Find the solution of a differential equation in the form of a power series 
Find the solution of the differential equation $y''(x)=y(x)$ with $y(0)=1$ and $y'(0)=0 $ in the form of the power series $y(x)=\sum_{j=0}^{\infty}a_jx^j$
Use the following method of determining the coefficients $a_j$. Differentiate the power series twice term by term, then apply the identity theorem for power series to the series for y and y''

So differentiating I get:
$y'(x)=\sum_{j=0}^{\infty}(j+1)a_{j+1}x^j$ and $y''(x)=\sum_{j=0}^{\infty}(j+1)j\,a_{j+1}x^{j-1}$
Using the identity theorem then $y''(x)=y$ if $a_j=(j+1)j\,a_{j+1}$
And $y(0)=1=a_0$ and $y'(0)=0=a_1$
However I am unsure where to go from here? Thanks

Continue with your answer, but corrected a little:
$$y'(x)=\sum_{j=1}^{\infty}j a_j x^{j-1}\\
y''(x)=\sum_{j=2}^{\infty}j(j-1) a_j x^{j-2}$$
The original equation then gives
$$\sum_{j=2}^{\infty}j(j-1) a_j x^{j-2}=\sum_{j=0}^{\infty} a_j x^{j}\Rightarrow\\
\sum_{m=0}^{\infty}(m+2)(m+1) a_{m+2} x^{m}=\sum_{m=0}^{\infty} a_m x^{m}$$
where I changed $m=j-2$ for the left hand side and just $m=j$ for the right hand side. 
This gives the formula
$$a_{m+2}=\frac{a_m}{(m+1)(m+2)}$$
Now with your initial condition $a_0=1, a_1=0$, setting $m=0,1,2,\dots$, you should be able to get the coefficients. Can you find the pattern then and write it as a sum?
