I'm having trouble completing the following exercise from my homework:
Let $y'' = f(x, y, y')$ be an ODE that fulfills the criteria for the existence and uniqueness theorem (Picard–Lindelöf theorem) in the interval [-1, 1].
$y_1(x)$ solves the IVP with initial conditions $y_1(0) = 1$ and $y_1'(0) = 0$.
$y_2(x)$ solves the IVP with initial conditions $y_2(0) = 2$ and $y_2'(0) = 0$
assume $y_1 \ne$ constant 1.
Prove that a 2nd order ODE that fulfills the criteria for the existence and uniqueness theorem so that $y_1(x)\cdot y_2(x)$ is a solution to it within [-1, 1] does not exist.
I'm not very familiar with ODEs, so it might be obvious, but I can't find a way to prove it...
I suspect it involves proving that $y_2(x)$ and $y_1(x)\cdot y_2(x)$ both solve the same ODE and since they have the same initial conditions it forces them to be equal (because of the existence and uniqueness theorem), contradicting the assumption that $y_1 \ne$ constant 1. I can't manage to find the ODE that both of them solve/a way to prove that they both do since I don't know any specific information about the ODEs.
Any help would be very welcome, thanks!