First Order ODE/x^2 y' = 3 (x^2 + y^2) arctan (y over x) + x y

Theorem
The first order ordinary differential equation:


 * $(1): \quad x^2 \dfrac {\d y} {\d x} = 3 \paren {x^2 + y^2} \arctan \dfrac y x + x y$

is a homogeneous differential equation with solution:


 * $y = x \tan C x^3$

Proof
Let:
 * $\map M {x, y} = 3 \paren {x^2 + y^2} \arctan \dfrac y x + x y$
 * $\map N {x, y} = x^2$

Put $t x, t y$ for $x, y$:

Thus both $M$ and $N$ are homogeneous functions of degree $2$.

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

By Solution to Homogeneous Differential Equation, its solution is:
 * $\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$

where:
 * $\map f {x, y} = \dfrac {3 \paren {x^2 + y^2} \arctan \dfrac y x + x y} {x^2}$

Thus:

Substituting $u = \arctan z$:

Hence: