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Question 278 of 293

Q.If y=emtan⁡−1xy = e^{m\tan^{-1}x}, then show that (1+x2)d2ydx2+(2x−m)dydx=0(1+x^2)\dfrac{d^2y}{dx^2} + (2x-m)\dfrac{dy}{dx} = 0

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2022Subjective· 3mImportance★★★★★
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Differentiate yy once to get (1+x2)y′=my(1+x^2)y'=my, then differentiate again.

y=emtan⁡−1xy=e^{m\tan^{-1}x}

y′=emtan⁡−1x⋅m1+x2=my1+x2y' = e^{m\tan^{-1}x}\cdot\dfrac m{1+x^2} = \dfrac{my}{1+x^2}

So: (1+x2)y′=my(1+x^2)y' = my ... (*)

Differentiating (*) with respect to xx (product rule on LHS): …

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