# What is the general solution of the differential equation ?  2xdy/dx = 10x^3y^5+y

Mar 28, 2017

$y = \frac{\sqrt{x}}{\sqrt{\left(C - 4 {x}^{5}\right)}}$

#### Explanation:

We have:

$2 x \frac{\mathrm{dy}}{\mathrm{dx}} = 10 {x}^{3} {y}^{5} + y$

We can rewrite as:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 5 {x}^{2} {y}^{5} + \frac{y}{2 x}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{y}{2 x} = 5 {x}^{2} {y}^{5}$

This is a Bernoulli equitation which has a standard method to solve. Let:

$u = {y}^{- 4} \implies \frac{\mathrm{du}}{\mathrm{dy}} = - 4 {y}^{- 5}$ and $\frac{\mathrm{dy}}{\mathrm{du}} = - {y}^{5} / 4$

By the chain rule we have;

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Substituting into the last DE we get;

$\left(- {y}^{5} / 4\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) - \frac{y}{2 x} = 5 {x}^{2} {y}^{5}$
$\therefore \frac{\mathrm{du}}{\mathrm{dx}} + 2 {y}^{- 4} / \left(x\right) = - 20 {x}^{2}$
$\therefore \frac{\mathrm{du}}{\mathrm{dx}} + \frac{2}{x} u = - 20 {x}^{2}$

So the substitution has reduced the DE into a first order linear differential equation of the form:

$\frac{d \zeta}{\mathrm{dx}} + P \left(x\right) \zeta = Q \left(x\right)$

We solve this using an Integrating Factor

$I = \exp \left(\setminus \int \setminus P \left(x\right) \setminus \mathrm{dx}\right)$
$\setminus \setminus = \exp \left(\int \setminus \frac{2}{x} \setminus \mathrm{dx}\right)$
$\setminus \setminus = \exp \left(2 \ln x\right)$
$\setminus \setminus = \exp \left(\ln {x}^{2}\right)$
$\setminus \setminus = {x}^{2}$

And if we multiply the last by this Integrating Factor, $I$, we will have a perfect product differential;

$\therefore \frac{\mathrm{du}}{\mathrm{dx}} + \frac{2}{x} u = - 20 {x}^{2}$
$\therefore {x}^{2} \frac{\mathrm{du}}{\mathrm{dx}} + 2 x u = - 20 {x}^{4}$
$\therefore \frac{d}{\mathrm{dx}} \left({x}^{2} u\right) = - 20 {x}^{4}$

Which is now a trivial separable DE, so we can "separate the variables" to get:

${x}^{2} u = \int \setminus - 20 {x}^{4} \setminus \mathrm{dx}$

And integrating gives us:

$\setminus \setminus \setminus \setminus \setminus {x}^{2} u = - 20 {x}^{5} / 5 + C$
$\therefore {x}^{2} u = - 4 {x}^{5} + C$

Restoring the substitution we get:

$\setminus {x}^{2} {y}^{- 4} = - 4 {x}^{5} + C$
$\therefore {y}^{- 4} = \frac{C - 4 {x}^{5}}{x} ^ 2$
$\therefore \setminus \setminus \setminus {y}^{4} = {x}^{2} / \left(C - 4 {x}^{5}\right)$
$\therefore \setminus \setminus \setminus \setminus \setminus y = {\left({x}^{2} / \left(C - 4 {x}^{5}\right)\right)}^{\frac{1}{4}}$
$\therefore \setminus \setminus \setminus \setminus \setminus y = \frac{\sqrt{x}}{\sqrt{\left(C - 4 {x}^{5}\right)}}$