# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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20

Section 2.4

v' + (1-n)p(t)y1 n = (1-n)q(t). Setting v = y1n then yields a linear D.E. for v.

-2 ë dv „ -3dy dy 1 3dv

28. n = 3 so v = y and --------- = -2y — or --- = -— y ------.

dt dt dt 2 dt Substituting this into the D.E. gives

1 3 dv 2 1 3

- — y — + —y = —y . Simplifying and setting

2 dt t t2

-2

y = v then gives the linear D.E.

4 2 1

v' - —v = -—, where m (t) = — and

t t2 t4

4 2 2+5ct5 5 1/2

v(t) = ct + ---- = --------------. Thus y = ±[5t/(2 + 5ct )] ' .

5t 5t

-1 dv 2 dv

29. n = 2 so v = y and - = -y —. Thus the D.E.

dt dt

2 dv 2 dv

becomes -y ------- - ry = -ky or — + rv = k. Hence

dt dt

rt -rt

m(t) = e and v = k/r + ce . Setting v = 1/y then yields the solution.

32. Since g(t) is continuous on the interval

0 < t < 1 we may solve the I.V.P.

y1 + 2y1 = 1, y1(0) = 0 on that interval to obtain

y1 = 1/2 - (1/2)e-2t, 0 < t < 1. g(t) is also continuous

/

for 1 < t; and hence we may solve y2 + 2y2 = 0 to obtain -2t

y2 = ce , 1 < t. The solution y of the original I.V.P.

must be continuous (since its derivative must exist) and hence we need c in y2 so that y2 at 1 has the same value as y1 at 1. Thus

-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 2t 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 + y0

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

22

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.

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