Solving the time-differential equation $\ddot{y}=y\exp(t)$ I want to solve $$\ddot{y}=y\exp(t)$$ with the initial conditions $y(0) = 0$, and $y(1) = 0$. How would I even go about doing this? I tried using mathematica but it could not find a general solution.
 A: Let's try - as suggested by @Luciano - a power series ansatz. Denote $y(t) = \sum_n a_n t^n$, then
$$ \dot y(t) = \sum_n na_n t^{n-1}, \quad \ddot y(t) = \sum_n n(n-1)a_n t^{n-2} 
  = \sum_n (n+2)(n+1)a_{n+2} t^n $$
On the other hand
$$ y(t)\exp(t) = \sum_n a_n t^n \sum_m \frac 1{m!} t^m = \sum_n \sum_{k=0}^n \frac {a_k}{(n-k)!}t^n  $$
If we compare the coefficients, we get
\begin{align*}
   2a_2 &= a_0\\
   6a_3 &= a_0 + a_1\\
   12a_4 &= \frac 12 a_0 + a_1 + a_2\\
         &\vdots 
\end{align*}
The initial conditions give us $a_0 = 0$ and $\sum_n a_n = 0$. The formulae above gives us, that the sign of $a_1$ determines the signs of all $a_n$. That is $\sum_n a_n = 0$ is only possible for $a_1 = 0$. But then (by induction) and the above $a_n = 0$ for all $n$. Hence $y = 0$.
A: Compare a hypothetical non-zero solution, switch its sign if necessary so that $y'(0)>0$, with the function $u(t)=e^{ct}$. Then the Wronskian-like expression
$$
h(t)=u(t)y'(t)-u'(t)y(t)\implies h(0)=y'(0)>0
$$
has the derivative
$$
h'(t)=(e^t-c^2)u(t)y(t).
$$
With $c=e^{1/3}$ this gives $h$ is positive and growing on the interval $[0,1]$ as long as $y>0$. However, a second root $t_1\in(0,1]$ of $y$ is impossible, for instance

*

*If $y(t_1)=0$, then $$0<h(t_1)=u(t_1)y'(t_1)\implies y'(t_1)>0,$$ however $y$ would have to fall towards the smallest positive root, so should have a negative derivative there.

*If $y(t_1)=0$, then there is some point $t_2\in(0,t_1)$ where $y'(t_2)=0$. This would give $$0<h(t_2)=-u'(t_2)y(t_2)\implies y(t_2)<0,$$ which is likewise a contradiction.

So any non-zero solution of the differential equation with a root at $t=0$ can not have a second root, at least not inside $[0,1]$ with the above construction.
