Question

If x = sin t, y = sin pt, prove that $\left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+p^{2}y=0$.

Answer 1

Here,
$$\begin{gathered} x=\sin t\mathrm{and}y=\sin pt \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}t,\mathrm{we}\mathrm{get} \\ \frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}=\cos t\mathrm{and}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{t}}=p\cos pt \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{p\cos pt}{\cos t} \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathit{}x,\mathrm{we}\mathrm{get} \\ \frac{{\mathit{d}}^{\mathit{2}}\mathit{y}}{\mathit{d}{\mathit{x}}^{\mathit{2}}}=\frac{-p^2\sin pt\cos t+p\cos pt\sin t}{{\cos }^{2}t}\times \frac{dt}{dx} \\ \Rightarrow \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=\frac{-p^2\sin pt\cos t+p\cos pt\sin t}{{\cos }^{3}t} \\ \Rightarrow \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=\frac{-p^2\sin pt\cos t}{{\cos }^{3}t}+\frac{p\cos pt\sin t}{{\cos }^{3}t} \\ \Rightarrow \frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=\frac{-p^{2}y}{{\cos }^{2}t}+\frac{x\frac{\mathit{d}y}{\mathit{d}x}}{{\cos }^{2}t} \\ \Rightarrow {\cos }^{2}t\frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=-p^{2}y+x\frac{\mathit{d}y}{\mathit{d}x} \\ \Rightarrow \left(1-{\sin }^{2}t\right)\frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=-p^{2}y+x\frac{\mathit{d}y}{\mathit{d}x} \\ \Rightarrow \left(1-x^2\right)\frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}-x\frac{\mathit{d}y}{\mathit{d}x}+p^{2}y=0 \end{gathered}$$.

Hence proved.