Prove the solution of $f''(x)-4f(x)=0$ is $f(x)=\sum_{p=0}^{\infty} \frac{4^{p+1}}{(2p)!}x^{2p}$ I'm wondering about this question :
We have the differential equation $f''(x)-4f(x)=0$ and we want to find $f$ as a power serie with $f(0)=4$ and $f'(0)=0$. I would like to prove the only solution is $f(x)=\sum_{p=0}^{\infty} \frac{4^{p+1}}{(2p)!}x^{2p}$ as well as express it using usual functions.
With the usual method to solve differential equations with power series I arrive to a recursive relation $(n+2)(n+1)a_{n+2}-4a_n=0$. How should we proceed to come to the final result ?
Thank you

Multiply by $n!$ to get
$$
(n+2)!a_{n+2}=4(n!a_n)
$$
Divide by $2^{n+2}$ to get
$$
2^{-(n+2)}(n+2)!a_{n+2}=2^{-n}n!a_n
$$
which has to be a constant expression for the odd and even indices.
