Q.$y=\sin(m\cos^{-1}x)$, show that $(1-x^2)\dfrac{d^2y}{dx^2}-x\dfrac{dy}{dx}+m^2y=0$

Given $y=\sin(m\cos^{-1}x)$. Let $\theta=\cos^{-1}x$, so $y=\sin(m\theta)$ and $x=\cos\theta$, with $\dfrac{d\theta}{dx}=-\dfrac{1}{\sqrt{1-x^2}}$.

Step 1: $y_1=m\cos(m\theta)\cdot\dfrac{d\theta}{dx}=-\dfrac{m\cos(m\theta)}{\sqrt{1-x^2}}$, so $\sqrt{1-x^2}\,y_1=-m\cos(m\theta)$.

Step 2 — differentiate both sides w.r.t. $x$. On the right: $\dfrac{d}{dx}\left[-m\cos(m\theta)\right]=m^2\sin(m\theta)\cdot\dfrac{d\theta}{dx}=m^2y\cdot\left(-\dfrac{1}{\sqrt{1-x^2}}\right)=-\dfrac{m^2y}{\sqrt{1-x^2}}$. …

Substitute $\theta=\cos^{-1}x$ so $y=\sin(m\theta)$; differentiate w.r.t. $\theta$, convert back using $d\theta/dx=-1/\sqrt{1- …

Sign error in $d\theta/dx=-1/\sqrt{1-x^2}$ — dropping the minus sign flips the sign of $y_1$ and cascad …