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Reducible Differential Equation in Exact Form | JEE

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

Question
The solution of dydx=y3e2x+y2\dfrac{dy}{dx} = \dfrac{y^3}{e^{2x} + y^2} is
Ae−2xy2+2ln⁡∣y∣=ce^{-2x}y^2 + 2\ln|y| = ccorrect
Be2xy2−2ln⁡∣y∣=ce^{2x}y^2 - 2\ln|y| = c
Cex+ln⁡∣y∣=ce^x + \ln|y| = c
DNone of these
Solution
Step 1: Divide numerator and denominator by e2xe^{2x}:
dydx=y3e−2x1+y2e−2x⇒dy+y2e−2x dy=y3e−2x dx.\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(e−2xy2)=2e−2x(y dy−y2 dx)d\big(e^{-2x}y^2\big) = 2e^{-2x}(y\,dy - y^2\,dx). Rearranging the equation and matching this form gives
2dyy+d(e−2xy2)=0.2\frac{dy}{y} + d\big(e^{-2x}y^2\big) = 0.
 Step 3: Integrate:
2ln⁡∣y∣+e−2xy2=k⇒e−2xy2+2ln⁡∣y∣=c.2\ln|y| + e^{-2x}y^2 = k \Rightarrow e^{-2x}y^2 + 2\ln|y| = c.
 (Check: differentiating e−2xy2+2ln⁡∣y∣=ce^{-2x}y^2 + 2\ln|y| = c got back same DE dydx=y3e2x+y2\dfrac{dy}{dx} = \dfrac{y^3}{e^{2x}+y^2}.) Correct answer: (1)
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