question_answer

Form the differential equation having \[y={{({{\sin }^{-1}}x)}^{2}}+A{{\cos }^{-1}}x+B,\] where A and B are arbitrary constants, as its general solution.

Answer:

Given, \[y={{({{\sin }^{-1}}x)}^{2}}+A{{\cos }^{-1}}x+B\] On differentiating w.r.t x, we get \[\frac{dy}{dx}=\frac{2{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}+\frac{(-A)}{\sqrt{1-{{x}^{2}}}}\] \[\Rightarrow \]   \[\sqrt{1-{{x}^{2}}}\frac{dy}{dx}=2{{\sin }^{-\,1}}x-A\] Again differentiating w.r.t. x, we get \[\sqrt{1-{{x}^{2}}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{dy}{dx}\cdot \frac{-\,2x}{2\sqrt{1-{{x}^{2}}}}=\frac{2}{\sqrt{1-{{x}^{2}}}}\] \[\Rightarrow \]\[(1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}-\frac{x}{\sqrt{1-{{x}^{2}}}}\cdot \sqrt{1-{{x}^{2}}}\frac{dy}{dx}=2\] \[\Rightarrow \]   \[(1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}-x\frac{dy}{dx}=2\] \[\Rightarrow \]   \[(1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}-x\frac{dy}{dx}-2=0\] Which is the required differential equation.