Solve the differential equation $\left(x + y\right)^2 \; \dfrac{dy}{dx} = a^2$
The given differential equation is $\;\;$ $\left(x + y\right)^2 \; \dfrac{dy}{dx} = a^2$ $\;\;\; \cdots \; (1)$
Let $x + y =u$ $\;\;\; \cdots \; (2a)$
Differentiating equation $(2a)$ w.r.t x gives
$1 + \dfrac{dy}{dx} = \dfrac{du}{dx}$
i.e. $\dfrac{dy}{dx} = \dfrac{du}{dx} - 1$ $\;\;\; \cdots \; (2b)$
$\therefore$ $\;$ In view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes
$u^2 \left(\dfrac{du}{dx} -1\right) = a^2$
i.e. $\dfrac{du}{dx} - 1 = \dfrac{a^2}{u^2}$
i.e. $\dfrac{du}{dx} = \dfrac{a^2}{u^2} + 1$
i.e. $\dfrac{du}{dx} = \dfrac{a^2 + u^2}{u^2}$
i.e. $\left(\dfrac{u^2}{u^2 + a^2}\right) du = dx$ $\;\;\; \cdots \; (3)$
Integrating equation $(3)$ gives
$\displaystyle \int \dfrac{u^2}{u^2 + a^2} \; du = \int dx$
i.e. $\displaystyle \int \dfrac{u^2 + a^2}{u^2 + a^2} \; du - \int \dfrac{a^2}{u^2 + a^2} \; du = \int dx$
i.e. $\displaystyle \int du - a^2 \int \dfrac{du}{u^2 + a^2} = \int dx$
i.e. $u - a^2 \times \dfrac{1}{a} \tan^{-1}\left(\dfrac{u}{a}\right) = x + c$ $\;\;\; \cdots \; (4)$
Substituting the value of $u$ from equation $(2a)$ gives
i.e. $x + y - a \tan^{-1} \left(\dfrac{x + y}{a}\right) = x + c$
i.e. $y - a \tan^{-1} \left(\dfrac{x + y}{a}\right) = c$ $\;\;\; \cdots \; (5)$
Equation $(5)$ is the general solution of the given differential equation.