TITLE: Does there exist continuous functions $P$ and $Q$ on [0,1]
QUESTION [6 upvotes]: Does there exist continuous functions  $P$ and $Q$ on $[0,1]$ such that $y(t)=sin(t^2)$  is a solution to $y''+Py'+Qy=0$ on $[\frac1n,1]$ for all $n\geq1$?
I find $y'$ and $y''$ and compare with the given differential equation and found that $P(t)= -1/t$ and which is not continuous on [0,1]. So such function $P$ can not exist.
Is my concept is correct . Please help me to solve this also if any other method .
Thanks in advance.

REPLY [2 votes]: Your differential equation gives
$$(2 t P(t)+2) \cos \left(t^2\right)+\left(Q(t)-4 t^2\right) \sin \left(t^2\right)=0$$
Demanding that $(2 t P(t)+2)$ and $\left(Q(t)-4 t^2\right)$ are 0 for all $t$ gives the solution that you already know, with $P(t)=-1/t$.
But what if they aren't zero for all $t$? Rearranging a bit
$$\frac{\cos \left(t^2\right)}{\sin \left(t^2\right)}=-\frac{Q(t)-4 t^2}{2 t P(t)+2 } $$
The l.h.s. has a singularity at $t=0$. Assuming $Q(t)$ is continuous in $[0,1]$, the denominator in the r.h.s. must have a 0 at $t=0$, which is not possible for continuous $P(t)$.