Solving 1st Order PDE with Initial Condition - Help Needed

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    First order Pde
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I'm trying to solve this equation:

Ux + Uy + U = e^-(x+y) with the initial condition that U(x,0)=0


I played around and and quickly found that U = -e^-(x+y) solves the equation, but does not hold for the initial condition. For the initial condition to hold, I think there needs to be some factor of y in the equation for U, but after trying a few equations, I can't find one that satisfies both the initial condition and the equation.

Can someone throw me a hint?

Thanks!
 
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You might try the substitution u(x,y)=v(x,y)e^{-(x+y)}...
 
You need to variables for the plane:

z=x+y, t=x-y

your equation only depends on one of them

2u_{,z}+u=e^{-z}
 
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There is the following linear Volterra equation of the second kind $$ y(x)+\int_{0}^{x} K(x-s) y(s)\,{\rm d}s = 1 $$ with kernel $$ K(x-s) = 1 - 4 \sum_{n=1}^{\infty} \dfrac{1}{\lambda_n^2} e^{-\beta \lambda_n^2 (x-s)} $$ where $y(0)=1$, $\beta>0$ and $\lambda_n$ is the $n$-th positive root of the equation $J_0(x)=0$ (here $n$ is a natural number that numbers these positive roots in the order of increasing their values), $J_0(x)$ is the Bessel function of the first kind of zero order. I...
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