Passage: Consider the differential equation e^ x+y (dy)/(dx)=e^ x-y . Question: What is the solution of the differential equation with y(0)=0 ?

Correct answer: C. 2y= (2x+1) The given differential equation is e^ x+y (dy)/(dx)=e^ x-y e^x e^y (dy)/(dx) = e^x e^ -y e^y (dy)/(dx) = e^ -y e^ 2y dy = dx Integrating both sides: ∫ e^ 2y dy = ∫ dx e^ 2y 2 = x + C Given y(0)=0 , substituting x=0 and y=0 : (e^0)/(2) = 0 + C C = (1)/(2) Substituting C back into the equation: e^ 2y 2 = x + (1)/(2) e^ 2y = 2x + 1 2y = (2x+1)

NDA 2026

Exam: NDA
Year: 2026
Subject: Mathematics
Chapter: Differential Equations

NDA 2026 · Mathematics · Differential Equations

Passage: Consider the differential equation e^ x+y (dy)/(dx)=e^ x-y...

Passage: Consider the differential equation

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

.

Question: What is the solution of the differential equation with

y(0)=0

Options

A $y=\ln(2x+1)$
B $y=\ln(2x-1)$
C $2y=\ln(2x+1)$
D $2y=\ln(2x-1)$

Correct answer

C. $2y=\ln(2x+1)$

Step-by-step solution

The given differential equation is

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

e^x e^y \dfrac{dy}{dx} = e^x e^{-y}

e^y \dfrac{dy}{dx} = e^{-y}

e^{2y} dy = dx

Integrating both sides:

\int e^{2y} dy = \int dx

\dfrac{e^{2y}}{2} = x + C

Given

y(0)=0

, substituting

x=0

and

y=0

\dfrac{e^0}{2} = 0 + C \Rightarrow C = \dfrac{1}{2}

Substituting

C

back into the equation:

\dfrac{e^{2y}}{2} = x + \dfrac{1}{2}

e^{2y} = 2x + 1

2y = \ln(2x+1)

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