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

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5. Use the ratio test:

| (2x+1) n+1/(n+1) 2| n2 ³ żż ³

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

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

ro

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

n=0

the ratio test it follows that p = ro.

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 = ro.

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

OO

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 + axx + a2x +...+aax + ..., so

y' = a! + 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 =

an, which yields the desired result for n = 0,1,2,3... .

Set m = n-1 on the right hand side of the equation. Then n = m+1 and when n = 1, m = 0. Thus the right hand side

becomes am(x-1) m+1, which is the same as the left hand

m=0

side when m is replaced by n.

Multiplying each term of the first series by x yields *x nanxn 1 = nanxn = nanxn, where the last

n=1 n=1 n=0

equality can be verified by writing out the first few terms. Changing the index from k to n (n=k) in the second series then yields

X nanxn + Xanxn = X (n+1)anxR.

nn n=0 n=0 n=0

m(m-1)amxm 2 + x^^ kakxk 1 =

m=2 k=1

X (n+2)(n+1)an+2xn + X kakxk =

n=0 k=1

X, [(n+2)(n+1)an+2 + nan]xn. In the first case we have

n=0

let n = m - 2 in the first summation and multiplied each term of the second summation by x. In the second case we have let n = k and noted that for n = 0, nan = 0.

28. If we shift the index of summation in the first sum by

Section 5.2

75

letting m = n-1, we have

nanxn-1 = (m+1)am+1xm. Substituting this into the

n=1 m=0

given equation and letting m = n again, we obtain:

1 (n+1)an+1 xn + 2 1 anxn = 0, or

n=0 n=0

1 [(n+1)an+1 + 2an]xn = 0.

n=0

Hence an+]_ = -2an/(n+1) for n = 0,1,2,3,... . Thus

2 3

a1 = -2a0, a2 = -2a1/2 = 2 a0/2, a3 = -2a2/3 = -2 a0/2-3 =

-23a0/3!... and an = (-1) n2na0/n!. Notice that for n = 0 this formula reduces to a0 so we can write

1 anxn = 1 (-1) n2n a0xn/n! = a01 (-2x) n/n! = a0e-2x

n=0 n=0 n=0

Section 5.2, Page 247

2. y = 1 anxn; y' = I nanxn 1 and since we must multiply

n=0 n=1

y' by x in the D.E. we do not shift the index; and

y" = n(n-1)anxn-2 = (n+2)(n+1)an+2xn. Substituting

n=2 n=0

in the D.E., we obtain

1 (n+2)(n+1)an+2xn - nanxn - anxn = 0. In order to

n=0 n=1 n=0

have the starting point the same in all three summations, we let n = 0 in the first and third terms to obtain the following

(2-1 a2 - a0)x0 + 1 [(n+2)(n+1)an+2 - (n+1)an]xn = 0.

n=1

Thus an+2 = an/(n+2) for n = 1,2,3,... . Note that the recurrence relation is also correct for n = 0. We show how to calculate the odd a's:

oo

oo

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