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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

(2xy^4e^y + 2xy^3 + y)\,dx + (x^2y^4e^y - x^2y^2 - 3x)\,dy = 0

is:

x^2e^y + \dfrac{x^2}{y} - \dfrac{x}{y^3} = K

x^2e^y - \dfrac{x^2}{y} + \dfrac{x}{y^3} = K

x^2e^y + \dfrac{x^2}{y} + \dfrac{x}{y^3} = K

correct

x^2e^y - \dfrac{x^2}{y} - \dfrac{x}{y^3} = K

Solution

Regroup the equation as

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

y^4

\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(x^2e^y) + d\left(\frac{x^2}{y}\right) + d\left(\frac{x}{y^3}\right) = 0.

Integrating,

x^2e^y + \frac{x^2}{y} + \frac{x}{y^3} = K.

Correct answer: (3)

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