Nonhomogeneous Power Series Solution

[Telemachus]

Hi. I have to solve: y''+xy'-2y=e^x
Using series. So, this is what I did:
y(x)=\sum_0^{\infty}a_n x^n
y'(x)=\sum_1^{\infty}n a_n x^{n-1}
y''(x)=\sum_2^{\infty}n(n-1) a_n x^{n-1}
And e^x=\sum_0^{\infty}\frac{x^n}{n!}
Then, using that m=n-2 for y'' and then replacing in the diff. eq:
\sum_0^{\infty}(n+2)(n+1)a_{n+2} x^n+x\sum_1^{\infty}n a_n x^{n-1}-2\sum_0^{\infty}a_n x^n=\sum_0^{\infty}\frac{x^n}{n!}
So:
2a_2-2a_0-1+\sum_1^{\infty}\left [(n+2)(n+1)a_{n+2}+n a_n -2a_n -\frac{1}{n!} \right ] x^n=0
Then: 2a_2-2a_0-1=0\rightarrow a_2=1/2+a_0
And: (n+2)(n+1)a_{n+2}+n a_n -2a_n -\frac{1}{n!}=0\rightarrow a_{n+2}=\frac{a_n(2-n)+\frac{1}{n!}}{(n+2)(n+1)}
Now I have to find the recurrence relation for a_n
Is this ok?
Now, I've tried some terms to find the recurrence relation, but I couldn't find it yet.
Even:
a_2=a_{0+2}=1/2+a_0
a_4=a_{2+2}=\frac{a_2(0)+1/2}{4.3}=1/4!
a_6=a_{4+2}=\frac{a_4(-2)+1/4!}{6.5}=-\frac{1}{6!}
a_8=a_{6+2}=\frac{a_6(-4)+1/6!}{8.7}=\frac{5}{8!}
a_{10}=a_{8+2}=\frac{a_8(-6)+1/8!}{10.9}=\frac{-29}{10!}

Odd:
a_3=a_{1+2}=\frac{a_1+1}{6}=\frac{a_1+1}{3!}
a_5=a_{3+2}=\frac{a_3(-1)+1/3!}{5.4}=-a_3+1/5!=-\frac{a_1+1}{5!}+1/5!=-\frac{a_1}{5!}
a_7=a_{5+2}=\frac{a_5(-2)+1/5!}{7.6}=\frac{-3a_1+1}{7!}
a_9=a_{7+2}=\frac{a_7(-5)+1/7!}{9.8}=\frac{15a_1-4}{9!}

 

[tiny-tim]

Hi Telemachus!

(n+2)(n+1)a_{n+2}+n a_n -2a_n -\frac{1}{n!}=0

Multiply throughout by n! ?

 

[Telemachus]

Oh, I think that's a good idea. Thank you tiny-tim.