Q244 / 270 · NDA 2025 (Phase 2) · Mathematics
If $k$ is an arbitrary constant, then what is the general solution of the equation $(x + y)^2 \dfrac{dy}{dx} = k^2$?
- A$y + x = \tan(x + c) + k$
- Correct answerB$x + y = k\tan\left(\dfrac{y - c}{k}\right)$
- C$x - y = k\tan\left(\dfrac{y - c}{k}\right)$
- D$y - x = \tan(x + c) + k$
Solution
Let $x + y = v$
Differentiating with respect to $x$, we get $1 + \dfrac{dy}{dx} = \dfrac{dv}{dx} \Rightarrow \dfrac{dy}{dx} = \dfrac{dv}{dx} - 1$
Substituting in the given differential equation:
$v^2 \left( \dfrac{dv}{dx} - 1 \right) = k^2$
$v^2 \dfrac{dv}{dx} = v^2 + k^2$
$\dfrac{v^2}{v^2 + k^2} dv = dx$
$\left( 1 - \dfrac{k^2}{v^2 + k^2} \right) dv = dx$
Integrating both sides:
$\int \left( 1 - \dfrac{k^2}{v^2 + k^2} \right) dv = \int dx$
$v - k \tan^{-1}\left(\dfrac{v}{k}\right) = x + c$
Substituting $v = x + y$:
$x + y - k \tan^{-1}\left(\dfrac{x + y}{k}\right) = x + c$
$y - c = k \tan^{-1}\left(\dfrac{x + y}{k}\right)$
$\dfrac{y - c}{k} = \tan^{-1}\left(\dfrac{x + y}{k}\right)$
$x + y = k \tan\left(\dfrac{y - c}{k}\right)$
Answer: $x + y = k\tan\left(\dfrac{y - c}{k}\right)$
Differentiating with respect to $x$, we get $1 + \dfrac{dy}{dx} = \dfrac{dv}{dx} \Rightarrow \dfrac{dy}{dx} = \dfrac{dv}{dx} - 1$
Substituting in the given differential equation:
$v^2 \left( \dfrac{dv}{dx} - 1 \right) = k^2$
$v^2 \dfrac{dv}{dx} = v^2 + k^2$
$\dfrac{v^2}{v^2 + k^2} dv = dx$
$\left( 1 - \dfrac{k^2}{v^2 + k^2} \right) dv = dx$
Integrating both sides:
$\int \left( 1 - \dfrac{k^2}{v^2 + k^2} \right) dv = \int dx$
$v - k \tan^{-1}\left(\dfrac{v}{k}\right) = x + c$
Substituting $v = x + y$:
$x + y - k \tan^{-1}\left(\dfrac{x + y}{k}\right) = x + c$
$y - c = k \tan^{-1}\left(\dfrac{x + y}{k}\right)$
$\dfrac{y - c}{k} = \tan^{-1}\left(\dfrac{x + y}{k}\right)$
$x + y = k \tan\left(\dfrac{y - c}{k}\right)$
Answer: $x + y = k\tan\left(\dfrac{y - c}{k}\right)$
Hence, the correct answer is option: 'B'.