Q248 / 270 · NDA 2026 (Phase 1) · Mathematics

Passage: Consider the differential equation $e^{x+y}\dfrac{dy}{dx}=e^{x-y}$.

Question: What is the solution of the differential equation with $y(0)=0$ ?

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

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:

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

$e^{2y} = 2x + 1$

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

Hence, the correct answer is option: 'C'.

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