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Exact Differential Equation Using d(x²+y²) | JEE

JEE Maths question with a full step-by-step solution.

Question
The solution of the differential equation x dx+y dy+x dy−y dxx2+y2=0x\,dx + y\,dy + \dfrac{x\,dy - y\,dx}{x^2 + y^2} = 0 is:
Ay=tan⁡(c−x−y2)y = \tan\left(\dfrac{c - x - y}{2}\right)
By=xtan⁡(c−x2−y22)y = x\tan\left(\dfrac{c - x^2 - y^2}{2}\right)correct
Cy=xtan⁡(c+x2+y22)y = x\tan\left(\dfrac{c + x^2 + y^2}{2}\right)
Dy=xtan⁡(c+x2−y22)y = x\tan\left(\dfrac{c + x^2 - y^2}{2}\right)
Solution
Recognise the two standard differentials:
x dx+y dy=12 d(x2+y2),x dy−y dxx2+y2=d(tan⁡−1yx).x\,dx + y\,dy = \frac{1}{2}\,d(x^2+y^2), \qquad \frac{x\,dy - y\,dx}{x^2+y^2} = d\left(\tan^{-1}\frac{y}{x}\right).
 So the equation is
12 d(x2+y2)+d(tan⁡−1yx)=0⇒12(x2+y2)+tan⁡−1yx=C1.\frac{1}{2}\,d(x^2+y^2) + d\left(\tan^{-1}\frac{y}{x}\right) = 0 \Rightarrow \frac{1}{2}(x^2+y^2) + \tan^{-1}\frac{y}{x} = C_1.
⇒tan⁡−1yx=C1−x2+y22⇒yx=tan⁡(c−x2−y22)⇒y=xtan⁡(c−x2−y22).\Rightarrow \tan^{-1}\frac{y}{x} = C_1 - \frac{x^2+y^2}{2} \Rightarrow \frac{y}{x} = \tan\left(\frac{c - x^2 - y^2}{2}\right) \Rightarrow y = x\tan\left(\frac{c - x^2 - y^2}{2}\right).
 Correct answer: (2)
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