 = a2 2. dy 1 1 y y2 a2 dx = cosh (a ) = x + c y(x) = acosh[(c x)] y = asinh[(c x)] and since y(0) = 0, c = 0; hence, y(x) = acosh(x).
Sec. 1.12
Some Higher Order DE
51
17.
d2y dy dy d2y du 2 + p(x)dx = q(x). Let u = dx so that 2 = dx . Substituting these results into the first dx dx dy du equation gives us the equivalent system: dx = u and dx + p(x)u = q(x). The second p(x)dx p(x)dx equation of the system has u = e e q(x)dx + c1 as its solution so from the p(x)dx p(x)dx dy first equation of the system it follows that dx = e e q(x)dx + c1 . Thus p(x)dx p(x)dx e e y = q(x)dx + c1dx + c2 is a solution of the original equation
where c1 and c2 are constants. 19. d2 g d 2 + L sin() = 0, (0) = 0 and dt (0) = 0. dt d2 g d d2 du du d du a) + L = 0. Let u = dt so that 2 = dt = d dt = u d . Substituting these 2 dt dt du g results into the last equation yields u d + L = 0 g d g u2 = L 2 + c12, but dt (0) = 0 and (0) = 0 so c12 = L 02 g u2 = L ( 02 2) u =
g L
02 2
sin1( ) =
0
(0) = 0 so c2 = 2 Sin1( ) = 2 0 = 0cos(
g L t = 0sin(2
g L t + c2, but g L t)
g L t). Yes, the predicted motion is reasonable.
b)
d2 g d d2 du du d du + L sin() = 0. Let u = dt so that 2 = dt = d dt = u d . Substituting 2 dt dt du g 2g these results into the last equation yields ud + L sin() = 0 u2 = L cos() + c. d 2g 2g 2g Since (0) = 0 and dt (0) = 0, c = L cos( 0) and so u2 = L cos() L cos( 0) d dt = d dt =
[
2g 2g L cos() L cos( 0) 2g 1/2 L cos() cos( 0)] . L d 2g [cos() cos( )]1/2 = dt. When the pendulum goes from 0
c)
From part b),
52
Ch. 1
First-Order Differential Equations
d = 0 to = 0 (which corresponds to one quarter of a period) dt is negative; hence, choose the negative sign. Thus, T =
L d 2g [cos() cos( )]1/2 d T = 0
0 0
0
L d 2g [cos() cos( )]1/2 d. 0
0
0
d)
T =
L d 2g [cos() cos( )]1/2 d T = 0
0
L d d 2g 2 0 2 1/2 [2sin ( 2 ) 2sin (2 )]
0
0
1 T = 2
L d d. Let k = sin( 20 ) so that 2g 2 0 2 1/2 [sin ( 2 ) sin (2 )]
0
0
1 T = 2
L d . g 2 [k sin2( )]1/2 2
0
0
(18.1)
Now let sin(/2) = ksin(u). When = 0, u = 0 and when = 0, then u = /2; moreover, d =
2kcos(u)du 2k du 1 sin2(u) d = cos(/2) 1 sin2(/2) 22 du 2 du k2 (ksin(u)) k2 sin2(/2) . Making a change of d = d = 1 k2sin2(u) 1 k2sin2(u) variable in equation (18.1) so that the upper and lower limits are changed as is the value of d, then (18.1) becomes 1 T = 2
L 2 du k2 sin2(/2) 2 g [k sin2(/2)]1/2 or 2 2 1 k sin (u)
/2
/2
0
T =
L du g where k = sin( 0/2). 2 2 1 k sin (u)
0
Solutions to 1.13
1. In the phase plane variables u = y, and v = y, the given DE is replaced by the system u = v, v dv u = 16u. To determine the trajectories we must solve du = 16 v. This separable DE has
Sec. 1.13 The Phase Plane
53
general solution v2 + 16u2 = c. Consequently the trajectories are ellipses. A maple generated phase portrait and direction field is given in the figure below.
3.
In the phase plane variables u = y, and v = y, the given DE is replaced by the system u = v, v = 3u2v. We see that each point on the uaxis is an equilibrium point. Hence no trajectory can cross the uaxis. dv To determine the trajectories we must solve du = 3u2. This DE has general solution v = u3 + c. A maple generated phase portrait and direction field is given in the figure below.
Initial conditions: (a) y(0) = 1, y(0) = 2. (b) y(0) = 1, y (0) = 1. (c) y(0) = 1, y(0) = 1. 5. y = 2yy, y(0) = 0, y(0) = 1. If we let u = y, and v = y, then as we have seen in the preceding problem, u and v are related by v = u2 + c. The initial conditions u(0) = 0, v(0) = 1, require that c = 1. Consequently, v = u2 + 1. Returning to the original variables we have dy 2 dt = y + 1. This separable DE has general solution tan1 y = t + c1, or equivalently, y(t) = tan(t + c1).
54
Ch. 1
First-Order Differential Equations
The initial condition y(0) = 0 requires that c1 = 0, so that the solution to the IVP is y(t) = tan t. Consequently, as t ranges from /2 to /2, y ranges from to . Hence, the corresponding trajectory (curve (b) in the preceding figure) is traversed in a finite time interval. 7. In the phase plane variables u = y, and v = y, the given DE is replaced by the system u = v, v = v(2v u). We see that each point on the uaxis is an equilibrium point. Hence no trajectory can cross the uaxis. A maple generated phase portrait and direction field is given in the figure below.
9.
(a) y + y + 0.5y 2 = 0 is a conservative DE with F(y) = y + 0.5y2. Consequently, the potential energy function is V(y) =
Introducing the phase plane variables u = y, and v = y yields the system 1 u = v, v = u 0.5u2 = 2 u(2 + u). We see that (0, 0) and (2, 0) are the only equilibrium points. The equation for the trajectories is 1 2 E 6 u2(u + 3) . A sketch of V(u) and the corresponding phase portrait is given below. (b) The maximum of V(u) occurs when u = 2, and the corresponding maximum value is V(2) = 2/3. Consequently the separatrix between the periodic and non-periodic regimes v=
0
(w + 0.5w ) dw = 1 y (y + 3). 6
2 2
y
Sec. 1.13 The Phase Plane
55
corresponds to an energy level E = 2/3. This constant energy line intersects the potential function again when u = 1. Hence the solutions will be periodic for 0 < < 1.
11.
(a) y + y 0.5y3 = 0 is a conservative DE with F(y) = y 0.5y3. Consequently, the potential energy function is V(y) =
Introducing the phase plane variables u = y, and v = y yields the system 1 u = v, v = u + 0.5u 3 = 2 u(u2 2). We see that (0, 0), ( 2, 0) and ( 2, 0) are the only equilibrium points. The equation for the trajectories is 1 2 E 8 u2(4 u2) . A sketch of V(u) and the corresponding phase portrait is given below. (b) V(u) has two local maximums at u = 2. and a local minimum at u = 0. The corresponding value of V at both of these maximums is V( 2) = 1/2. Consequently the initial condition u(0) = , v(0) = 0 places us onto a periodic trajectory. v=
0
(w 0.5w ) dw = 1 y (4 y ). 8
3 2 2
y
56
Ch. 1
First-Order Differential Equations