# Question #20d0b

Mar 18, 2017

This is a Bernoulli Differential Equation of the form :

$y ' + p \left(x\right) y = q \left(x\right) {y}^{n}$

#### Explanation:

Rewrite, $2 x \frac{\mathrm{dy}}{\mathrm{dx}} = 10 {x}^{3} {y}^{5} + y$, in the above form:

$y ' - \frac{1}{2 x} y = 5 {x}^{2} {y}^{5}$

Divide both sides by y^5

$y ' {y}^{-} 5 - \frac{1}{2 x} {y}^{-} 4 = 5 {x}^{2}$

Let $u = {y}^{-} 4$, then $u ' = 4 {y}^{-} 5 y ' \mathmr{and} y ' {y}^{-} 5 = \frac{u '}{4}$:

$\frac{u '}{4} - \frac{1}{2 x} u = 5 {x}^{2}$

$u ' - \frac{2}{x} u = 20 {x}^{2} \text{ }$

The integrating factor for equation  is $I = {e}^{\int - \frac{2}{x} \mathrm{dx}}$ which is a trivial integration:

$I = {e}^{- 2 \ln \left(x\right)}$

$I = {e}^{\ln} \left(\frac{1}{x} ^ 2\right)$

$I = \frac{1}{x} ^ 2$

Divide equation  by ${x}^{2}$

$\frac{u '}{x} ^ 2 - \frac{2}{{x}^{3}} u = 20$

We know that the left side is the reverse of the product rule so integration is simple:

$\frac{u \left(x\right)}{x} ^ 2 = 20 x + C$

Multiply both sides by ${x}^{2}$:

$u \left(x\right) = 20 {x}^{3} + C {x}^{2}$

Reverse the substitution:

${y}^{-} 4 = 20 {x}^{3} + C {x}^{2}$

$y = {\left(20 {x}^{3} + C {x}^{2}\right)}^{- \frac{1}{4}}$