Solving ODE with Frobenius Method

[roughwinds]

Homework Statement

Solve for
xy'' + y' +αy + βxy = 0

α and β are constants

The Attempt at a Solution

What I initially had in mind was:
xy'' + y' +αy + βxy = x²y'' + xy' +αxy + βx²y = 0
y = \sum_{n=0}^\infty a_n x^{n}
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}
x²y = \sum_{n=0}^\infty a_n x^{n+2} = \sum_{n=2}^\infty a_{n-2} x^{n}
y' = \sum_{n=1}^\infty na_n x^{n-1}
xy' = \sum_{n=1}^\infty na_n x^{n} = a_1x + \sum_{n=2}^\infty na_n x^{n}
y'' = \sum_{n=2}^\infty n(n-1)a_n x^{n-2}
xy'' = \sum_{n=2}^\infty n(n-1)a_n x^{n}
a_0x + a_1x +\sum_{n=2}^\infty[n(n-1)a_n + na_n + αa_{n-1} + βa_{n-2}]x^{n} = 0
a_1 = - a_0
Recurrence relation for n ≥ 2:
a_n = \frac {-(αa_{n-1}+βa_{n-2})}{n(n-1)+n}
a_n = \frac {-(αa_{n-1}+βa_{n-2})}{n²}
a_2 = \frac {-(αa_{1}+βa_{0})}{4} = \frac {αa_{0}-βa_{0}}{4} = \frac {(α-β)a_{0}}{4}
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}
a_4 = \frac {-(αa_{3}+βa_{2})}{16} =\frac {(9αβ - 9β^{2} + 4αβ - α^{3} + α^{2}β)a_0}{576}
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.

 

[roughwinds]

Frobenius attempt:
y = \sum_{n=0}^\infty a_n x^{n+r}
y' = \sum_{n=0}^\infty a_n(n+r) x^{n+r-1}
y'' = \sum_{n=0}^\infty a_n(n+r)(n+r-1) x^{n+r-2}
x²y'' = \sum_{n=0}^\infty a_n(n+r)(n+r-1) x^{n+r} = a_0r(r-1)x^{r} + a_1r(r+1)x^{r+1} + \sum_{n=2}^\infty a_n(n+r)(n+r-1) x^{n+r}
xy' = \sum_{n=0}^\infty a_n(n+r) x^{n+r} = a_0rx^{r} + a_1(r+1)x^{r+1} + \sum_{n=2}^\infty a_n(n+r) x^{n+r}
xy = \sum_{n=0}^\infty a_n x^{n+r+1} = a_0x^{r+1} + \sum_{n=2}^\infty a_{n-1} x^{n+r}
x²y = \sum_{n=0}^\infty a_n x^{n+r+2} = \sum_{n=2}^\infty a_{n-2} x^{n+r}
a_0r(r-1)x^{r} + a_1r(r+1)x^{r+1} + a_0rx^{r} + a_1(r+1)x^{r+1} + αa_0x^{r+1} + \sum_{n=2}^\infty[a_n(n+r)(n+r-1) + a_n(n+r) + αa_{n-1} + βa_{n-2}]x^{n+r}
a_n = \frac {-(αa_{n-1}+βa_{n-2})}{(n+r)^{2}}
a_0r(r-1)x^{r} + a_1r(r+1)x^{r+1} + a_0rx^{r} + a_1(r+1)x^{r+1} + αa_0x^{r+1} = 0
a_0r(r-1) + a_1r(r+1)x + a_0r + a_1(r+1)x + αa_0x = 0
a_0[r(r-1) + r+αx ] + a_1[r(r+1)x +(r+1)x] = 0
a_0[r²+αx ] + a_1x[(r+1)^{2}] = 0
Then I get stuck. Is there any other method that can solve this kind of ODE?

 

[vela]

$$a_0[r²+αx ] + a_1x[(r+1)^{2}] = 0$$

You want to collect powers of $x$:
$$a_0 r^2 + [\alpha a_0 +(r+1)^2 a_1]x = 0.$$ For this to hold true for all values of $x$, you require each term to vanish. That let's you solve for $r$ and $a_1$. (I didn't check your algebra, but I didn't see any obvious mistakes.)

 

[roughwinds]

You want to collect powers of $x$:
$$a_0 r^2 + [\alpha a_0 +(r+1)^2 a_1]x = 0.$$ For this to hold true for all values of $x$, you require each term to vanish. That let's you solve for $r$ and $a_1$. (I didn't check your algebra, but I didn't see any obvious mistakes.)

$$a_0 r^2 + [\alpha a_0 +(r+1)^2 a_1]x = 0$$
$$a_0r^2 = 0$$
$$r = 0$$
$$\alpha a_0 +(r+1)^2 a_1 = 0$$
$$a_1 = \frac{-\alpha a_0}{(r+1)^2} = -\alpha a_0$$
$$a_n = \frac {-(αa_{n-1}+βa_{n-2})}{(n+r)^{2}}$$
$$a_n = \frac {-(αa_{n-1}+βa_{n-2})}{n^{2}}$$
$$a_2 = \frac {(α^{2}+β)a_0}{4}$$
$$a_3 = \frac {(5αβ -α^{3})a_0}{9*4}$$
$$a_4 = \frac {(α^{4} -6α^{2}β -β^{2})a_0}{16*9*4}$$
The denominators seems to follow the pattern:
$$n^{2}*(n-1)^{2}*(n-2)^{2}*...*9*4$$
But I can't see any with the numerators.
I'll double check everything.