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## Homework Statement

Solve for

[tex]xy'' + y' +αy + βxy = 0[/tex]

α and β are constants

## The Attempt at a Solution

What I initially had in mind was:

[tex]xy'' + y' +αy + βxy = x²y'' + xy' +αxy + βx²y = 0[/tex]

[tex]y = \sum_{n=0}^\infty a_n x^{n}[/tex]

[tex]xy = \sum_{n=0}^\infty a_n x^{n+1} = \sum_{n=1}^\infty a_{n-1} x^{n} = a_0x + \sum_{n=2}^\infty a_{n-1} x^{n}[/tex]

[tex]x²y = \sum_{n=0}^\infty a_n x^{n+2} = \sum_{n=2}^\infty a_{n-2} x^{n}[/tex]

[tex]y' = \sum_{n=1}^\infty na_n x^{n-1}[/tex]

[tex]xy' = \sum_{n=1}^\infty na_n x^{n} = a_1x + \sum_{n=2}^\infty na_n x^{n}[/tex]

[tex]y'' = \sum_{n=2}^\infty n(n-1)a_n x^{n-2}[/tex]

[tex]xy'' = \sum_{n=2}^\infty n(n-1)a_n x^{n}[/tex]

[tex]a_0x + a_1x +\sum_{n=2}^\infty[n(n-1)a_n + na_n + αa_{n-1} + βa_{n-2}]x^{n} = 0 [/tex]

[tex]a_1 = - a_0[/tex]

Recurrence relation for n ≥ 2:

[tex]a_n = \frac {-(αa_{n-1}+βa_{n-2})}{n(n-1)+n}[/tex]

[tex]a_n = \frac {-(αa_{n-1}+βa_{n-2})}{n²}[/tex]

[tex]a_2 = \frac {-(αa_{1}+βa_{0})}{4} = \frac {αa_{0}-βa_{0}}{4} = \frac {(α-β)a_{0}}{4}[/tex]

[tex]a_3 = \frac {-(αa_{2}+βa_{1})}{9} = \frac {-α\frac {(α-β)a_{0}}{4}+βa_{0}}{9} = \frac {(4β + αβ - α^{2})a_0}{36}[/tex]

[tex]a_4 = \frac {-(αa_{3}+βa_{2})}{16} =\frac {(9αβ - 9β^{2} + 4αβ - α^{3} + α^{2}β)a_0}{576} [/tex]

Sadly I can't see how to proceed from here. Did I mess anything up?

Apparently this has to be solved with Frobenius, I'll edit this thread later with another attempt at this.

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