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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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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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