Bernoulli's Equation/y' - (1 over x + 2 x^4) y = x^3 y^2

Theorem
The first order ODE:
 * $(1): \quad y' - \paren {\dfrac 1 x + 2 x^4} y = x^3 y^2$

has the general solution:
 * $y = \dfrac {2 x} {C \, \map \exp {-\dfrac {2 x^5} 5} - 1}$

Proof
It can be seen that $(1)$ is in the form:
 * $\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$

where:
 * $\map P x = -\paren {\dfrac 1 x + 2 x^4}$
 * $\map Q x = x^3$
 * $n = 2$

and so is an example of Bernoulli's equation.

By Solution to Bernoulli's Equation it has the general solution:
 * $(3): \quad \displaystyle \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \, \map \mu x \rd x + C$

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

Thus $\map \mu x$ is evaluated:

and so substituting into $(3)$:

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


 * $y = \dfrac {2 x} {C \, \map \exp {-\dfrac {2 x^5} 5} - 1}$