Introduction : In this solution we explain step by step the method How to solve pde ut+aux=e^(-2x), u(x,0)=f(x).
Question: solve pde
\[u_{t}+au_{x}=e^{-2x}\]
subject to initial condition \[u(x,0)=f(x)\]
Solution:
Given pde
\[u_{t}+au_{x}=e^{-2x}\]
\[u(x,0)=f(x)\]
So the parametrizing the curve as
\[Γ=(s,0,f(s))\]
Now characteristics equations of the pde are
\[\frac{dx}{dt}=a,\frac{dy}{dt}=1,\frac{dz}{dt}=e^{-2x}\]
integrate first equation
\[x=at+c_{1}\]
using initial condition
\[s=0+c_{1}=>c_{1}=s\]
\[x=at+s\]
And integrating second equation
\[ y=t\]
similarly integrating third equation
\[\frac{dz}{dt}=e^{-2x}=e^{-2\left(at+s\right)}\]
\[z=-\frac{1}{2c}e^{-2(at+s))}+c_{2}\]
using initial condition
\[f(s)=-\frac{1}{2a}e^{-2a}+c_{2}=>c_{2}=f(s)+\frac{1}{2a}\]
\[z(s,t)=-\frac{1}{2a}e^{-2(at+s)}+f(s)+\frac{1}{2a}e^{-2s}\]
\[u(x,y)=-\frac{1}{2a}e^{-2(ay+x-at)}+f(x-at)+\frac{1}{2a}e^{\left(-2(x-at\right))}\]
\[=\frac{1}{2a}(e^{-2(x-at)}-e^{-2(ay+x-at)})+f(x-at)\]
since t= y
\[u(x,y)=\frac{1}{2a}e^{-2(x-ay)}(1-e^{-2ay})+(x-ay)\]
This is the required solution of the the given pde.