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Primitive of an Exact Differential Equation | JEE

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

Question
The primitive of the differential equation (2xy4ey+2xy3+y) dx+(x2y4ey−x2y2−3x) dy=0(2xy^4e^y + 2xy^3 + y)\,dx + (x^2y^4e^y - x^2y^2 - 3x)\,dy = 0 is:
Ax2ey+x2y−xy3=Kx^2e^y + \dfrac{x^2}{y} - \dfrac{x}{y^3} = K
Bx2ey−x2y+xy3=Kx^2e^y - \dfrac{x^2}{y} + \dfrac{x}{y^3} = K
Cx2ey+x2y+xy3=Kx^2e^y + \dfrac{x^2}{y} + \dfrac{x}{y^3} = Kcorrect
Dx2ey−x2y−xy3=Kx^2e^y - \dfrac{x^2}{y} - \dfrac{x}{y^3} = K
Solution
Regroup the equation as
y4(2xey dx+x2ey dy)+y2(2xy dx−x2 dy)+(y dx−3x dy)=0,y^4\big(2xe^y\,dx + x^2e^y\,dy\big) + y^2\big(2xy\,dx - x^2\,dy\big) + \big(y\,dx - 3x\,dy\big) = 0,
 then divide throughout by y4y^4:
(2xey dx+x2ey dy)+2xy dx−x2 dyy2+y dx−3x dyy4=0.\big(2xe^y\,dx + x^2e^y\,dy\big) + \frac{2xy\,dx - x^2\,dy}{y^2} + \frac{y\,dx - 3x\,dy}{y^4} = 0.
 Each group is an exact differential:
d(x2ey)+d(x2y)+d(xy3)=0.d(x^2e^y) + d\left(\frac{x^2}{y}\right) + d\left(\frac{x}{y^3}\right) = 0.
 Integrating,
x2ey+x2y+xy3=K.x^2e^y + \frac{x^2}{y} + \frac{x}{y^3} = K.
 Correct answer: (3)
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