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
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.7$: Homogeneous Equations: Problem $1 \ \text{(d)}$