If y = (\sin^{-1}x)^2, prove that (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - 2 =0.

 

 

 

 
 
 
 
 

Answers (1)

 y = (\sin^{-1}x)^2, prove that (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - 2 =0

Given y = (\sin^{-1}x)^2

        \frac{dy}{dx} = \frac{2\sin^{-1}x}{\sqrt{1-x^2}}

        \frac{d^2y}{dx^2} = \frac{\frac{2x}{\sqrt{1-x^2}}-2\sin^{-1}x\times\frac{1}{2\sqrt{1-x^2}}\cdot-2x}{1-x^2}

        (1-x^2)\frac{d^2y}{dx^2} =2 + \frac{2x\sin^{-1}x}{\sqrt{1-x^2}}

        (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - 2 =0

Hence Proved.

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