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Homogeneous Differential Equation – Substitution x=vy | JEE

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Question
The solution of the differential equation y2 dx+(x2−xy+y2) dy=0y^2\,dx + (x^2 - xy + y^2)\,dy = 0 is:
Atan⁡−1(xy)+ln⁡y+C=0\tan^{-1}\left(\dfrac{x}{y}\right) + \ln y + C = 0correct
B2tan⁡−1(xy)+ln⁡x+C=02\tan^{-1}\left(\dfrac{x}{y}\right) + \ln x + C = 0
Cln⁡(y+x2+y2)+ln⁡y+C=0\ln\left(y + \sqrt{x^2 + y^2}\right) + \ln y + C = 0
Dln⁡(x+x2+y2)+C=0\ln\left(x + \sqrt{x^2 + y^2}\right) + C = 0
Solution
Step 1: Treat xx as a function of yy:
dxdy+x2−xy+y2y2=0⇒dxdy+x2y2−xy+1=0.\frac{dx}{dy} + \frac{x^2 - xy + y^2}{y^2} = 0 \Rightarrow \frac{dx}{dy} + \frac{x^2}{y^2} - \frac{x}{y} + 1 = 0.
 Step 2: Put x=vyx = vy, so dxdy=v+ydvdy\dfrac{dx}{dy} = v + y\dfrac{dv}{dy}:
v+ydvdy+v2−v+1=0⇒ydvdy=−(1+v2)⇒dv1+v2=−dyy.v + y\frac{dv}{dy} + v^2 - v + 1 = 0 \Rightarrow y\frac{dv}{dy} = -(1 + v^2) \Rightarrow \frac{dv}{1+v^2} = -\frac{dy}{y}.
 Step 3: Integrate:
tan⁡−1v=−ln⁡y+(const)⇒tan⁡−1(xy)+ln⁡y+C=0.\tan^{-1}v = -\ln y + (\text{const}) \Rightarrow \tan^{-1}\left(\frac{x}{y}\right) + \ln y + C = 0.
 Correct answer: (1)
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