question_answer

If \[y=\frac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}},\] then show that \[(1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}-3x\frac{dy}{dx}-y=0.\]

Answer:

Given, \[y=\frac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}\]                                      ?(i) \[\Rightarrow \]   \[(\sqrt{1-{{x}^{2}}})y=si{{n}^{-1}}x\] On squaring both sides, we get             \[(1-{{x}^{2}}){{y}^{2}}={{(si{{n}^{-1}}x)}^{2}}\] On differentiating both sides w.r.t. x, we get \[(1-{{x}^{2}})2y\cdot \frac{dy}{dx}+{{y}^{2}}(-\,2x)=(2si{{n}^{-1}}x)\cdot \frac{1}{\sqrt{1-{{x}^{2}}}}\] \[\Rightarrow \] \[2y\left[ (1-{{x}^{2}})\frac{dy}{dx}-xy \right]=2y\]           [from Eq. (i)] \[\Rightarrow \] \[(1-{{x}^{2}})\frac{dy}{dx}-xy=1\]                    [divided by 2y] Again differentiating w.r.t. x, we get  \[(1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{dy}{dx}(-\,2x)-\left( x\frac{dy}{dx}+1\cdot y \right)=0\]\[\Rightarrow \] \[(1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{dy}{dx}-\,2x\frac{dy}{dx}-x\frac{dy}{dx}-y=0\] \[\therefore \]      \[(1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}-\,3x\frac{dy}{dx}-y=0\]          Hence proved.