Solve for power series $y'' - 9y = 0$ I really need help with this, the solution from this equation is $y(x) = c_1 e^{3x} + c_2 e^{-3x}$. But I can't get to it, I obtain the next:
$$y(x) = \sum_{n=0}^{\infty}a_nx^n$$
$$y''(x) = \sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}$$
Then the coefficients must be $a_{2m} = \frac{9^m a_0}{(2m)!}$ and $a_{2m+1}= \frac{9^ma_1}{(2m+1)!}$
Substituting in the first equation I have:
$$y(x) = a_0 \sum_{m=0}^{\infty} \frac{(3x)^{2m}}{(2m)!} + a_1\sum_{m=0}^{\infty} \frac{3^{2m}x^{2m+1}}{(2m+1)!} $$
Since $e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n!}$, I think its obvious that the first part of the last equation is $a_0e^{3x}$, but in the second part I dont really know how to get $a_1e^{-3}$. I am wrong? 
 A: Note that $a_0=y(0)=c_1+c_2$ and $a_1=y'(0)=3c_1-3c_2$. Then
\begin{align}
y(x) &= a_0 \sum_{m=0}^{\infty} \frac{(3x)^{2m}}{(2m)!} + a_1\sum_{m=0}^{\infty} \frac{3^{2m}x^{2m+1}}{(2m+1)!}\\ \ \\
&= (c_1+c_2) \sum_{m=0}^{\infty} \frac{(3x)^{2m}}{(2m)!} + (3c_1-3c_2)\sum_{m=0}^{\infty} \frac{3^{2m}x^{2m+1}}{(2m+1)!}\\ \ \\
&= c_1\,\left(\sum_{m=0}^{\infty} \frac{(3x)^{2m}}{(2m)!} + 3\sum_{m=0}^{\infty} \frac{3^{2m}x^{2m+1}}{(2m+1)!}\right)+c_2\,\left(\sum_{m=0}^{\infty} \frac{(3x)^{2m}}{(2m)!} - 3\sum_{m=0}^{\infty} \frac{3^{2m}x^{2m+1}}{(2m+1)!}\right)\\ \ \\
&= c_1\,\left(\sum_{m=0}^{\infty} \frac{(3x)^{2m}}{(2m)!} + \sum_{m=0}^{\infty} \frac{3^{2m+1}x^{2m+1}}{(2m+1)!}\right)+c_2\,\left(\sum_{m=0}^{\infty} \frac{(-3x)^{2m}}{(2m)!}  \sum_{m=0}^{\infty} \frac{(-3x)^{2m+1}}{(2m+1)!}\right)\\ \ \\
&=c_1\,e^{3c}+c_2\,e^{-3x}.
\end{align}
A: From 
$$y(x) = a_0 \sum_{m=0}^{\infty} \frac{(3x)^{2m}}{(2m)!} + a_1\sum_{m=0}^{\infty} \frac{3^{2m}x^{2m+1}}{(2m+1)!} $$
it can be seen that $y(x)$ can be reduced to
$$y(x) = a_0 \, \cosh(3x) + \frac{a_1}{3} \, \sinh(3x). $$
This can be developed using
\begin{align}
\cosh(x) &= \frac{e^{x} + e^{-x}}{2} = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \\
\sinh(x) &= \frac{e^{x} - e^{-x}}{2} = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}.
\end{align}
