Suppose $p(x)$,$q(x)$,$r(x)$,$f(x)$ are continuous everywhere. Is it possible that $y=2$, $y=x^3+2$, $y=x^4+2$ are all solutions of the differential equation $$y''' + p(x)y'' + q(x)y' + r(x)y = f(x)$$
I first verified that the 3 given solutions are linearly independent, and they are. I'm not sure if that's enough to say that they can be solution.
I also noticed that $y(0)=2$ for all three solutions. So if we take an IVP with $y(0)=2$,$y'(0)=0$ and $y''(0)=0$, since the coefficient of $y'''$ is never zero and $p(x),q(x),r(x),f(x)$ are all continuous, by the existence uniqueness theorem, there is only one solution to the IVP on $\mathbb{R}$, therefore we cannot have these three solutions to the DE. Is my reasoning correct?
Yes: since these three functions all have the same values of $y(0)$, $y'(0)$, and $y''(0)$, at most one of them can be a solution.