Show that it is impossible to find continuous functions $ a_1, a_0 : \mathbb{R} \rightarrow \mathbb{R} $ so that the differential equation $ x''(t) + a_1(t)x'(t) + a_0(t)x(t)=0 $ has the solution $x(t) = t^2$.
I think its possible to rearrange the equation and show that it won't always be continuous but I think you can also show that the solution(s) violate uniqueness (existence and uniqueness theorem) but I'm not exactly sure how to show this? TIA
$x''(t)+a_1(t)x'(t)+a_0 x(t)=0$
$x(t)=t^2$.
so from solution $x(0)=0$.
so if substituted back to equation:
$2+2a_1(t)t+a_0(t)t^2=0$
at $t=0$ you get:
$2=0$
so $a_0(t)=O(\frac{1}{t^2})$ for the solution to be valid at $0$.