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

ISBN 0-471-31999-6

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5. Replace sect in Problem 7 with e tsint.

7. Since et, cost and sint are solutions of the related

homogenous equation we have

Y(t) = u1et + u2cost + u3sint. Eqs. (10) then are

u/1et + u'2cost + u'3sint = 0

u/1et - u^sint + u'3cost = 0

u/1et - u^cost - u'3sint = sect.

Using Abel's identity, W(t) = cexp(-lpL(t)dt) = cet.

Section 4.4

71

Using the above equations, W(0) = 2, so c = 2 and

sect W1(t)

W(t) = 2et. From Eq.(11), we have u' 1(t) = -

2e

where W1 =

0 cost sint

0 -sint cost

1 -cost -sint

= 1 and thus

1 -t

u, (t) = —e /cost. Likewise 12

sect W2(t) 1

u, = -------------- = - — sect(cost - sint) and

t

2e

sect W3(t)

ui = —

2e

1 ft e-sds

t

2

= -—sect(sint + cost). Thus 2

t e ds 1 1 1 1

, u2 = - —t - —ln(cost) and u3 = - —t + —ln(cost)

2 t0 coss 2 2 2 3 2 2

which, when substituted into the assumed form for Y, yields

the desired solution.

u3 =

11. Since the D.E. is the same as in Problem 7, we may use the complete solution from that, with t0 = 0. Thus

11

y(0) = c1 + c2 = 2, y (0) = c1 + c3 - — + — = -1 and 11

y (0) = c - c2 + — - 1 + — = 1. Again, a computer 1 2 2 2

algebra system may be used to yield the respective

derivatives.

14. Since a fundamental set of solutions of the homogeneous D.E. is y1 = et, y2 = cost, y3 = sint, a particular

solution is of the form Y(t) = etu1(t) + (cost)u2(t) +

(sint)u3(t). Differentiating and making the same

assumptions that lead to Eqs.(10), we obtain t

u1e + u2cost + u3sint = 0 t

u1e - u2sint + u3cost = 0 t

u1e - u2cost - u3sint = g(t)

Solving these equations using either determinants or by

/

1

u2 = (1/2)(sint - cost)g(t),u3 = -(1/2)(sint + cost)g(t) Integrating these and substituting into Y yields

elimination, we obtain u1 = (1/2)e tg(t),

72

Section 4.4

Y(t) = —{et[ e sg(s)ds + cost l (sins - coss)g(s)ds 2 Jto Jto

-sintl (sins + coss)g(s)ds}.

Jto

This can be written in the form

i* t

i t s

Y(t) = (—/2)l (e + costsins - costcoss

to

-sintsins - sintcoss)g(s)ds.

If we use the trigonometric identities sin(A-B) = sinAcosB - cosAsinB and cos(A-B) = cosAcosB + sinAsinB, we obtain the desired result. Note: Eqs.(——) and (—2) of this section give the same result, but it is not recommended to memorize these equations.

— 6. The particular solution has the form Y = etu—(t) +

t 2 t

te u2(t) + t e u3(t). Differentiating, making the same assumptions as in the earlier problems, and solving the three linear equations for u—, u^, and u"3 yields

u— =(—/2)t2e-tg(t), u'2 = -te-tg(t) and u"3 = (—/2)e-tg(t).

Integrating and substituting into Y yields the desired solution. For instance

t — l"t -n-„„ (t-s),

tetu2 = -tet[ se sg(s)ds = [ 2tse(t s)g(s)ds, and

2 Jt0 ^t

likewise for u3 and u3. If g(t) = t 2et then g(s) = es/s2 and the integration is accomplished using the power rule. Note that terms involving t0 become part of the

complimentary solution.

73

CHAPTER 5

Section 5.1, Page 237

2. Use the ratio test:

I (n + 1)xn+1/2n+1| n+1 1|i |x|

lim ------------------- = lim — | x | = .

n ^ » |nxn/2n| n ^ “ n 2 2

Therefore the series converges absolutely for |x| < 2.

For x = 2 and x = -2 the nth term does not approach zero as n ^ ^ so the series diverge. Hence the radius of convergence is p = 2.

5. Use the ratio test:

| (2x+1) n+1/(n+1) 2| n2 I || |

lim ---------------------- = lim | 2x+1 | = 12x+1 | .

n ^ ~ | (2x+1) n/n2| n ^ ~ (n+1) 2

Therefore the series converges absolutely for |2x+1| < 1, or |x+1/2| < 1/2. At x = 0 and x = -1 the series also converge absolutely. However, for |x+1/2| > 1/2 the series diverges by the ratio test. The radius of convergence is p = 1/2.

9. For this problem f(x) = sinx. Hence f'(x) = cosx, f"(x) = -sinx, f'"(x) = -cosx,... . Then f(0) = 0,

f'(0) = 1, f"(0) = 0, f'"(0) = -1,... . The even terms in the series will vanish and the odd terms will alternate

in sign. We obtain sinx = (-1)n x2n+1/(2n+1)!. From

n=0

the ratio test it follows that p = ^.

2

12. For this problem f(x) = x . Hence f (x) = 2x, f (x) = 2,

and f(n)(x) = 0 for n > 2. Then f(-1) = 1, f'(-1) = -2,

22 f (-1) = 2 and x = 1 - 2(x+1) + 2(x+1) /2! = 1 - 2(x+1) +

(x+1) . Since the series terminates after a finite number of terms, it converges for all x. Thus p = ^.

13. For this problem f(x) = lnx. Hence f'(x) = 1/x,

f"(x) = -1/x2, f"' (x) = 1-2/x3,... , and f (n)(x) =

(-1) n+1(n-1)!/xn. Then f(1) = 0, f'(1) = 1, f"(1) = -1,

f"'(1) = 1-2,... , f(n)(1) = (-1) n+1(n-1)! The Taylor

23

series is lnx = (x-1) - (x-1) /2 + (x-1) /3 - ... =

I

n=1

(-1) n+1(x-1) n/n. It follows from the ratio test that

CO

the series converges absolutely for |x-l| < 1. However, the series diverges at x = 0 so p = 1.

18.

19.

23.

25.

Writing the individual terms of y, we have

2 n

y = ao + a1x + a2x +...+aax + ..., so

y' = a1 + 2a2x + 3a3x2 + ... + (n+1)an+1xn + ..., and

y" = 2a2 + 3'2a3x + 4-3a4x2 + ... + (n+2)(n+1)an+2xn + ... .

If y" = y, we then equate coefficients of like powers of x to obtain 2a2 = a0, 3-2a3 = a1, 4-3a4 = a2, ... (n+2)(n+1)an+2 =

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