Problem with solving the ODE equation $ty'=3y+t^5y^\frac{1}{3}$. I have problem with solving the following equation:


  $$ty'=3y+t^5y^\frac{1}{3}$$


I know it's easy without the $y^\frac{1}{3}$ term, but I'm confused now.

Any help would be appreciated.
It's a Bernoulli's ODE, first $y=0 $ is a solution , now by dividing by $y^{1/3} $ we find $ \frac{y'}{y^{1/3}} - 3 \frac{ty}{y^{1/3}} = t^4  $ , Now we put $z= \frac{1}{y^{\frac{-2}{3}}}$ we will find by replacing in the ODE $ \frac{2}{3}z'-\frac{3}{t}z=4t^4 $ by solving it you will find easily $z$ then $y$.
for your Control the solution is given by $$\left( y \left( t \right)  \right) ^{2/3}-2/9\,{t}^{5}-{t}^{2}{\it 
\_C1}=0
$$