2x^3(y''')-3x^2(y'')+5xy'=0
Let's suppose to search as solution of the DE...
$\displaystyle 2x^{3}\cdot y^{'''} - 2x^{2}\cdot y^{''} + 5x\cdot y^{'}=0$ (1)
... function of the type $\displaystyle y=x^{\alpha}$. Is...
$\displaystyle y^{'} = \alpha\cdot x^{\alpha-1} $
$\displaystyle y^{''} = \alpha\cdot (\alpha-1)\cdot x^{\alpha-2} $
$\displaystyle y^{'''} = \alpha\cdot (\alpha-1)\cdot (\alpha-2)\cdot x^{\alpha-3} $ (2)
... and the substitution of (2) in (1) leads us to write...
$\displaystyle 2\cdot \alpha\cdot (\alpha-1)\cdot (\alpha-2) - 3\cdot \alpha\cdot (\alpha-1) + 5\cdot \alpha =0 $ (3)
If $\displaystyle \alpha_{1}$, $\displaystyle \alpha_{1}$ and $\displaystyle \alpha_{3}$ are the roots ot the algebraic third order equation (3), then the general solution of (1) will be...
$\displaystyle y= c_{1}\cdot x^{\alpha_{1}} + c_{2}\cdot x^{\alpha_{2}} + c_{3}\cdot x^{\alpha_{3}}$ (4)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$