Consider the IVP $$y'=h(t)y(t)$$ with $y(0)=1$ initial condition and $h(t)=1$ for $ t\geq 0 $ and $0$ elsewhere. Prove that it does not have a solution.
Sollution: Let $y$ be a sollution then i get $y(t)=e^t$, $t \geq 0$ and 1 else. but that $y$ is not differentiable zero. hence no sollution.
But isn't the $$y=1+ \int_{0}^{t}h(s)y(s)ds $$ a solution, or equivalent to the problem? And now the function $y(t)=e^t$ $t \geq 0$ and 1 else is a solution to the
$y=1+ \int_{0}^{t}h(s)y(s)ds $ I think im confused . Can someone explain me the difference between the integral form and the ODE??
The solution of the differential equation has to be differentiable; this is impossible in this case, as you have shown. But solutions of the integral equation do not need to be differentiable. The integral equation makes perfect sense for continuous functions.
A solution of the differential equation will always be a solution of the integral equation. But if the right hand side of the differential equation is not continuous, it may not be equivalent to the integral equation, as this example shows.
Think of the simpler example $y'=h(t)$. There is not a differentiable function whose derivative is $h$. But $\int_0^xh(t)\,dt$ is well defined and is a continuous (but not differentiable at $x=0$) function.