First Order ODE/(x^2 y^3 + y) dx = (x^3 y^2 - x) dy

Theorem

The first order ODE:

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

has the general solution:

$-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$

Proof

Let $(1)$ be expressed as:

$(2): \quad \paren {y + x^2 y^3} \rd x + \paren {x - x^3 y^2} \rd y = 0$

We note that $(2)$ is in the form:

$\map M {x, y} \rd x + \map N {x, y} \rd y = 0$

but is not exact.

So, let:

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

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

Let:

$\map P {x, y} = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$

Thus:

\(\ds \map P {x, y}\)

\(=\)

\(\ds \paren {1 + 3 x^2 y^2} - \paren {1 - 3 x^2 y^2}\)

\(\ds \)

\(=\)

\(\ds 6 x^2 y^2\)

It can be observed that:

\(\ds \frac {\map P {x, y} } {\map N {x, y} y - \map M {x, y} x}\)

\(=\)

\(\ds \frac {6 x^2 y^2} {\paren {x y - x^3 y^3} - \paren {x y + x^3 y^3} }\)

\(\ds \)

\(=\)

\(\ds -\frac {6 x^2 y^2} {2 x^3 y^3}\)

\(\ds \)

\(=\)

\(\ds -\frac 3 {x y}\)

Thus $\dfrac {\map P {x, y} } {\map M {x, y} }$ is a function of $x y$.

So Integrating Factor for First Order ODE: Function of Product of Variables can be used.

Let $z = x y$.

Then:

$\map \mu {x y} = \map \mu z = e^{\int \map g z \rd z}$

Hence:

\(\ds \int \map g z \rd z\)

\(=\)

\(\ds \int -\frac 3 z \rd z\)

\(\ds \)

\(=\)

\(\ds - 3 \ln z\)

\(\ds \)

\(=\)

\(\ds \map \ln {z^{-3} }\)

\(\ds \leadsto \ \ \)

\(\ds e^{\int \map g z \rd z}\)

\(=\)

\(\ds \frac 1 {z^3}\)

\(\ds \)

\(=\)

\(\ds \frac 1 {x^3 y^3}\)

Thus an integrating factor for $(1)$ has been found:

$\mu = \dfrac 1 {x^3 y^3}$

which yields, when multiplying it throughout $(2)$:

$\paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \paren {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y = 0$

which is now exact.

By First Order ODE: $\paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \paren {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y = 0$, its general solution is:

$-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Problem $6$