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Find the Curve from a Second-Order Differential Equation | JEE

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Question
If a curve y=f(x)y = f(x) satisfies y′′x2+x(y′)2=2xy′y''x^2 + x(y')^2 = 2xy', with f(0)=0f(0) = 0 and f′(1)=1f'(1) = 1, then f(x)f(x) is:
A2(x−tan⁡−1x)2(x - \tan^{-1}x)correct
B2(x−ln⁡(1+x)2)2\big(x - \ln(1+x)^2\big)
C2(ln⁡(1+x2)−x)2\big(\ln(1+x^2) - x\big)
DNone of these
Solution
Step 1: Rearrange so a known derivative appears:
y′′x2−2xy′=−x(y′)2⇒2xy′−x2y′′(y′)2=x.y''x^2 - 2xy' = -x(y')^2 \Rightarrow \frac{2xy' - x^2y''}{(y')^2} = x.
 The left side is exactly ddx(x2y′)\dfrac{d}{dx}\left(\dfrac{x^2}{y'}\right), so
ddx(x2y′)=x⇒x2y′=x22+c.\frac{d}{dx}\left(\frac{x^2}{y'}\right) = x \Rightarrow \frac{x^2}{y'} = \frac{x^2}{2} + c.
 Step 2: Solve for y′y' and use f′(1)=1f'(1)=1:
y′=2x2x2+2c,1=21+2c⇒c=12.y' = \frac{2x^2}{x^2 + 2c}, \qquad 1 = \frac{2}{1+2c} \Rightarrow c = \frac{1}{2}.
∴y′=2x2x2+1=2(1−11+x2).\therefore y' = \frac{2x^2}{x^2+1} = 2\left(1 - \frac{1}{1+x^2}\right).
 Step 3: Integrate and use f(0)=0f(0)=0:
y=2(x−tan⁡−1x)+c′,f(0)=0⇒c′=0.y = 2\big(x - \tan^{-1}x\big) + c', \qquad f(0)=0 \Rightarrow c' = 0.
∴f(x)=2(x−tan⁡−1x).\therefore f(x) = 2\big(x - \tan^{-1}x\big).
 Correct answer: (1)
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