Using integral transforms.

Solve \[ {x}^{2}{\frac {\partial ^{2}}{\partial {x}^{2}}}u \left ( x,y \right ) + x{\frac {\partial }{\partial x}}u \left ( x,y \right ) +{\frac { \partial ^{2}}{\partial {y}^{2}}}u \left ( x,y \right ) =0 \] With boundary conditions \begin {align*} u \left ( x,1 \right ) &=\begin {cases} 1& 0\leq x \hspace {5 pt} \text {and} \hspace {5pt} x<1\\ 0&1<x\\ \end {cases}\\ u(x, 0) &= 0 \end {align*}

Mathematica ✓

ClearAll["Global`*"]; 
pde = x^2*D[u[x, y], {x, 2}] + x*D[u[x, y], x] + D[u[x, y], {y, 2}] ==0; 
bc = {u[x, 0] == 0, u[x, 1] == Piecewise[{{1, 0 <= x < 1}, {0, 1 < x}}]}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}], 60*10]];

\[\left \{\left \{u(x,y)\to \frac {\text {Integrate}\left [\frac {x^{-\frac {149}{33}-i K[1]} \csc \left (\frac {149}{33}+i K[1]\right ) \sin \left (y \left (\frac {149}{33}+i K[1]\right )\right )}{\frac {149}{33}+i K[1]},\{K[1],-\infty ,\infty \},\text {Assumptions}\to K[1]\in \mathbb {R}\right ]}{2 \pi }\right \}\right \}\]

Maple ✓

restart; 
pde := x^2*diff(u(x, y), x, x) + x*diff(u(x, y), x) + diff(u(x, y), y, y) = 0; 
iv := u(x, 0) = 0, u(x, 1) = piecewise(0 <= x and x < 1, 1, 1 < x, 0); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,iv],u(x,y))),output='realtime'));

\[u \left (x , y\right ) = \]

2.15.10 Using integral transforms.