Originally Posted by

**seerTneerGevoLI** Alright, Jhevon. How does this look?

$\displaystyle \displaystyle t^2 y' + 2ty - y^3 = 0 \iff \frac{{y'}}{{y^3 }}t^2 + \frac{{2t}}{{y^2 }} = 1$

$\displaystyle u=\frac1{y^2}\implies u'=-\frac{2y'}{y^3}$

$\displaystyle -\frac{1}{2}u't^2+2tu=1$

$\displaystyle \displaystyle u' - \frac{4u}{t} = \frac{-2}{t^2}$

Integrating factor, M(t) is:

$\displaystyle \displaystyle M(t) = e^{\int{-\frac{4}{t}\,dt}} = e^{-4\ln{|t|}} = \frac{1}{t^4}$

Multiply each term by M(t)

$\displaystyle \displaystyle \frac{1}{t^4}u' - \frac{4u}{t}\cdot \frac{1}{t^4} = -\frac{2}{t^2}\cdot \frac{1}{t^4} \implies \frac{u'}{t^4}-\frac{4u}{t^5} = \frac{-2}{t^6}$

Use product rule in reverse:

$\displaystyle \displaystyle \left(\frac{u}{t^4}\right)' = \frac{-2}{t^6}$

Take integral

$\displaystyle \displaystyle \frac{u}{t^4} = \frac{2}{5t^5} + C$

Subt. back in for u

$\displaystyle \frac{\frac{1}{y^2}}{t^4} = \frac{2}{5t^5} + C \implies \frac{1}{t^4y^2} = \frac{2}{5t^5} + C$

Solve for y:

$\displaystyle \displaystyle \frac{1}{t^4y^2} = \frac{2}{5t^5} + C \implies {\color{red}t^4y^2 = \frac{2}{5t^5} + \frac{1}{C} \implies y^2 = \frac{2}{5t^6} + \frac{1}{t^4C}}$