To obtain a second linearly independent solution to

Chapter 8, Problem 44E

(choose chapter or problem)

To obtain a second linearly independent solution to equation (20):

(a) Substitute $w(r, x)$ given in (21) into (20) and conclude that the coefficients

$a_{k}, k \geq 1$, must satisfy the recurrence relation

$(k+r-1)(2 k+2 r-1) a_{k}$

$+[(k+r-1)(k+r-2)+1] a_{k-1}=0$

(b) Use the recurrence relation with $r=1 / 2$to derive the second series solution

$w\left(\frac{1}{2}, x\right)=$

$a_{0}\left(x^{1 / 2}-\frac{3}{4} x^{3 / 2}+\frac{7}{32} x^{5 / 2}-\frac{133}{1920} x^{7 / 2}+\ldots\right)$

Equation Transcription:

$w(r,x)$

${a}_{k},\ k\geq{}1$

$(k+r-1)(2k+2r-1){a}_{k}$

$+[(k+r-1)(k+r-2)+1]{a}_{k-1}=$ 0

$r=1/2$

$w(\frac{1}{2},x)=$

${a}_{0}({x}^{1/2}-\frac{3}{4}{x}^{3/2}+\frac{7}{32}{x}^{5/2}-\frac{133}{1920}{x}^{7/2}+...)$

$r=1$

$w(1,x)$

Text Transcription:

w(r,x)

ak, k \geq 1

(k+r-1)(2k+2r-1)ak

+[(k+r-1)(k+r-2)+1]ak-1=0

r=1/2

w(12,x)=

a0(x1/2-34x3/2+732x5/2-1331920x7/2+...)

r=1

w(1,x)

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