Question

$\mathrm{If}y=\frac{{\sin }^{-1}x}{1-x^2},\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}\left(1-x^2\right)\frac{d^{2}y}{{\mathrm{dx}}^2}-3x\frac{\mathrm{dy}}{\mathrm{dx}}-y=$

Answer 1

$\mathrm{We}\mathrm{have},y=\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}\Rightarrow y\sqrt{1-x^2}={\sin }^{-1}x\mathrm{Diffferentiating}\mathrm{both}\mathrm{sides}w.r.t.x,\mathrm{we}\mathrm{get}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}\sqrt{1-x^2}-\frac{x}{\sqrt{1-x^2}}y=\frac{1}{\sqrt{1-x^2}}\Rightarrow \left(1-x^2\right)\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{xy}=1\mathrm{Diffferentiating}\mathrm{again}\mathrm{both}\mathrm{sides}w.r.t.x,\mathrm{we}\mathrm{get}\Rightarrow \left(1-x^2\right)\frac{d^{2}y}{{\mathrm{dx}}^2}-2x\frac{\mathrm{dy}}{\mathrm{dx}}-x\frac{\mathrm{dy}}{\mathrm{dx}}-y=0\Rightarrow \left(1-x^2\right)\frac{d^{2}y}{{\mathrm{dx}}^2}-3x\frac{\mathrm{dy}}{\mathrm{dx}}-y=0$