Lets say that i have a differential equation
$y'(x)+y(x)=x $ the answer will be pretty straightforward right?, $e^{\int(1)*dt}=e^t$, then:
$y*e^t=\int(x*e^x)dx $ (we have integration by parts)
$y*e^t=x*e^x-e^x+C1 $
$ y*e^x=e^x(x-1)+C1 $
$ y=x-1+C1e^{-x} $ Well, then what will happen if instead of x, we have (u(t)) where u(t) is the step function?
if we are using u(t), that means that, x is not the independent variable, so lets create another example
$y'(t)+y(t)=x(t)*u(t)$, where x(t) can be any function of t (lets say $e^ {-t}$, as an example or whatever)
Well... what i should do then?, u(t) means that x(t) is only valid if t>0, so i should answer the problem skipping the u(t) and then add that the answer for y(t) is only valid for t>0??
What if we have an impulse instead of u(t), or a ramp(t)... ?
You get a solution for $t<0$ and one for $t>0$ and have to join them continuously at $t=0$.
There are other piecewise ODE that are even less compatible to the case of continuous ODE functions. These will in general have more complicated solutions, as the local existence theorems do only apply inside the continuous pieces, not at their boundaries.
Demonstrative examples are $y'(t)=-{\rm sign}(y(t))$ or $y''(t)+{\rm sign}(y'(t))+y(t)=0$.
Again, as a solution strategy, you determine the solution for each domain piece as if the piece were extended to the full domain, determine where the solution meets the boundary of the piece, and then try to assemble a global solution by continuous matching of the solution pieces. This matching can become impossible at some boundary, the point being still an inner point of the overall domain.