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Exact Differential Equation – Dividing by x³y³ | JEE

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

Question
The solution of the differential equation y(xy+2x2y2) dx+x(xy−x2y2) dy=0y(xy + 2x^2y^2)\,dx + x(xy - x^2y^2)\,dy = 0 is given by:
A2log⁡∣x∣−log⁡∣y∣−1xy=c2\log|x| - \log|y| - \dfrac{1}{xy} = ccorrect
B2log⁡∣y∣−log⁡∣x∣−1xy=c2\log|y| - \log|x| - \dfrac{1}{xy} = c
C2log⁡∣x∣+log⁡∣y∣+1xy=c2\log|x| + \log|y| + \dfrac{1}{xy} = c
D2log⁡∣y∣+log⁡∣x∣+1xy=c2\log|y| + \log|x| + \dfrac{1}{xy} = c
Solution
Step 1: Expand and divide by x3y3x^3y^3:
(1x2y+2x)dx+(1xy2−1y)dy=0.\left(\frac{1}{x^2y} + \frac{2}{x}\right)dx + \left(\frac{1}{xy^2} - \frac{1}{y}\right)dy = 0.
 Step 2: Combine the first terms of each bracket:
1x2y dx+1xy2 dy=y dx+x dyx2y2=d(xy)(xy)2=−d(1xy).\frac{1}{x^2y}\,dx + \frac{1}{xy^2}\,dy = \frac{y\,dx + x\,dy}{x^2y^2} = \frac{d(xy)}{(xy)^2} = -d\left(\frac{1}{xy}\right).
 So the equation becomes
−d(1xy)+2x dx−1y dy=0.-d\left(\frac{1}{xy}\right) + \frac{2}{x}\,dx - \frac{1}{y}\,dy = 0.
 Step 3: Integrate:
−1xy+2log⁡∣x∣−log⁡∣y∣=c⇒2log⁡∣x∣−log⁡∣y∣−1xy=c.-\frac{1}{xy} + 2\log|x| - \log|y| = c \Rightarrow 2\log|x| - \log|y| - \frac{1}{xy} = c.
 Correct answer: (1)
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