Answer
See below
Work Step by Step
Given: $y''-y=0$
Let $y=\sum a_nx^n$, obtain:
$y''-y$$=\sum k(k-1)(k+2)a_{k}x^{k-2}-\sum a_{k}x^k\\
=(k+1)(k+2)a_{k+2}x^{k}-\sum a_{k}x^k\\
=\sum [(k+1)(k+2)a_{k+2}-a_k]x^k$
Comparing $x^k$ on both sides:
$a_{k+2}=\frac{a_{k}}{(k+1)(k+2)}$
It gives:
$a_2=\frac{1}{1.2}a_0\\
a_4=\frac{1}{3.4}a_2=\frac{1}{1.2.3.4}a_0\\
a_{2k}=\frac{a_0}{(2k)!}$
and:
$a_3=\frac{1}{2.3}a_1\\
a_4=\frac{1}{4.5}a_3=\frac{1}{1.2.3.4.5}a_1\\
a_{2k+1}=2[\frac{a_1}{(2k+1)!}]$
Therefore: $f(x)=a_0\sum \frac{x^{2i}}{(2i)!}+a_1 \sum \frac{x^{2i+1}}{(2i+1)!}$
Assume that: $c_1+c_2=a_0\\
c_1-c_2=a_1$
we have: $y(x)=(c_1+c_2)\sum \frac{x^{2i}}{(2i)!}+(c_1-c_2)\sum \frac{x^{2i+1}}{(2i+1)!} \\
=c_1\sum \frac{x^i}{i!}+c_2\sum \frac{(-x)^i}{i!}\\
=c_1e^x+c_2e^{-x}$
