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

Theorem
The linear first order ODE:
 * $(1): \quad x y' + y = x^4 y^3$

has the solution:
 * $\dfrac 1 {y^2} = - x^4 + C x^2$

Proof
Let $(1)$ be rearranged as:
 * $(2): \quad \dfrac {\mathrm d y} {\mathrm d x} + \dfrac 1 x y = x^3 y^3$

It can be seen that $(2)$ is in the form:
 * $\dfrac {\mathrm d y}{\mathrm d x} + P \left({x}\right) y = Q \left({x}\right) y^n$

where:
 * $P \left({x}\right) = \dfrac 1 x$
 * $Q \left({x}\right) = x^3$
 * $n = 3$

and so is an example of Bernoulli's equation.

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

where:
 * $\mu \left({x}\right) = e^{\int \left({1 - n}\right) P \left({x}\right) \, \mathrm d x}$

Thus $\mu \left({x}\right)$ is evaluated:

and so substituting into $(3)$:

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


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