Do solutions of $\dot{x} = \frac{x}{t^2} + t$ exist satisfying $x(0) =0$ Suppose we have the 1-dimensional ODE 
\begin{equation}
\dot{x} = \frac{x}{t^2} + t
\end{equation}
Do there exist solution curves with initial condition $x(0)=0$?  If you proceed in a standard way then you would get as solution formula
\begin{equation}
x= C {\rm e}^{-1/t} + {\rm e}^{-1/t}  \int^{t}_0  \tau {\rm e}^{1/\tau}  d \tau
\end{equation} 
but then the expression in the integral explodes! This does not imply that solution curves with $x(0)=0$ do not exist. How to proceed with such existence results?

Set $x(t)=t^2 y(t)$, then the original equation becomes:
$$t^2 \dot y(t)+2t y(t)=y(t)+t\tag{1}$$
The $y(0)=0$ solution is:
$$y(t)=\frac{1}{2}+\frac{1}{2t}-\frac{e^{-1/t}}{2t^2}Ei(1/t)\tag{2}$$
When $t\to 0$, we have
$$y(t)\approx -t - 3 t^2 - 12 t^3 - 60 t^4...\tag{3}$$
