First Order ODE/(sine x sine y - x e^y) dy = (e^y + cosine x cosine y) dx

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Theorem

The first order ordinary differential equation:

$(1): \quad \paren {\sin x \sin y - x e^y} \rd y = \paren {e^y + \cos x \cos y} \rd x$


is an exact differential equation with solution:

$\sin x \cos y + x e^y = C$


This can also be presented as:

$\dfrac {\d y} {\d x} = \dfrac {e^y + \cos x \cos y} {\sin x \sin y - x e^y}$


Proof

First express $(1)$ in the form:

$(2): \quad -\paren {e^y + \cos x \cos y} + \paren {\sin x \sin y - x e^y} \dfrac {\d y} {\d x}$

Let:

$\map M {x, y} = -\paren {e^y + \cos x \cos y}$
$\map N {x, y} = \sin x \sin y - x e^y$

Then:

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


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:

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

and:

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


Thus:

$\map f {x, y} = -\sin x \cos y - x e^y$

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

$\sin x \cos y + x e^y = C$

$\blacksquare$


Sources

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