Show that $\sin (x^2)$ and $\cos (x^2)$ can't both solve a second order linear homogenous ODE I want to show that there does not exist $p(x)$ and $q(x)$ that are continuous on $\mathbb{R}$ such that $\sin (x^2)$ and $\cos (x^2)$ are both solutions for $$\ \ \ L[y]=\dfrac{d^2 y}{d x^2}+p(x)\dfrac{d y}{d x}+q(x)y=0.$$
I don't think I can use uniqueness here since we only know that $p(x)$ and $q(x)$ are continuous, nothing about their derivatives. I see that the Wronskian of $\sin x^2$ and $\cos x^2$ is $-2t$, so they are linearly independent when $t \neq 0$.
I tried contradicting that $c_1\sin t^2 + c_2 \cos t^2$, but I didn't make any progress and am stuck. I'm looking for some pointers on how to proceed.  
 A: The Wronskian is $0$ at $t=0$. Therefore there can't be a $p$ and $q$ that are continuous at $t=0$.
EDIT: Abel's theorem says if $y_1(x)$ and $y_2(x)$ are solutions of the DE $y'' + p(x) y' + q(x) y = 0$ on some interval where $p(x)$ and $q(x)$ are continuous, their Wronskian $W(x)$ satisfies
$$ W(x) = c \exp \left( - \int p(x)\; dx \right) $$
on that interval for some constant $c$.  In particular, this is either $0$ for all $x$ (if $c = 0$) or nonzero for all $x$ (if $c \ne 0$).  
A: The solution space to a linear homogeneous ODE is a linear space, so if $\sin^2 x$ and $\cos^2 x$ are solutions, then so is $\sin^2 x + \cos^2 x = 1.$ Plugging in, we see that $q(x) \equiv 0.$ 
That means that $\sin^2 x$ and $\cos^2$ satisfy the ODE with $q(x) = 0,$ which is a first order ode in the derivative, which is $\pm 2\sin x cos (x) = \pm \sin(2 x).$ So your question reduces to $\sin$ not satisfying a first order ODE with continuous coefficients. However, since we know the derivative of $\sin,$ this is equivalent to the assertion that $\cot$ is not continuous on $\mathbb{R}.$ 
UPDATE As pointed out in the comments, these were not the right functions. However, do not despair, plugging in the two functions, we get two linear equations for $p, q,$ solving which obtains:
$$
p=\frac{-3 + \cos(2 x^2) - 2 x^2 \sin(2 x^2)}{4 x}.
$$
This obviously blows up at $x=0.$
