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

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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 $x, y$:

\(\ds \map M {t x, t y}\) \(=\) \(\ds 3 \paren {\paren {t x}^2 + \paren {t y}^2} \arctan \dfrac {t y} {t x} + t x t y\)
\(\ds \) \(=\) \(\ds t^2 \paren {3 \paren {x^2 + y^2} \arctan \dfrac y x + x y}\)
\(\ds \) \(=\) \(\ds t^2 \, \map M {x, y}\)
\(\ds \map N {t x, t y}\) \(=\) \(\ds \paren {t x}^2\)
\(\ds \) \(=\) \(\ds t^2 \paren {x^2}\)
\(\ds \) \(=\) \(\ds t^2 \, \map N {x, y}\)


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:

\(\ds \ln x\) \(=\) \(\ds \int \frac {\d z} {\dfrac {3 \paren {1^2 + z^2} \arctan \frac z 1 + z} {1^2} - z} + C_1\)
\(\ds \) \(=\) \(\ds \int \frac {\d z} {3 \paren {1 + z^2} \arctan z} + C_1\)


Substituting $u = \arctan z$:

\(\ds \dfrac {\d u} {\d x}\) \(=\) \(\ds \frac 1 {z^2 + 1}\) Derivative of Arctangent Function
\(\ds \leadsto \ \ \) \(\ds \dfrac {\d z} {\d u}\) \(=\) \(\ds z^2 + 1\)


Hence:

\(\ds \ln x\) \(=\) \(\ds \int \frac {\d z} {3 \paren {1 + z^2} \arctan z} + C_1\)
\(\ds \) \(=\) \(\ds \int \frac {\d u} {3 u} + C_1\)
\(\ds \) \(=\) \(\ds \frac 1 3 \ln u + C_1\) Primitive of Reciprocal
\(\ds \leadsto \ \ \) \(\ds \ln x^3 + \ln C\) \(=\) \(\ds \ln u\) putting $\ln C = - C_1$
\(\ds \leadsto \ \ \) \(\ds C x^3\) \(=\) \(\ds u\)
\(\ds \leadsto \ \ \) \(\ds C x^3\) \(=\) \(\ds \arctan \frac y x\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds x \tan C x^3\)

$\blacksquare$


Sources