# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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-2 -2 2 ce = 1/2 - e /2 or c = (1/2)(e -1) and we obtain

y =

y' =

1/2 - (1/2)e 2t 0 < t < 1

2 -2t and 1/2(e2-1)e 1 < t

e-2t 0 < t < 1

2 -2t

(1-e )e 1 < t.

Evaluating the two parts of y' at t0 = 1 we see that they are different, and hence y' is not continuous at t0 = 1.

Section 2.5

21

Section 2.5, Page 84

Problems 1 through 13 follow the pattern illustrated in Fig.2.5.3 and the discussion following Eq.(11).

dy

3. The critical points are found by setting — equal to

dt

zero. Thus y = 0,1,2 are the critical points. The graph

of y(y-1)(y-2) is positive for O < y < 1 and 2 < y and

negative for 1 < y < 2. Thus y(t) is increasing

dy

(— > 0) for 0 < y < 1 and 2 < y and decreasing dt

dy

(— < 0) for 1 < y < 2. Therefore 0 and 2 are unstable dt

critical points while 1 is an asymptotically stable critical point.

dy dy

6. — is zero only when arctany is zero (ie, y = 0). > 0

dt dt

dy

for y < 0 and — < 0 for y > 0. Thus y = 0 is an dt

asymptotically stable critical point.

dy

7c. Separate variables to get --------------- = kt. Integration

(1-y)2

1 1 kt + c - 1

yields ------ = kt + c, or y = 1 - ------------- = .

1-y kt + c kt + c

c-1

Setting t = 0 and y(0) = y0 yields y0 = ------------- or

c

1 (1-y0)kt + y 0

c = ------. Hence y(t) = ---------------------. Note that for

1-y0 (1-y0)kt + 1

y0 < 1 y ^ (1-y0)k/(1-y0)k = 1 as t ^ •. For y0 > 1 notice that the denominator will have a zero for some value of t, depending on the values chosen for y0 and k. Thus the solution has a discontinuity at that point.

dy

9. Setting — = 0 we find y = 0, ± 1 are the critical dt

dy dy

points. Since — > 0 for y < -1 and y > 1 while ------------------------------------------------------ < 0

dt dt

for -1 < y < 1 we may conclude that y = -1 is

asymptotically stable, y = 0 is semistable, and y = 1 is

unstable.

22

Sect ion

2 . 5

2 2

11. y = b /a and y = 0 are the only critical points. For 2 . 2 dy

0 < y < b /a , — < 0 and thus y = 0 is asymptotically dt

22

stable. For y > b /a , dy/dt > 0 and thus y = b2/a2 is unstable.

14. If f/(y1) < 0 then the slope of f is negative at y1 and thus f(y) > 0 for y < y1 and f(y) < 0 for y > y1 since f(y1) = 0. Hence y1 is an asysmtotically stable critical point. A similar argument will yield the result for f/(y1) > 0.

16b. By taking the derivative of y ln(K/y) it can be shown that dy

the graph of — vs y has a maximum point at y = K/e. Thus dt

dy

— is positive and increasing for 0 < y < K/e and thus y(t) dt

dy

is concave up for that interval. Similarly — is positive

dt

and decreasing for K/e < y < K and thus y(t) is concave down for that interval.

16c. ln(K/y) is very large for small values of y and thus

(ry)ln(K/y) > ry(1 - y/K) for small y. Since ln(K/y) and (1 - y/K) are both strictly decreasing functions of y and since ln(K/y) = (1 - y/K) only for y = K, we may conclude dy

that — = (ry)ln(K/y) is never less than dt

dy

— = ry(1 - y/K). dt

u dy ,,du

17a. If u = ln(y/K) then y = Ke and — = Ke — so that the

dt dt

D.E. becomes du/dt = -ru.

2

18a. The D.E. is dV/dt = k - apr . The volume of a cone of

height L and radius r is given by V = pr L/3 where

L = hr/a from symmetry. Solving for r yields the desired solution.

2

18b.Equilibrium is given by k - apr = 0.

18c. The equilibrium height must be less than h.

Section 2.6

23

20b. Use the results of Problem 14.

20d. Differentiate Y with respect to E. dy

21a. Set — = 0 and solve for y using the quadratic formula. dt

21b. Use the results of Problem 14.

21d. If h > rK/4 there are no critical points (see part a) and dy

— < 0 for all t. dt

1 dx x dn

24a. If z = x/n then dz/dt = — — - — —. Use of

n dt n2 dt

Equations (i) and (ii) then gives the I.V.P. (iii).

vdz

24b. Separate variables to get -------------- = - pdt. Using

z(1-vz)

dz dz

partial fractions this becomes — + ------------ = -pdt.

z 1-z

Integration and solving for z yields the answer.

24c. Find z(20).

26a. Plot dx/dt vs x and observe that x = p and x = q are critical points. Also note that dx/dt > 0 for x < min(p,q) and x > max(p,q) while dx/dt < 0 for x between min(p,q) and max(p,q). Thus x = min(p,q) is an asymptotically stable point while x = max(p,q) is unstable. To solve the D.E., separate variables and use

1 dx dx

partial fractions to obtain --------- [--------- - -] = adt.

q-p q-x p-x

Integration and solving for x yields the solution.

dx

26b. x = p is a semistable critical point and since — > 0,

dt

x(t) is an increasing function. Thus for x(0) = 0, x(t) approaches p as t —— •. To solve the D.E., separate variables and integrate.

Section 2.6, Page 95

2 2 2 3. M(x,y) = 3x -2xy+2 and N(x,y) = 6y -x +3, so My = -2x = Nx

and thus the D.E. is exact. Integrating M(x,y) with

24

Section

2 . 6

32

respect to x we get y(x,y) = x - xy + 2x + H(y).

Taking the partial derivative of this with respect to y

and setting it equal to N(x,y) yields -x2+h'(y) =

6y2-x2+3, so that h'(y) = 6y2 + 3 and h(y) = 2y3 + 3y. Substitute this h(y) into y(x,y) and recall that the equation which defines y(x) implicitly is y(x,y) = c.

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