Differential EquationsmediumFree

Reducible Differential Equation in Exact Form | JEE

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

Question

The solution of

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

e^{-2x}y^2 + 2\ln|y| = c

correct

e^{2x}y^2 - 2\ln|y| = c

e^x + \ln|y| = c

DNone of these

Solution

Step 1: Divide numerator and denominator by

e^{2x}

\frac{dy}{dx} = \frac{y^3 e^{-2x}}{1 + y^2 e^{-2x}} \Rightarrow dy + y^2 e^{-2x}\,dy = y^3 e^{-2x}\,dx.

Step 2: Recognise the combination

d\big(e^{-2x}y^2\big) = 2e^{-2x}(y\,dy - y^2\,dx)

. Rearranging the equation and matching this form gives

2\frac{dy}{y} + d\big(e^{-2x}y^2\big) = 0.

Step 3: Integrate:

2\ln|y| + e^{-2x}y^2 = k \Rightarrow e^{-2x}y^2 + 2\ln|y| = c.

(Check: differentiating

e^{-2x}y^2 + 2\ln|y| = c

got back same DE

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

.) Correct answer: (1)

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.