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

Theorem
The first order ordinary differential equation:


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

is a homogeneous differential equation with solution:


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

Proof
Let:
 * $M \left({x, y}\right) = 3 \left({x^2 + y^2}\right) \arctan \dfrac y x + x y$
 * $N \left({x, y}\right) = x^2$

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

where:
 * $f \left({x, y}\right) = \dfrac {3 \left({x^2 + y^2}\right) \arctan \dfrac y x + x y} {x^2}$

Thus:

Substituting $u = \arctan z$:

Hence: