If k is an arbitrary constant, then what is the general solution of the equation (x+y)^{2}\frac{dy}{dx}=k^{2}?

Explanation

Let v = x+y. Then \frac{dv}{dx} = 1 + \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{dv}{dx} - 1. The equation becomes v^2\left(\frac{dv}{dx} - 1\right) = k^2 \implies \frac{dv}{dx} = \frac{k^2+v^2}{v^2}. Separating variables gives \int \left(1 - \frac{k^2}{k^2+v^2}\right) \,dv = \int dx. This integrates to v - k\tan^{-1}\left(\frac{v}{k}\right) = x + c. Substituting v=x+y yields y - c = k\tan^{-1}\left(\frac{x+y}{k}\right), which means x+y = k\tan\left(\frac{y-c}{k}\right).

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