First Order ODE/(y + y cosine x y) dx + (x + x cosine x y) dy = 0

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Theorem

The first order ordinary differential equation:

$(1): \quad \paren {y + y \cos x y} \rd x + \paren {x + x \cos x y} \rd y = 0$


is an exact differential equation with solution:

$x y + \sin x y = C$


This can also be presented as:

$\dfrac {\d y} {\d x} + \dfrac {y + y \cos x y} {x + x \cos x y} = 0$


Proof

Let:

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

Then:

\(\ds \dfrac {\partial M} {\partial y}\) \(=\) \(\ds 1 + \cos x y - x \sin x y\)
\(\ds \dfrac {\partial N} {\partial x}\) \(=\) \(\ds 1 + \cos x y - x \sin x y\)


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:

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

and:

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


Thus:

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

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

$x y + \sin x y = C$

$\blacksquare$


Sources

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