# Solve y'+ty=t^3y^3?

Apr 13, 2018

See below.

#### Explanation:

Making the substitution

$y = \frac{t}{z} \Rightarrow \frac{1}{2} t \frac{d}{\mathrm{dt}} \left({z}^{2}\right) - \left({t}^{2} + 1\right) {z}^{2} + {t}^{6} = 0$ now considering

$\xi = {z}^{2}$ we have the linear differential equation

$\frac{1}{2} t \xi ' - \left({t}^{2} + 1\right) \xi + {t}^{6} = 0$

now solving for $\xi$

$\xi = {t}^{2} \left({C}_{0} {e}^{{t}^{2}} + {t}^{2} + 1\right)$ and then

$z = \left\mid t \right\mid \sqrt{{C}_{0} {e}^{{t}^{2}} + {t}^{2} + 1}$ and finally

$y = \frac{t}{\left\mid t \right\mid \sqrt{{C}_{0} {e}^{{t}^{2}} + {t}^{2} + 1}}$