$\displaystyle y^2u_{yy} + 2yu_y - 2u = 0$
The bad thing here is that the variable with respect to which partial differentiation is being done also appears in the PDE. How to get this one done?
Since the independent variable $\displaystyle x$ is not at all present, the PDE equation is equivalent to the linear homogeneus DE...
$\displaystyle y^{2}\cdot u^{''} + 2\cdot y\cdot u^{'} - 2\cdot u =0$ (1)
... the solution of which is...
$\displaystyle u(x,y) = c_{1} (x) \cdot u_{1} (y) + c_{2} (x) \cdot u_{2} (y)$ (2)
... where $\displaystyle u_{1}(y)$, $\displaystyle u_{2} (y)$ are two linearly independent solutions of (1) and $\displaystyle c_{1} (x)$, $\displaystyle c_{2} (x)$ two arbitrary functions of the $\displaystyle x$ only...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$