# Solve: dy/dx+ y = xy^3 .??

Jul 5, 2018

$y = \pm \frac{1}{\sqrt{x + \frac{1}{2} + C {e}^{2 x}}}$

#### Explanation:

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

This is non-linear but we can make it linear using a Bernoulli substitution. Here we let:

• $z = \frac{1}{y} ^ 2 q \quad q \quad z ' = - \frac{2}{y} ^ 3 y '$

Multiply both sides of the DE by $- \frac{2}{y} ^ 3$:

${\underbrace{\frac{\mathrm{dy}}{\mathrm{dx}} \cdot \left(- \frac{2}{y} ^ 3\right)}}_{= z '} + {\underbrace{y \cdot \left(- \frac{2}{y} ^ 3\right)}}_{= - 2 z} = {\underbrace{x {y}^{3} \cdot \left(- \frac{2}{y} ^ 3\right)}}_{= - 2 x}$

For this, the integrating factor is: $\exp \left(\int \left(- 2\right) \mathrm{dx}\right) = {e}^{- 2 x}$

${e}^{- 2 x} \left(z ' - 2 z\right) \left[= {\left({e}^{- 2 x} z\right)}^{'}\right] = - 2 x {e}^{- 2 x}$

Integrating both sides, using IBP on RHS:

${e}^{- 2 x} z = \int \setminus x \setminus d \left({e}^{- 2 x}\right)$

${e}^{- 2 x} z = \setminus x \setminus {e}^{- 2 x} - \int \mathrm{dx} \setminus {e}^{- 2 x}$

${e}^{- 2 x} z = x {e}^{- 2 x} + \frac{1}{2} \setminus {e}^{- 2 x} + C$

$z = x + \frac{1}{2} + C {e}^{2 x} q \quad = \frac{1}{y} ^ 2$

$y = \pm \frac{1}{\sqrt{x + \frac{1}{2} + C {e}^{2 x}}}$