Given Solutions of Second-Order Inhomogeneous Linear DE, Show Origin is Singular Point

Question

I've come across this problem while working on some MIT OpenCourseWare:

Problem: Show that any second-order linear inhomogeneous equation that has $x^2$ and $\sin^2 x$ as solutions must have a singular point at the origin.

My Work: Currently, the best reasoning I have is to say that $x^2 - \sin^2 x$ is a solution of the homogeneous equation. If we know one non-vanishing solution, then we can find the other using the formula,

$$cf\int \frac{e^{-\int p(t)\, dt}}{f^2(t)}\, dt$$

where $p$ is the coefficient function of $y'$ in the DE ($y'' + py' + qy = r$) and $f$ is the known non-vanishing solution. Clearly, $x^2 - \sin^2 x = 0$ only when $x = 0$, and so the second solution from this formula would fail at this point. However, this just tells me that this method fails here and not that there is necessarily a singular point at $x = 0$.

Any help would be much appreciated.

Answer 1

$y(x)=x^2-\sin^2x$ is a solution of the homogeneous equation satisfying $y(0)=y'(0)=0$. If $x=0$ were a regular point, the unique solution with those initial conditions would be $y(x)\equiv0$. Thus, $x$ is a singular point.

Answer 2

I came up with a possible solution. I'm not entirely sure how well this works, but it's better than my previous attempt. We simply substitute into our general equation ($y'' + p(x)y' + q(x)y = r(x)$) to come up with a system:

\begin{align*} 2 + 2p(x)x + x^2 q(x) &= r(x)\\ 2\cos^2 x - 2\sin^2 x + 2p(x)\cos x \sin x + q(x)\sin^2 x &= r(x) \end{align*}

We then set these equal to each other.

$$2\cos^2 x - 2\sin^2 x + 2p(x)\cos x \sin x + q(x)\sin^2 x = 2 + 2p(x)x + x^2 q(x)$$

We simplify and rearrange,

$$p(x) = \frac{2\sin^2 x - 2\cos^2 x + 2 + (x^2 - \sin^2 x)q(x)}{2\cos x \sin x - 2x}$$

This is defined at every point except for when $x = 0$ where we get $\frac{0}{0}$. We use L'hospital's rule to get:

$$\frac{8\sin x \cos x + (2x - 2\sin x \cos x)q + (x^2 - \sin^2 x)q'}{2\cos^2 x - 2\sin^2 x - 2}$$

This still gives $\frac{0}{0}$, but if we use L'hospital's rule again, we get,

$$\frac{8\cos^2 x - 8\sin^2 x + (2 - 2\cos^2 x + 2 \sin^2 x)q + (4x - 4\sin x \cos x)q' + (x^2 - \sin^2 x)q''}{-8\cos x \sin x}$$

This is infinite when $x = 0$. Thus, $p(x)$ is discontinuous at $x = 0$ and our DE has a singular point at the origin.