First Order ODE/x sine (y over x) y' = y sine (y over x) + x

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Theorem

The first order ordinary differential equation:

$(1): \quad x \sin \dfrac y x \dfrac {\d y} {\d x} = y \sin \dfrac y x + x$


is a homogeneous differential equation with solution:

$\cos \dfrac y x + \ln C x = 0$


Proof

Let:

$\map M {x, y} = y \sin \dfrac y x + x$
$\map N {x, y} = x \sin \dfrac y x$


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

\(\ds \map M {t x, t y}\) \(=\) \(\ds t y \sin \dfrac t y t x + t x\)
\(\ds \) \(=\) \(\ds t \paren {y \sin \dfrac y x + x}\)
\(\ds \) \(=\) \(\ds t \, \map M {x, y}\)
\(\ds \map N {t x, t y}\) \(=\) \(\ds t x \sin \dfrac {t y} {t x}\)
\(\ds \) \(=\) \(\ds t \paren {x \sin \dfrac y x}\)
\(\ds \) \(=\) \(\ds t \, \map N {x, y}\)


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, its solution is:

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

where:

$\map f {x, y} = \dfrac {y \sin \dfrac y x + x} {x \sin \dfrac y x}$


Thus:

\(\ds \ln x\) \(=\) \(\ds \int \frac {\d z} {\dfrac {z \sin z + 1} {\sin z} - z} + C_1\)
\(\ds \) \(=\) \(\ds \int \sin z \rd z + C_1\)
\(\ds \) \(=\) \(\ds -\cos z + C_1\) Primitive of $\sin z$
\(\ds \leadsto \ \ \) \(\ds \ln x + \ln C\) \(=\) \(\ds -\cos z\) putting $\ln C = - C_1$
\(\ds \leadsto \ \ \) \(\ds \cos \frac y x + \ln C x = 0\) \(=\) \(\ds 0\)

$\blacksquare$


Sources