I'm stuck with finding the recursion relation of this differential equation using the power series method. So I started by setting: $$y(x)=\sum\limits_{n=0}^{\infty}{a_n x^n} \\ e^x=\sum\limits_{n=0}^{\infty}{\dfrac{x^n}{n!}}$$

But when I implemented this in the equation, I got stuck with finding the recursion relation because of the double summation. This is what I have:

$\sum\limits_{n=2}^\infty (n)(n-1) a_n x^{n-2}+\left(\sum\limits_{n=0}^\infty\dfrac{x^n}{n!}\right)\left(\sum\limits_{n=1}^\infty n a_nx^{n-1}\right) - \sum\limits_{n=0}^{\infty}{a_n x^n} =0$

I know there exist the Cauchy product but I don't know how to use it in this case.

PS: I know it is also possible to set $y=\sum\limits_{n=0}^\infty\dfrac{a_nx^n}{n!}$ , but I chose to use $y(x)=\sum\limits_{n=0}^{\infty}{a_n x^n}$.


$$\sum\limits_{n=2}^\infty (n)(n-1) a_n x^{n-2}+\left(\sum\limits_{j=0}^\infty\dfrac{x^j}{j!}\right)\left(\sum\limits_{n=1}^\infty n a_nx^{n-1}\right) - \sum\limits_{n=0}^{\infty}{a_n x^n} =0$$ So the term with coefficient $x^n$ is $$(n+2)(n+1)a_{n+2}-a_n+\left((n+1)a_{n+1}+na_n+\frac{(n-1)a_{n-1}}2+\frac{(n-2)a_{n-2}}{6}+\cdots\right)$$

Starting with $n=0$, we get $$2a_2=a_0-a_1$$

$n=1$ gives $$6a_3=a_1-2a_2-a_1$$

And so the general recursion relation is: $$(n+2)(n+1)a_{n+2}=a_n-\sum_{k=1}^{n+1}\frac{ka_k}{(k-1)!}$$


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