I'm solving the following boundary value problem $$ y \frac{\partial u}{\partial y}+\frac{\partial u}{\partial x}=1, \quad u(x, 1)=1=u(0, y) . $$
I've derived that $\bar{u}(p, y)=p^{-2}+p^{-1}-p^{-2} y^{-p}$, however, I have no idea how to invert the last term. Thanks a lot in advance.
Inverse Laplace transform of $\bar{u}(p, y)=p^{-2}+p^{-1}-p^{-2} y^{-p}$ with $p \to x$ gives
$$u \left(x,y\right)\mapsto 1+x-\left(x-\ln\! \left(y\right)\right)\cdot \mathrm{H}\! \left(x-\ln\! \left(y\right)\right), x \ge 0, y \ge 1$$
and $H(z)$ the Heaviside step function.
Visualization: