First Order ODE/(x + y) dx = (x - y) dy

Theorem
The first order ordinary differential equation:


 * $(1): \quad \left({x + y}\right) \mathrm d x = \left({x - y}\right) \mathrm d y$

is a homogeneous differential equation with solution:


 * $\arctan \dfrac y x = \ln \sqrt{x^2 + y^2} + C$

Proof
Let:
 * $M \left({x, y}\right) = x + y$
 * $N \left({x, y}\right) = x - y$

We have that:


 * $M \left({t x, t y}\right) = t x + t y = t \left({x + y}\right) = t M \left({x, y}\right)$
 * $N \left({t x, t y}\right) = t x - t y = t \left({x - y}\right) = t N \left({x, y}\right)$

Thus both $M$ and $N$ are homogeneous functions of degree $1$.

Thus by definition $(1)$ is a homogeneous differential equation.

By Solution to Homogeneous Differential Equation:


 * $\displaystyle \ln x = \int \frac {\mathrm d z} {f \left({1, z}\right) - z} + C$

where:
 * $f \left({1, z}\right) = \dfrac {1 + z} {1 - z}$

Hence:

Substituting $y / x$ for $z$ reveals the solution:


 * $\arctan \dfrac y x = \ln \sqrt{x^2 + y^2} + C$