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\centerline{Mathematics 373 -- Applied Mathematics (PDE)}
\centerline{Solution for Problem C, PS 1}
\bigskip
\noindent
The problem was to solve the Cauchy problem 
$$\cases{u_x + 2u_y + u = e^{3x-2y}&\cr
         u(x,0) = 0&\cr}$$
The solution of this inhomogeneous linear 1st order PDE has the 
form $u = u^p + u^h$, where $u^h$ is the general
solution of the associated homogeneous equation
$u_x + 2u_y + u = 0$, and $u^p$ is any particular solution.

We find $u^h$ by changing variables:
$$\cases{r = 2x - y &\cr
         s = x + 2y &\cr}$$
The homogeneous equation becomes $5u_s + u = 0$, so 
$$u = g(r) e^{-s/5} = g(2x - y) e^{-(x+2y)/5}$$
where $g$ is an arbitrary function of one variable.

To find $u^p$, we can use the idea of undetermined 
coefficients.  $e^{3x-2y}$ is a solution of the 
homogeneous equation, so we must try a function
such as $u^p = A(3x - 2y)e^{3x-2y}$.  Then
$$\eqalign{u^p_x + 2u^p_y + u^p &= 
e^{3x-2y}(3A + 3A(3x-2y) -4A -4A(3x-2y) + A(3x-2y))\cr
     &= -Ae^{3x-2y} = e^{3x-2y}\cr}$$
when $A = -1$.  This gives a general solution
$$u = g(2x - y) e^{-(x+2y)/5} - (3x - 2y)e^{3x-2y}.$$

Finally, we want $u(x,0) = 0$, so 
$$0 = g(2x) e^{-x/5} - 3xe^{3x} \Rightarrow g(2x) = 3xe^{16x/5}$$
This says 
$$g(x) = {3\over 2}xe^{8x/5}$$
so 
$$\eqalign{u &= {3\over 2} (2x-y)e^{(16x-8y)/5}e^{(-x+2y)/5} - (3x - 2y)e^{3x-2y}\cr
&= \left(3x - {3y\over 2}\right)e^{3x-2y} - (3x - 2y)e^{3x-2y}\cr
&= {y\over 2}e^{3x-2y}\cr}$$

\bye