Question about PDE solution by power series of an IVP

I have this problem: find by power series with respect to $y$ the solution of the Cauchy problem

\begin{cases} u_{yy}=u_{xx}+u\\ u(x,0)=e^x\\ u_y(x,0)=0 \end{cases} I know that the solution has the form $\sum_{n=0}^{\infty}a_n(x)y^n$ and by initial conditions I found that $a_0(x)=e^x$ and that $a_1(x)=0$, but I cannot continue and conclude...

(P.S.: differentiating $u(x,0)$ with respect to $x$ I found that $a^{(k)}_0(x)=e^x$ with $k\geq0$ and it's ok. Moreover differentiating $u_y(x,0)$ w.r.t. $y$, we found that $u_{yy}(x,0)=u_{yyy}(x,0)=\ldots=0$ but I cannot take advantage of this..)

Thanks :)

Differentiating term by term, you got: $$\begin{split} u(x,y) &= \sum_{n=0}^\infty a_n(x)\ y^n\\ u_{xx}(x,y) &= \sum_{n=0}^\infty a_n^{\prime \prime} (x)\ y^n\\ u_{y}(x,y) &= \sum_{n=0}^\infty (n+1)a_{n+1}(x)\ y^n\\ u_{yy}(x,y) &= \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}(x)\ y^n\end{split}$$ and plugging these relations both in the PDE and in the initial conditions yields: $$\begin{cases} \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}(x)\ y^n = \sum_{n=0}^\infty \big( a_n^{\prime \prime} (x) +a_n(x)\big)\ y^n\\ a_0(x) = e^x \\ a_1(x) = 0\; ,\end{cases}$$ which translates into a system of recurrence equations (involving also second derivatives) for the $a_n$s, namely: $$\tag{R} \begin{cases} (n+2)(n+1)a_{n+2}(x) = a_n^{\prime \prime} (x) +a_n(x)\\ a_0(x) = e^x \\ a_1(x) = 0\; .\end{cases}$$ Recurrence (R) is second order with the RHside which does not depend on $a_{n+1}$; therefore, it separates the $a_n$s depending on odd indices from the $a_n$s depending on even indices. In particular, if $n=2h+1$ you get: $$\begin{cases} (2h+3)(2h+2)a_{2h+3}(x) = a_{2h+1}^{\prime \prime} (x) +a_{2h+1}(x)\\ a_1(x) = 0\; ,\end{cases}$$ from which it follows: $$a_{2h+1}(x) = 0\quad \text{for all } h\in \mathbb{N};$$ on the other hand, if $n=2h$ you get: $$\begin{cases} (2h+2)(2h+1)a_{2h+2}(x) = a_{2h}^{\prime \prime} (x) +a_{2h}(x)\\ a_0(x) = e^x\; ,\end{cases}$$ from which it follows: $$\begin{split} a_0(x) &= e^x\\ a_{2}(x) &= \frac{1}{2!}\ \big(a_0^{\prime \prime} (x) +a_0(x)\big)\\ &= \frac{2}{2!}\ e^x\\ a_{4}(x) &= \frac{1}{4\cdot 3}\ \big(a_{2}^{\prime \prime} (x) +a_{2}(x)\big) \\ &= \frac{4}{4!}\ e^x\\ a_{6}(x) &= \frac{1}{6\cdot 5} \big( a_{4}^{\prime \prime} (x) +a_{4}(x)\big)\\ &= \frac{8}{6!}\ e^x\end{split}$$ and in general: $$a_{2h}(x) = \frac{2^h}{(2h)!}\ e^x\quad \text{for all } h\in \mathbb{N}.$$
Hence: $$u(x,y) = \sum_{h=0}^\infty \frac{2^h}{(2h)!}\ e^x\ y^{2h} = e^x\ \cosh(\sqrt{2}\ y)$$ is the unique solution of your IVP.