I want to find a solution for the following equation
$$ \partial_y u(x,y) + \partial_x u(x,y) + \int_{0}^{x} r(x-x') u(x',y) dx' = 0 $$
$$ u(x,0)=0 \quad u(0,y)=\delta(y) $$
in $(x,y) \in (0,+\infty)\times(0,+\infty)$
Formally this is how I would go:
Laplace transforming in $x$ $$ \partial_y \hat{u}(s,y) + (\:s + \hat{r}(s) \:)\; \hat{u}(s,y) = \delta(y) $$
then
$$ \hat{u}(s,y) = e^{-sy} * e^{-r(s)\:y} H(y) $$
and (this is related to this other question)
$$ u(x,y) = \mathcal{L}^{-1}\{ e^{-r(s)\:y}H(y) \}(x-y,y) $$
where $\mathcal{L}^{-1}$ is the inverse laplace transform and $H(y)$ the Heaviside step function
Is this correct ? I'm not very sure about solving the ODE in $\hat{u}$ in that way...