First Order ODE/(2 x y^3 + y cosine x) dx + (3 x^2 y^2 + sine x) dy

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Theorem

The first order ordinary differential equation:

$(1): \quad \paren {2 x y^3 + y \cos x} \d x + \paren {3 x^2 y^2 + \sin x} \d y = 0$


is an exact differential equation with solution:

$x^2 y^3 + y \sin x = C$


This can also be presented as:

$\dfrac {\d y} {\d x} = -\dfrac {2 x y^3 + y \cos x} {3 x^2 y^2 + \sin x}$


Proof

Let:

$\map M {x, y} = 2 x y^3 + y \cos x$
$\map N {x, y} = 3 x^2 y^2 + \sin x$

Then:

\(\ds \frac {\partial M} {\partial y}\) \(=\) \(\ds 2 x \cdot 3 y^2 + \cos x\)
\(\ds \) \(=\) \(\ds 6 x y^2 + \cos x\)
\(\ds \frac {\partial N} {\partial x}\) \(=\) \(\ds 3 x^2 \cdot 2 y^2 + \cos x\)
\(\ds \) \(=\) \(\ds 6 x y^2 + \cos x\)


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:

\(\ds f\) \(=\) \(\ds \map M {x, y} \rd x + \map g y\)
\(\ds \) \(=\) \(\ds \int \paren {2 x y^3 + y \cos x} \rd x + \map g y\)
\(\ds \) \(=\) \(\ds x^2 y^3 + y \sin x + \map g y\)


and:

\(\ds f\) \(=\) \(\ds \int \map N {x, y} \rd y + \map h x\)
\(\ds \) \(=\) \(\ds \int \paren {3 x^2 y^2 + \sin x} \rd y + \map h x\)
\(\ds \) \(=\) \(\ds x^2 y^3 + y \sin x + \map h x\)


Thus:

$\map f {x, y} = x^2 y^3 + y \sin x$

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

$x^2 y^3 + y \sin x = C$

$\blacksquare$


Sources

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