Does the solution x=$\phi$(t) of an ordinary differential equation
$f(t)=\frac{dx}{dt}$, where $$ f(t)=\begin{cases} 1, & \text{for $t{\geqslant}0$}\\ 0, & \text{for t <0} \end{cases} $$ exists $\forall$ t $\in$ $\mathbb{R}$
My answer is yes. If I take a piecewise solution defined here such as -$t^{3}$ for $t<0$ and $t^{3}$ for $t\geqslant$0 . In both cases the result is 0 however I'm not sure if I'm thinking about this problem correctly. Am i allowed to assume an arbitrary solution and conclude a result? Or do I have to use x=$\phi$(t) ?$$ $$I was given the solution here and asked if it exists which is where my confusion is. I'm accustomed to doing the reverse in which i was given the pdf and then solved. What is the methodology behind approaching this questions?
Simply solve the equation separately for $t \geq 0$ and $t < 0$. In other words you have to solve the equations $$1 = \dfrac{dx}{dt} \quad \text{ and } \quad 0=\dfrac{dx}{dt}.$$