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Exact Differential Equation with Initial Condition | JEE

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

Question
The differential equation ey dx+(eyx+2y) dy=0e^y\,dx + (e^y x + 2y)\,dy = 0 has the particular solution y(0)=1y(0) = 1. The value of xx when y=0y = 0 is:
A11correct
B00
C22
D−1-1
Solution
Group as exact differentials:
ey dx+xey dy+2y dy=0⇒d(xey)+d(y2)=0⇒xey+y2=C.e^y\,dx + x e^y\,dy + 2y\,dy = 0 \Rightarrow d(xe^y) + d(y^2) = 0 \Rightarrow xe^y + y^2 = C.
 Apply y(0)=1y(0) = 1 (i.e. x=0x = 0 at y=1y = 1): 0⋅e+1=C⇒C=10\cdot e + 1 = C \Rightarrow C = 1, so xey+y2=1xe^y + y^2 = 1. When y=0y = 0: x⋅e0+0=1⇒x=1x\cdot e^0 + 0 = 1 \Rightarrow x = 1. Correct answer: (1)
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