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EXERCISE 1.5 · Q191

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

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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Given y=emtan⁡−1xy=e^{m\tan^{-1}x}.

Step 1 — differentiate once by the chain rule: y1=emtan⁡−1x⋅m1+x2=my1+x2y_1=e^{m\tan^{-1}x}\cdot\dfrac{m}{1+x^2}=\dfrac{my}{1+x^2}.

Step 2 — clear the denominator: (1+x2)y1=my(1+x^2)y_1=my.

Step 3 — differentiate both sides of this relation w.r.t. xx, using the product rule on the left side: (1+x2)y2+2x y1=m y1(1+x^2)y_2+2x\,y_1=m\,y_1. …

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