First Order ODE/(x + y) dx = (x - y) dy/Proof 1

Theorem
The first order ordinary differential equation:


 * $(1): \quad \paren {x + y} \rd x = \paren {x - y} \rd y$

is a homogeneous differential equation with solution:


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

Proof
Let:
 * $\map M {x, y} = x + y$
 * $\map N {x, y} = x - y$

We have that:


 * $\map M {t x, t y} = t x + t y = t \paren {x + y} = t \map M {x, y}$
 * $\map N {t x, t y} = t x - t y = t \paren {x - y} = t \map N {x, y}$

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:


 * $\displaystyle \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$

where:
 * $\map f {1, z} = \dfrac {1 + z} {1 - z}$

Hence:

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


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