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

ISBN 0-471-31999-6

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This is a first order linear D.E. for v with integrating

50

Section 3 . 5

factor m(x) = exp[J (1-N/x)dx] = x Nex. Hence

, -N x /, / _ , / N -x , . ,

(x e v ) = 0, and v = cx e which gives v(x) = cJxNe xdx + k. On taking k = 0 we obtain as the second solution y2(x) = cexJ xNe-xdx. The integral can be evaluated by using the method of integration by parts.

At each stage let u = xN or xN 1, or whatever the power of x that remains, and let dv = e x. Note that this dv is not related to the v(x) in y2(x). For N = 2 we have

-x -x

y2(x) = cexJ x2e-xdx = cex[x2—— - J 2x——dx] 2 -1 -1

-x -x

2 x e c e

= -cx + ce [2x - J 2 dx]

-1 -1

22 = c(-x - 2x - 2) = -2c(1 + x + x /2!).

Choosing c = -1/2! gives the desired result. For the

general case c = - 1/N!

33. (y2/y1)' = (y1y,2 - y1y2)/y12 = W(y1,y2)/y12. Abel's identity

is W(y1,y2) = c exp[-l tp(r)dr]. Hence

Jt0

(y2/y1)' = cy1-2 exp[-lt p(r)dr]. Integrating and setting

c = 1 (since a solution y2 can be multiplied by any constant) and taking the constant of integration to be zero we obtain

exp[-J sp(r)dr]

y2(t) = y1(t) Jt----------—---------- ds.

[y1(s) ] 2

35. From Problem 33 and Abel's formula we have

y2 exp[J (1/t)dt] elnt 2 2

(—) = = = tcsc (t ). Thus

y ? 2/,2\ i 2 / , 2»

y1 sin (t ) sin (t )

22 y2/y1 = -(1/2)cot(t ) and hence we can choose y2 = cos(t )

22

since y1 = sin (t ).

r t r t

The general solution of the D.E. is y = c1e 1 + c2e 2

V2 2

b -4ac )/2a provided b - 4ac n 0.

2

In this case there are two possibilities. If b - 4ac > 0 2 1/2

then (b - 4ac) < b and r1 and r2 are real and

r t r t

negative. Consequently e 1 ^ 0 and e 2 ^ 0; and hence

Section 3.6 51

2

y ^ 0, as t ^ • . If b - 4ac < 0 then r1 and r2 are complex conjugates with negative real part. Again

r t r t

e 1 ^ 0 and e 2 ^ 0; and hence y ^ 0, as t ^ • .

2 r t r t

Finally, if b - 4ac = 0, then y = c1e 1 + c2te 1 where

r1 = -b/2a < 0. Hence, again y ^ 0 as t ^ • . This

conclusion does not hold if either b = 0 (since y(t) = c1coswt + c2sinwt) or c = 0 (since y^t) = c1).

42. Substituting z = lnt into the D.E. gives

d2y dy

+ — + 0.25y = 0, which has the solution

dz2 dz

I \ -z/2 -z/2 . ... 4.-1/2 4.-1/2-, 4_

y(z) = c1e + c2ze so that y(t) = c1t + c2t lnt

Section 3.6, Page 178

1. First we find the solution of the homogeneous D.E., which

2

has the characteristic equation r -2r-3 = (r-3)(r+1) = 0.

3t -t 2t

Hence yc = c1e + c2e and we can assume Y = Ae for the

2t 2t

particular solution. Thus Y = 2Ae and Y" = 4Ae and substituting into the D.E. yields

2t 2t 2t 2t

4Ae - 2(2Ae ) - 3(Ae ) = 3e . Thus -3A = 3 and

3t -t 2t

A = -1, yielding y = c e + c e - e .

12

4. Initially we assume Y = A + Bsin2t + Ccos2t. However,

since a constant is a solution of the related homogeneous

D.E. we must modify Y by multiplying the constant A by t

and thus the correct form is Y = At + Bsin2t + Ccos2t.

-t -t 2 -t

6. Since yc = c1e + c2te we must assume Y = At e , so

-t 2 -t -t -t 2 -t

that Y = 2Ate - At e and y" = 2Ae - 4Ate + At e .

2 -t

Substituting in the D.E. gives (At -4At+2A)e +

2 -t 2 -t -t

2(-At +2At)e + At e = 2e . Notice that all terms on

2

the left involving t and t add to zero and we are left

-t -t 2 -t

with 2A = 2, or A = 1. Hence y = c1e + c2te + t e .

8. The assumed form is Y = (At + B)sin2t + (Ct + D)cos2t, which is appropriate for both terms appearing on the right side of the D.E. Since none of the terms appearing

52

Section 3 . 6

in Y are solutions of the homogeneous equation, we do not need to modify Y.

rt

11. First solve the homogeneous D.E. Substituting y = e

gives r2 + r + 4 = 0. Hence yc = e [c1cos^ 15 t/2) +

I t -t

c2sin(y 15 t/2)]. We replace sinht by (e - e )/2 and

t -t t -t

then assume Y(t) = Ae + Be . Since neither e nor e are solutions of the homogeneous equation, there is no need to modify our assumption for Y. Substituting in the

t -t t -t

D.E., we obtain 6Ae + 4Be = e - e . Hence, A = 1/6

and B = -1/4. The general solution is

-t/2 ,--- , t -t

y = e [c1cos^ 15 t/2) + c2sin(y 15 t/2)] + e /6 - e /4.

[For this problem we could also have found a particular solution as a linear combination of sinht and cosht:

Y(t) = Acosht + Bsinht. Substituting this in the D.E. gives (5A + B)cosht + (A + 5B)sinht = 2sinht. The solution is A = -1/12 and B = 5/12. A simple calculation

t -t

shows that -(1/12)cosht + (5/12)sinht = e /6 - e /4.]

-2t t

13. yc = c1e + c2e so for the particular solution we

assume Y = At + B. Since neither At or B are solutions of the homogeneous equation it is not necessary to modify the original assumption. Substituting Y in the D.E. we obtain 0 + A -2(At+B) = 2t or -2A = 2 and A-2B = 0.

-2t t

Solving for A and B we obtain y = c1e + c2e - t - 1/2

as the general solution. y(0) = 0 ^ c1 + c2 - 1/2 = 0

and y'(0) = 1 ^ -2c1 + c2 - 1 = 1, which yield c1 = -1/2

t -2t

and c2 = 1. Thus y = e - (1/2)e - t - 1/2.

16. Since the nonhomogeneous term is the product of a linear polynomial and an exponential, assume Y of the same form:

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