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?

The problem is that your original problem is not a Cauchy problem
The usual setup for solving differential equations is that you are given a function $f: [a,b] \times \mathbb R \to \mathbb R$ and a point $(x_0, y_0)$ and you want to find a function $y:[a,b]\to\mathbb R$ which satisfies two conditions:


*

*$f(x,y(x)) = y'(x)$ for all values $x\in[a,b]$

*$y(x_0)=y_0$


In your case, you only defined the function $f$ on $(a,b]\times\mathbb R$, since $f(t, x)=\frac{x}{t^2}+t$ in your case. Your equation has many solutions. For example, taking $x(t)=0$ creates a function that satisfies $f(t, x(t))=\dot x(t)$ for every $t\in (0,1]$.
