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

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u1 + u2cost + u3sint = 0 (a).

Continuing this process we obtain Y" = -u2cost - u3sint,

... / /

Y = u2sint - u3cost - u2cost - u3sint and

-u^sint + u^cost = 0 (b).

Substituting Y and its derivatives, as given above, into

the D.E. we obtain the third equation:

-u^cost - u^sint = tant (c).

Equations (a), (b) and (c) constitute Eqs.(10) of the

text for this problem and may be solved to give

2

u1 = tant, u2 = -sint, and u3 = -sin t/cost. Thus

u1 = -lncost, u2 = cost and u3 = sint - ln(sect + tant) and Y = -lncost + 1 - (sint)ln(sect + tant). Note that the constant 1 can be absorbed in c1.

4. Replace tant in Eq. (c) of Prob. 1 by sect and use Eqs. (a) and (b) as in Prob. 1 to obtain u1 = sect, u'2 = -1 and u'3 = -sint/cost.

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(-lp1(t)dt) = cet.

Section 4.4

71

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

sect W—(t)

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

2e

where W1 =

0 cost sint

0 -sint cost

1 -cost -sint

= 1 and thus

r — -t u— (t) = —e /cost. Likewise

12

sect W2(t) —

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

t

2e

sect W3(t)

u— = —

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) = c— + c2 = 2, y(0) = c— + 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 y— = 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

u—e + u2cost + u3sint = 0 t

u —e - u2sint + u 3cost = 0 t

u —e - u2cost - u 3sint = 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 u— = (1/2)e tg(t),

72

Section 4.4

Y(t) = —{etf e sg(s)ds + cost f (sins - coss)g(s)ds 2 t0 t0

-sintl (sins + coss)g(s)ds}. t0

This can be written in the form t

t-s

Y(t) = (1/2)l (e + costsins - costcoss t0

-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.(11) and (12) of this section give the same result, but it is not recommended to memorize these equations.

16. The particular solution has the form Y = etu1(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 u1, u2, and u’3 yields

u1 =(1/2)t2e-tg(t), u'2 = -te-tg(t) and u’3 = (1/2)e-tg(t). Integrating and substituting into Y yields the desired solution. For instance

t 1 l"t (t-s) _

tetu2 = -tetf se sg(s)ds = -— f 2tse(t s)g(s)ds, and 2 t0 2 t

likewise for u1 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 ³ ³ |x|

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

n ^ ro |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 ^ ro so the series diverge. Hence the radius of convergence is p = 2.

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