Solve for power series $y'' - 9y = 0$
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}