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

From 
$$
(n+2)(n+1)a_{n+2}-4a_n=0 \tag1
$$ you deduce, for $n=0,1,2,\ldots$,
$$
a_{n+2}=\frac4{(n+2)(n+1)}a_n \tag2
$$ giving, with $n:=2p$,
$$
\begin{align}
a_{2(p+1)}&=\frac4{(2p+2)(2p+1)}\:a_{2p}
\\\\&=\frac4{(2p+2)(2p+1)}\cdot \frac4{2p(2p-1)}\:a_{2(p-1)}
\\\\&= \cdots
\\\\&=\frac4{(2p+2)(2p+1)}\cdot \frac4{2p(2p-1)}\cdots \frac4{4\times 3}\:\frac4{2\times 1}\:a_0
\\\\&=\frac{4^{p+2}}{(2(p+1))!}
\end{align}
$$ that is 

$$
a_{2p}=\frac{4^{p+1}}{(2p)!}
$$

as announced. 
Similarly, with $n=2p+1$, one gets 

$$a_{2p+1}=0$$ 

due to $a_1=f'(0)=0$.
