If k is an arbitrary constant, then what is the general solution of...

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$<br><br>Integrating both sides:<br><br>$\int \left( 1 - \dfrac{k^2}{v^2 + k^2} \right) dv = \int dx$<br><br>$v - k \tan^{-1}\left(\dfrac{v}{k}\right) = x + c$<br><br>Substituting$v = x + y$:<br><br>$x + y - k \tan^{-1} \left(\dfrac{x + y}{k}\right) =$x + c$

$y - c = k \tan^{-1}\left(\dfrac{x + y}{k}\right)<br><br>\dfrac{y - c}{k} = \tan^{-1}$\left(\dfrac{x + y}{k}\right)$<br><br>$x + y = k \tan \left(\dfrac{y - c}{k}\right)

Answer: $x + y = k\tan \left(\dfrac{y - c}{k}\right)