January 27, 2005 11:46
L24-ch09
Sheet number 1 Page number 395
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CHAPTER 9
Mathematical Modeling with Differential Equations EXERCISE SET 9.1 1. y = 9x2 ex = 3x2 y and y(0) = 3 by inspection. 3
2. y = x3 − 2 sin x, y(0) = 3 by inspection. dy dy = c; (1 + x) = (1 + x)c = y dx dx (b) second order; y = c1 cos t − c2 sin t, y + y = −c1 sin t − c2 cos t + (c1 sin t + c2 cos t) = 0
3. (a) first order;
c dy + y = 2 − e−x/2 + 1 + ce−x/2 + x − 3 = x − 1 dx 2 t (b) second order; y = c1 e − c2 e−t , y − y = c1 et + c2 e−t − c1 et + c2 e−t = 0
4. (a) first order; 2
5.
y3 1 dy dy dy dy = y 2 + 2xy , (1 − 2xy 2 ) = y 3 , = y dx dx dx dx 1 − 2xy 2
6. 2x + y 2 + 2xy
dy = 0, by inspection. dx
d 3x ye = 0, ye3x = C, y = Ce−3x dx dy separation of variables: = −3dx, ln |y| = −3x + C1 , y = ±e−3x eC1 = Ce−3x y including C = 0 by inspection
7. (a) IF: µ = e3
dx
= e3x ,
d [ye−2t ] = 0, ye−2t = C, y = Ce2t dt dy = 2dt, ln |y| = 2t + C1 , y = ±eC1 e2t = Ce2t separation of variables: y including C = 0 by inspection
(b) IF: µ = e−2
8. (a) IF: µ = e−4
dt
= e−2t ,
x dx
= e−2x , 2
2 d −2x2 ye = 0, y = Ce2x dx
2 2 dy = 4x dx, ln |y| = 2x2 + C1 , y = ±eC1 e2x = Ce2x y including C = 0 by inspection
separation of variables:
d t ye = 0, y = Ce−t dt dy = −dt, ln |y| = −t + C1 , y = ±eC1 e−t = Ce−t separation of variables: y including C = 0 by inspection
9. µ = e 4dx = e4x , e4x y = ex dx = ex + C, y = e−3x + Ce−4x
(b) IF: µ = e
10. µ = e2
x dx
dt
2
= ex ,
= et ,
2 2 2 d x2 1 2 1 ye = xex , yex = ex + C, y = + Ce−x dx 2 2
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