First Order ODE/exp x sine y dx + exp x cos y dy = y sine x y dx + x sine x y dy

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Theorem

The first order ordinary differential equation:

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

is an exact differential equation with solution:

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


Proof

Let $(1)$ be expressed as:

$\paren {e^x \sin y - y \sin x y} \rd x + \paren {e^x \cos y - x \sin x y} \rd y = 0$

Let:

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

Then:

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


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:

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

and:

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

Thus:

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

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

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

$\blacksquare$


Sources

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