# Bernoulli's Equation/x y' + y = x^4 y^3

## Theorem

The first order ODE:

$(1): \quad x y' + y = x^4 y^3$

has the general solution:

$\dfrac 1 {y^2} = - x^4 + C x^2$

## Proof

Let $(1)$ be rearranged as:

$(2): \quad \dfrac {\d y} {\d x} + \dfrac 1 x y = x^3 y^3$

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

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

where:

$\map P x = \dfrac 1 x$
$\map Q x = x^3$
$n = 3$

and so is an example of Bernoulli's equation.

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

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

where:

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

Thus $\map \mu x$ is evaluated:

 $\ds \paren {1 - n} \int \map P x \rd x$ $=$ $\ds \paren {1 - 3} \int \dfrac 1 x \rd x$ $\ds$ $=$ $\ds -2 \ln x$ $\ds$ $=$ $\ds \map \ln {\frac 1 {x^2} }$ $\ds \leadsto \ \$ $\ds \map \mu x$ $=$ $\ds e^{\map \ln {\frac 1 {x^2} } }$ $\ds$ $=$ $\ds \frac 1 {x^2}$

and so substituting into $(3)$:

 $\ds \frac 1 {x^2} \frac 1 {y^2}$ $=$ $\ds \paren {1 - 3} \int x^3 \frac 1 {x^2} \rd x$ $\ds$ $=$ $\ds -2 \int x \rd x$ $\ds \leadsto \ \$ $\ds \frac 1 {x^2 y^2}$ $=$ $\ds -2 \dfrac {x^2} 2 + C$

Hence the general solution to $(1)$ is:

$\dfrac 1 {y^2} = - x^4 + C x^2$

$\blacksquare$