What role do constant solutions play in the existance and uniqueness theorem? For instance, consider the IVP
$$\frac{dy}{dx} = x$$ $$ y(0) = 0 $$
Clearly, this IVP has a solution in the form of $y =\frac{x^2}{2}$. Now the existance and uniqueness theorem tells us that for $$\frac{dy}{dx} = f(x,y)$$ if $f(x,y)$ and $\frac{df}{dy}$ are both continuous around $(x_0,y_0)$ then there will be only one solution in some interval around $x_0$. And since both $x$ and the constant function $\frac{df}{dy} = 0$ are continuous around $(0,0)$ it should be unique. But the constant function $y = 0$ is also a solution if $\frac{dy}{dx} = x = 0$
Meaning, that $y = 0$ solves the IVP at that one point (a rather weak solution, I admit). But doesn't this contradict the uniqueness theorem of IVPs, since we in effect have two different functions that solve the same IVP?
The zero function does not satisfy that differential equation, since
$$\frac{dy}{dx} = 0 \neq x$$
on all open intervals that include $0$.
It satisfies the differential equation $$\frac{dy}{dx} = y$$
but that's a whole other story...