First Order ODE/(y^2 exp x y + cosine x) dx + (exp x y + x y exp x y) dy = 0

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Theorem

The first order ordinary differential equation:

$(1): \quad \paren {y^2 e^{x y} + \cos x} \rd x + \paren {e^{x y} + x y e^{x y} } \rd y = 0$

is an exact differential equation with solution:

$y e^{x y} + \sin x = C$


Proof

Let:

$\map M {x, y} = y^2 e^{x y} + \cos x$
$\map N {x, y} = e^{x y} + x y e^{x y}$

Then:

\(\ds \dfrac {\partial M} {\partial y}\) \(=\) \(\ds 2 y e^{x y} + x y^2 e^{x y}\)
\(\ds \dfrac {\partial N} {\partial x}\) \(=\) \(\ds y e^{x y} + x y^2 e^{x y} + y e^{x y}\)
\(\ds \) \(=\) \(\ds 2 y e^{x y} + x y^2 e^{x y}\)


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:

\(\ds f\) \(=\) \(\ds \map M {x, y} \rd x + \map g y\)
\(\ds \) \(=\) \(\ds \int \paren {y^2 e^{x y} + \cos x} \rd x + \map g y\)
\(\ds \) \(=\) \(\ds y e^{x y} + \sin x + \map g y\)

and:

\(\ds f\) \(=\) \(\ds \int \map N {x, y} \rd y + \map h x\)
\(\ds \) \(=\) \(\ds \int \paren {e^{x y} + x y e^{x y} } \rd y + \map h x\)
\(\ds \) \(=\) \(\ds \frac {e^{x y} } x + y e^{x y} - \frac {e^{x y} } x + \map h x\) Primitive of $x e^{a x}$
\(\ds \) \(=\) \(\ds y e^{x y} + \map h x\)

Thus:

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

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

$y e^{x y} + \sin x = C$

$\blacksquare$


Sources