Question

if y=tan^-1x, show that (1+x^2)y''+(2x-1)y'=0​

Answer 1

Question :-

$\sf if \: y = { \tan}^{ - 1} x \\ \sf show \: that \: (1 + {x}^{2} )y'' +( 2x - 1)y' = 0$

Proof :-

Used formula :-

$\star \: \sf{ \frac{d}{dx} ( { \tan}^{ - 1} x) = \frac{1}{1 + {x}^{2} } }$

Explanation :-

Here y' means differentiate of y with respect to x and y'' means double differentiation of y

$\star \: \sf \frac{dy}{dx} = y' \\ \\ \star \: \sf \: \frac{ {d}^{2}y }{d {x}^{2} } = y''$

firstly we will find y' and y''

$\to \sf\: y = { \tan}^{ - 1} x \\ \bf differentiate \: it \: \\ \\ \to \sf \frac{dy}{dx} = \frac{1}{1 + {x}^{2} } \\ \\ \sf again \: differentiate \: \: it \\ \\ \to \sf \frac{ {d}^{2}y }{d {x}^{2} } = \frac{ - 1(2x)}{ {(1 + {x}^{2}) }^{2} }$

Now taking which we will prove

$\to \sf (1 + {x}^{2} )y'' +( 2x + 1)y' = 0 \\ \: \\ \to \sf (1 + {x}^{2}) \frac{ - 2x}{ {(1 + {x}^{2}) }^{2} } + (2x - 1) \frac{1}{1 + {x}^{2} } \\ \\ \to \sf\frac{ - 2x}{1 + {x}^{2} } + \frac{2x - 1}{1 + {x}^{2} } \\ \\ \to \: \sf \frac{ - 2x + 2x - 1}{1 + {x}^{2} } \\ \\ \to \: \sf \frac{ - 1}{1 + {x}^{2} }$

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