First Order ODE/-1 over y sine x over y dx + x over y^2 sine x over y dy

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Theorem

The first order ordinary differential equation:

$(1): \quad -\dfrac 1 y \sin \dfrac x y \rd x + \dfrac x {y^2} \sin \dfrac x y \rd y = 0$


is an exact differential equation with solution:

$\dfrac x y = C$


This can also be presented as:

$\dfrac {\d y} {\d x} = \dfrac {\dfrac 1 y \sin \dfrac x y} {\dfrac x {y^2} \sin \dfrac x y}$


Proof

Let:

$\map M {x, y} = -\dfrac 1 y \sin \dfrac x y$
$\map N {x, y} = \dfrac x {y^2} \sin \dfrac x y$

Then:

\(\ds \frac {\partial M} {\partial y}\) \(=\) \(\ds \frac 1 {y^2} \sin \frac x y + \paren {-\frac 1 y} \paren {-\frac 1 {y^2} } x \cos \frac x y\)
\(\ds \) \(=\) \(\ds \frac 1 {y^2} \sin \frac x y + \frac x {y^3} \cos \frac x y\)
\(\ds \frac {\partial N} {\partial x}\) \(=\) \(\ds \frac 1 {y^2} \sin \frac x y + \frac x {y^2} \frac 1 y \cos \frac x y\)
\(\ds \) \(=\) \(\ds \frac 1 {y^2} \sin \frac x y + \frac x {y^3} \cos \frac x y\)


Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact.


By Solution to Exact Differential Equation, the solution to $(1)$ is:

$\map f {x, y} = C$

where:

\(\ds \dfrac {\partial f} {\partial x}\) \(=\) \(\ds \map M {x, y}\)
\(\ds \dfrac {\partial f} {\partial y}\) \(=\) \(\ds \map N {x, y}\)


Hence:

\(\ds f\) \(=\) \(\ds \map M {x, y} \rd x + \map g y\)
\(\ds \) \(=\) \(\ds \int \paren {-\dfrac 1 y \sin \dfrac x y} \rd x + \map g y\)
\(\ds \) \(=\) \(\ds -\dfrac 1 y y \cos \dfrac x y + \map g y\)
\(\ds \) \(=\) \(\ds -\cos \dfrac x y + \map g y\)


and:

\(\ds f\) \(=\) \(\ds \int \map N {x, y} \rd y + \map h x\)
\(\ds \) \(=\) \(\ds \int \paren {\dfrac x {y^2} \sin \dfrac x y} \rd y + \map h x\)


Set $z = \dfrac x y$:

\(\ds z\) \(=\) \(\ds \frac x y\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds -\frac x {y^2}\)


Thus by Integration by Substitution:

\(\ds \int \frac x {y^2} \sin \frac x y \rd y\) \(=\) \(\ds \int \frac x {y^2} \frac {-y^2} x \sin z \rd z\)
\(\ds \) \(=\) \(\ds -\int \sin z \rd z\)
\(\ds \) \(=\) \(\ds \cos z\)
\(\ds \) \(=\) \(\ds \cos \frac x y\)


Thus:

$\map f {x, y} = -\cos \dfrac x y$

and by Solution to Exact Differential Equation, the solution to $(1)$, after simplification, is:

\(\ds \cos \dfrac x y\) \(=\) \(\ds C_1\)
\(\ds \leadsto \ \ \) \(\ds \map \arccos {\cos \dfrac x y}\) \(=\) \(\ds \arccos C_1\)
\(\ds \leadsto \ \ \) \(\ds \dfrac x y\) \(=\) \(\ds C\) where $C = \arccos C_1$

$\blacksquare$


Sources

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