First Order ODE/(exp x - 3 x^2 y^2) y' + y exp x = 2 x y^3

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Theorem

The first order ordinary differential equation:

$(1): \quad \paren {e^x - 3 x^2 y^2} y' + y e^x = 2 x y^3$

is an exact differential equation with solution:

$y e^x - x^2 y^3 = C$


Proof

Let $(1)$ be expressed as:

$\paren {y e^x - 2 x y^3} \rd x + \paren {e^x - 3 x^2 y^2} \rd y = 0$


Let:

$\map M {x, y} = y e^x - 2 x y^3$
$\map N {x, y} = e^x - 3 x^2 y^2$

Then:

\(\ds \dfrac {\partial M} {\partial y}\) \(=\) \(\ds e^x - 6 x y^2\)
\(\ds \dfrac {\partial N} {\partial x}\) \(=\) \(\ds e^x - 6 x y^2\)


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:

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

and:

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

Thus:

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

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

$y e^x + x^2 y^3 = C$

$\blacksquare$


Sources

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