 Solution

The functionis continuous, satisfies the equation away from $\tilde{t}=y_0$, and has unequal one-sided derivatives there.

It is unique in the stated class. While a solution is positive its derivative is $-1$, so continuity forces it to follow $y_0-t$ until it first reaches zero at $t=y_0$. It cannot subsequently make a positive excursion: on any connected interval $(a,b)$ where $y>0$ and $y(a)=0$, integration of $y^{\prime }=-1$ gives $y(t)=-(t-a)<0$. A negative excursion similarly contradicts $y^{\prime }=1$. Hence the solution remains zero. This is the sign-decay differential equation.