Deducing the existence of a PDE by constructing it inductively via its Taylor series expansion Consider the differential equation 
$$u_{t t} = c^2 u_{x x} - k u, \qquad k, u > 0$$
together with the Cauchy data 
$$u(x, 0) = e^x, u_t(x, 0) = 0$$
I now want to find a solution $u: \mathbb R^2 \to \mathbb R$ for this equation which is real-analytic with respect to $t$.
Now I was given the tip that I might construct the desired solution $u$ via Taylor expansion in a point $(x, 0)$ (with respect to $t$) and see what comes out for the first few Taylor coefficients, and then prove a formula for all of them per induction. (And afterwards of course check that it actually solves the equation, but I think that will be the easier matter.)
Therefore, I started this way: let's say we write the desired solution $u$ in its Taylor expansion $u(x, t) = \sum_{n=0}^\infty \frac{u^{(n)}}{n!} t^n$ around $(x, 0)$. Then because of the Cauchy data, we would immediately have $u^{(0)} = u(x, 0) = e^x$ and $u^{(1)} = u_t(x, 0) = 0$. Putting that into our equation, we get $u^{(2)} = u_{t t}(x, 0) = c^2 e^x - k e^x = (c^2 - k) e^x$ I think. So we got the first few coefficients already.
But that's where my wisdom ends... how would I be able to get the rest of the coefficients? I don't even know how I can get the third one.
In other words, if I have the coefficients $u^{(0)}, \dots, u^{(n)}$ of the Taylor series expansion of the desired solution, how can I deduce the $n+1$-the one using only the given PDE and Cauchy data? As for a general formula for them, it would seem plausible that each $u^{(n)}$ is a product of $e^x$ and some constant term, given what the first few look like, but I'm not sure how I can actually prove that.
 A: Let $p=\dfrac{\sqrt kx}{c}$ ,
Then $u_{tt}=ku_{pp}-ku$ with $u(p,0)=e^\frac{cp}{\sqrt k}$ and $u_t(p,0)=0$
Let $q=\sqrt kt$ ,
Then $u_{qq}=u_{pp}-u$ with $u(p,0)=e^\frac{cp}{\sqrt k}$ and $u_q(p,0)=0$
Similar to Solve initial value problem for $u_{tt} - u_{xx} - u = 0$ using characteristics,
Let $u(p,q)=\sum\limits_{m=0}^\infty\dfrac{q^m}{m!}\dfrac{\partial^mu(p,0)}{\partial q^m}$ ,
Then $u(p,q)=\sum\limits_{m=0}^\infty\dfrac{q^{2m}}{(2m)!}\dfrac{\partial^{2m}u(p,0)}{\partial q^{2m}}+\sum\limits_{m=0}^\infty\dfrac{q^{2m+1}}{(2m+1)!}\dfrac{\partial^{2m+1}u(p,0)}{\partial q^{2m+1}}$
$u_{qqqq}=u_{ppqq}-u_{qq}=u_{pppp}-u_{pp}-u_{pp}+u=u_{pppp}-2u_{pp}+u$
$u_{qqqqqq}=u_{ppppqq}-2u_{ppqq}+u_{qq}=u_{pppppp}-u_{pppp}-2u_{pppp}+2u_{pp}+u_{pp}-u=u_{pppppp}-3u_{pppp}+3u_{pp}-u$
Similarly, $\dfrac{\partial^{2m}u}{\partial q^{2m}}=\sum\limits_{n=0}^m(-1)^{m-n}C_n^m\dfrac{\partial^{2n}u}{\partial p^{2n}}$
$u_{qqq}=u_{ppq}-u_q$
$u_{qqqqq}=u_{ppqqq}-u_{qqq}=u_{ppppq}-u_{ppq}-u_{ppq}+u_q=u_{ppppq}-2u_{ppq}+u_q$
$u_{qqqqqqq}=u_{ppppqqq}-2u_{ppqqq}+u_{qqq}=u_{ppppppq}-u_{ppppq}-2u_{ppppq}+2u_{ppq}+u_{ppq}-u_q=u_{ppppppq}-3u_{ppppq}+3u_{ppq}-u_q$
Similarly, $\dfrac{\partial^{2m+1}u}{\partial q^{2m+1}}=\sum\limits_{n=0}^m(-1)^{m-n}C_n^m\dfrac{\partial^{2n+1}u}{\partial p^{2n}\partial q}$
$\therefore u(p,q)=e^\frac{cp}{\sqrt k}\sum\limits_{m=0}^\infty\sum\limits_{n=0}^m\dfrac{(-1)^{m-n}C_n^mc^{2n}q^{2m}}{k^n(2m)!}$
Hence $u(x,t)=e^x\sum\limits_{m=0}^\infty\sum\limits_{n=0}^m\dfrac{(-1)^{m-n}C_n^mc^{2n}k^{m-n}t^{2m}}{(2m)!}$
