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^{2}-1)y'' + 3xy' + xy = 0

## Homework Equations

let y = [tex]\Sigma[/tex] (from [tex]\infty[/tex] to n=0) C

_{n}x

^{n}

let y' = [tex]\Sigma[/tex] (from [tex]\infty[/tex] to n=1) nC

_{n}x

^{n-1}

let y'' = [tex]\Sigma[/tex] (from [tex]\infty[/tex] to n=2) n(n-1)C

_{n}x

^{n-2}

## The Attempt at a Solution

I wrote the differential eq as: x

^{2}y''-y''+3xy'+xy=0

Substituting back into the differential eq and multiplying the x

^{2}gives:

[tex]\Sigma[/tex] (from [tex]\infty[/tex] to n=2) n(n-1)C

_{n}x

^{n}- [tex]\Sigma[/tex] (from [tex]\infty[/tex] to n=2) n(n-1)C

_{n}x

^{n-2}+ 3[tex]\Sigma[/tex] (from [tex]\infty[/tex] to n=1) nC

_{n}x

^{n}+ [tex]\Sigma[/tex] (from [tex]\infty[/tex] to n=0) C

_{n}x

^{n+1}= 0

For the first 2 terms I let k=n-2 , n=k+2 which would give:

[tex]\Sigma[/tex] (from [tex]\infty[/tex] to k=0) (k+2)(k+1)C

_{k+2}x

^{k+2}- [tex]\Sigma[/tex] (from [tex]\infty[/tex] to k=0) (k+2)(k+1)C

_{k+2}x

^{k}

For the 3rd and 4th term I let k=n which would give:

3[tex]\Sigma[/tex] (from [tex]\infty[/tex] to k=1) kC

_{k}x

^{k}+ [tex]\Sigma[/tex] (from [tex]\infty[/tex] to k=0) C

_{k}x

^{k+1}

The new series would be:

[tex]\Sigma[/tex] (from [tex]\infty[/tex] to k=0) (k+2)(k+1)C

_{k+2}x

^{k+2}- [tex]\Sigma[/tex] (from [tex]\infty[/tex] to k=0) (k+2)(k+1)C

_{k+2}x

^{k}+ 3[tex]\Sigma[/tex] (from [tex]\infty[/tex] to k=1) kC

_{k}x

^{k}+ [tex]\Sigma[/tex] (from [tex]\infty[/tex] to k=0) C

_{k}x

^{k+1}

I don't know how to make the starting value of each index the same (from [tex]\infty[/tex] to k=0) and how would I get to the general solution?