Bernoulli's Equation/2 x y dx + (x^2 + 2 y) dy = 0

Theorem

The first order ODE:

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

has the solution:

$y \paren {x^2 + y} = C$

Proof

\(\ds 2 x y \rd x + \paren {x^2 + 2 y} \rd y\)

\(=\)

\(\ds 0\)

\(\text {(2)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac {\d x} {\d y} + \frac 1 {2 y} x\)

\(=\)

\(\ds -\frac 1 x\)

It can be seen that $(2)$ is in the form:

$\dfrac {\d x} {\d y} + \map P y x = \map Q y x^n$

where:

$\map P y = \dfrac 1 {2 y}$

$\map Q y = -1$

$n = -1$

and so is an example of Bernoulli's equation.

By Solution to Bernoulli's Equation it has the general solution:

$(3): \quad \ds \frac {\map \mu y} {x^{n - 1} } = \paren {1 - n} \int \map Q y \, \map \mu y \rd y + C$

where:

$\map \mu y = e^{\paren {1 - n} \int \map P y \rd y}$

Thus $\map \mu x$ is evaluated:

\(\ds \paren {1 - n} \int \map P y \rd y\)

\(=\)

\(\ds \paren {1 - \paren {-1} } \int \dfrac 1 {2 y} \rd y\)

\(\ds \)

\(=\)

\(\ds 2 \frac 1 2 \ln y\)

\(\ds \)

\(=\)

\(\ds \ln y\)

\(\ds \leadsto \ \ \)

\(\ds \map \mu y\)

\(=\)

\(\ds e^{\ln y}\)

\(\ds \)

\(=\)

\(\ds y\)

and so substituting into $(3)$:

\(\ds \frac y {x^{\paren {-1} - 1} }\)

\(=\)

\(\ds \paren {1 - \paren {-1} } \int \paren {-1} y \rd y\)

\(\ds \)

\(=\)

\(\ds -2 \int y \rd y\)

\(\ds \leadsto \ \ \)

\(\ds y x^2\)

\(=\)

\(\ds -2 \frac {y^2} 2 + C\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds C\)

$\blacksquare$