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Solving a Differential Equation by d(1/x−1/y) | JEE

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

Question
The solution of x2 dy−y2 dx+xy2(x−y) dy=0x^2\,dy - y^2\,dx + xy^2(x - y)\,dy = 0 is:
Ay=x+cxy ey2/2y = x + cxy\,e^{y^2/2}correct
By=x+cxy ex2/2y = x + cxy\,e^{x^2/2}
Cy=x+cxy e−y2/2y = x + cxy\,e^{-y^2/2}
Dy=x+cxy e−x2/2y = x + cxy\,e^{-x^2/2}
Solution
Step 1: Divide by x2y2x^2y^2:
dyy2−dxx2+(1−yx)dy=0.\frac{dy}{y^2} - \frac{dx}{x^2} + \left(1 - \frac{y}{x}\right)dy = 0.
 Step 2: Note dyy2−dxx2=d(1x−1y)\dfrac{dy}{y^2} - \dfrac{dx}{x^2} = d\left(\dfrac{1}{x} - \dfrac{1}{y}\right) and (1−yx)dy=−(1x−1y)y dy\left(1 - \dfrac{y}{x}\right)dy = -\left(\dfrac{1}{x} - \dfrac{1}{y}\right)y\,dy. The equation becomes
d(1x−1y)1x−1y=y dy.\frac{d\left(\frac{1}{x} - \frac{1}{y}\right)}{\frac{1}{x} - \frac{1}{y}} = y\,dy.
 Step 3: Integrate:
ln⁡(1x−1y)=y22+c⇒1x−1y=Key2/2⇒y−xxy=Key2/2.\ln\left(\frac{1}{x} - \frac{1}{y}\right) = \frac{y^2}{2} + c \Rightarrow \frac{1}{x} - \frac{1}{y} = K e^{y^2/2} \Rightarrow \frac{y - x}{xy} = K e^{y^2/2}.
∴y=x+Kxy ey2/2.\therefore y = x + Kxy\,e^{y^2/2}.
 Correct answer: (1)
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