Prove that x2d2ydx2 - 2xdydx - 1 1 - x2

x^2y”+2xy'=11 x^2 ⇒d [x^2d yd x]d x=11 x^2 on variable seperable, and integrating , we get required ans. ...

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Question

prove that $x^{2} \frac{d^{2} y}{d x^{2}} + 2 x \frac{d y}{d x} = \frac{1}{(1 - x^{2})}$

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x = tan (\frac{1}{a} l o g y)

(1 + x^{2}) \frac{d^{2} y}{d x^{2}} + 2 x \frac{d y}{d x} - a 0

y = {[log (x + \sqrt{x^{2} + 1})]}^{2}

(1 + x^{2}) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} = 2

y = {({sin}^{- 1} x)}^{2}

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

y = e^{m . {cos}^{- 1} x}

(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = m^{2} y

y = ({sin}^{- 1} x)^{2}

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

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