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

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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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 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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