- #1

cjc0117

- 94

- 1

## Homework Statement

Solve [itex]x(1-x)\frac{d^{2}y}{dx^{2}}-2\frac{dy}{dx}+2y=0[/itex] using the Frobenius Method.

## Homework Equations

[itex]R(x)\frac{d^{2}y}{dx^{2}}+\frac{1}{x}P(x)\frac{dy}{dx}+\frac{1}{x^{2}}V(x)y=0[/itex]

[itex]R_{0}s(s-1)+P_{0}s+V_{0}=0[/itex]

[itex]y=\sum^{∞}_{m=0}a_{m}x^{m+s}[/itex]

[itex]y'=\sum^{∞}_{m=0}a_{m}(m+s)x^{m+s-1}[/itex]

[itex]y''=\sum^{∞}_{m=0}a_{m}(m+s)(m+s-1)x^{m+s-2}[/itex]

## The Attempt at a Solution

First, I divided everything by x. The diff. eq. becomes:

[itex](1-x)\frac{d^{2}y}{dx^{2}}-\frac{2}{x}\frac{dy}{dx}+\frac{2}{x}y=0[/itex]

It follows that [itex]R(x)=1-x[/itex], [itex]P(x)=-2[/itex], and [itex]V(x)=2x[/itex]. Thus, [itex]R_{0}=1[/itex], [itex]P_{0}=-2[/itex], and [itex]V_{0}=0[/itex]. The indicial roots are then [itex]s=0[/itex] and [itex]s=3[/itex].

I then plug [itex]y=\sum^{∞}_{m=0}a_{m}x^{m+s}[/itex] and its derivatives into the original diff. eq. and find the recurrence relation:

[itex](1-x)\sum^{∞}_{m=0}a_{m}(m+s)(m+s-1)x^{m+s-2}-\frac{2}{x}\sum^{∞}_{m=0}a_{m}(m+s)x^{m+s-1}+\frac{2}{x}\sum^{∞}_{m=0}a_{m}x^{m+s}=0[/itex]

[itex]\sum^{∞}_{m=0}a_{m}(m+s)(m+s-1)x^{m+s-2}-\sum^{∞}_{m=0}a_{m}(m+s)(m+s-1)x^{m+s-1}-2\sum^{∞}_{m=0}a_{m}(m+s)x^{m+s-2}+2\sum^{∞}_{m=0}a_{m}x^{m+s-1}=0[/itex]

[itex]\sum^{∞}_{m=0}a_{m+1}(m+s+1)(m+s)x^{m+s-1}-\sum^{∞}_{m=0}a_{m}(m+s)(m+s-1)x^{m+s-1}-2\sum^{∞}_{m=0}a_{m+1}(m+s+1)x^{m+s-1}+2\sum^{∞}_{m=0}a_{m}x^{m+s-1}=0[/itex]

[itex]a_{m+1}(m+s+1)(m+s-2)-a_{m}[(m+s)(m+s-1)-2]=0[/itex]

[itex]a_{m+1}=a_{m}[/itex]

I don't understand what this result means (if I even did it right) and how I would continue from here.