Find the particular solution of $u_x+2u_y-4u=e^{x+y}$ satisfying the following side condition $u(x,-x) = x$ 2.Find the particular solution of $u_x+2u_y-4u=e^{x+y}$ satisfying the following side condition $u(x,-x) = x$
I know that under that condition $y = -x$ which is the reflection of the $x$ graph. I have problems with taking the integration with respect to z because my change of variables is causing me to have a wicked exponential. 
Anyway, here is my attempt
The slope is $\frac{2}{1}$ which is is 2, so my characteristic lines must be positive.
The characteristic lines are in the form of $bx-ay=d$ which is for the pde $au_x+bu_y+cu=f(x,y)$
Using my change of variables, I get the following lines
$ 2x - y = w \rightarrow x=\frac{w+z}{2}$
$z = y \rightarrow y = z$
Now I have to take the partial derivatives and the chain rule.
$W_x = 2, W_y = -1, Z_x = 0, Z_y = 1$
$V_wW_x+V_zZ_x+2(V_wW_y+V_zZ_y) -4(v) = e^{\frac{w+z}{2}+z}$
$2V_w+2(-V_w+V_z) -4(v) = e^{\frac{w+z}{2}+z}$
$2V_w-2V_w+V_z -4(v) = e^{\frac{w+z}{2}+z}$
$V_z -4(v) = e^{\frac{w+z}{2}+z}$
Let $p(a) = -4$, and $q(a) = e^{\frac{w+z}{2}+z}$. Then my integrating factor would be. $e^{\int -4} \rightarrow e^{-4z}$
Multiplying the integrating factor, I get
$e^{-4z}V_z -e^{-4z}4(v) = e^{-4z}e^{\frac{w+z}{2}+z}$
$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w+z}{2}+z}e^{-4z}$
$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w+z}{2}+z-4z}$
$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w+z}{2}-3z}$
$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w+z-6z}{2}}$
$e^{-4z}V_z -e^{-4z}4(v) = e^{\frac{w-5z}{2}}$
By reverse product rule
$e^{-4z}(v) = \int e^{\frac{w-5z}{2}}$
Integrating the right hand side, I get
$e^{-4z}(v) = \frac{-2}{5}e^{\frac{w-5z}{2}}+C(w)$
$ v = \frac{\frac{-2}{5}e^{\frac{w-5z}{2}}}{e^{-4z}}$
$ v = \frac{-2}{5}e^{\frac{w-5z+4z}{2}}$
$ v = \frac{-2}{5}e^{\frac{w-z}{2}} + C(w)$
Substituting back I get 
$ u = \frac{-2}{5}e^{\frac{2x-y-y}{2}} +C(2x-y)$
$ u = \frac{-2}{5}e^{\frac{2x-2y}{2}}+C(2x-y)$
from side condition of $u(x,-x) = x$, we have
$x = \frac{-2}{5}e^{\frac{2x-2(-x)}{2}}+C(x)$
$x = \frac{-2}{5}e^{\frac{2x+2x}{2}}+C(x)$
$x = \frac{-2}{5}e^{\frac{4x}{2}}+C(x)$
$x = \frac{-2}{5}e^{2x}+C(x)$
The problem is... that's not right because the answer is $u(x,y)=-e^{x+y}+(\frac{2}{3}(x-\frac{1}{2}y+1)e^{\frac{4}{3}(x-\frac{1}{2}y)}e^{2y}$
What happened?
Edit: even if I did rewrite it correctly, the $\frac{-2}{5}$ is still there!!!
$v = \frac{-2}{5}e^{\frac{w-5z}{2}}e^{4z}+C(w)e^{4z}$
Applying change of variables 
$u = \frac{-2}{5}e^{\frac{2x-y-5y}{2}}e^{4y}+C(2x-y)e^{4y}$
$u = \frac{-2}{5}e^{\frac{2x-6y}{2}}e^{4y}+C(2x-y)e^{4y}$
$u = \frac{-2}{5}e^{x-3y+4y}+C(2x-y)e^{4y}$
$u = \frac{-2}{5}e^{x+y}+C(2x-y)e^{4y}$
 A: I spotted the following mistake, which invalidates all steps after it: from 
$$e^{-4z}(v) = \frac{-2}{5}e^{\frac{w-5z}{2}}+C(w)$$
you should have gotten 
$$v = \frac{-2}{5}e^{\frac{w-5z}{2}}e^{4z}+C(w)e^{4z}$$
Instead, you omitted $C$, then multiplied by $e^{4z}$, then brought $C$ back in, un-multiplied. 

Why don't you do this in a reasonable way? 


*

*Particular solution: try $u=ce^{x+y}$, find $ce^{x+y}+2ce^{x+y}-4ce^{x+y}=e^{x+y}$, so $c=-1$. 

*Homogeneous equation: $u_x+2u_y-4u=0$. Write $u=Ce^v$, so that $v_x+2v_y-4=0$. 

*Solve the equation for $v$ as usual: $v(x,y) = 4x+f(2x-y)$ with arbitrary $f$. 

*Collect and clean up: $u(x,y)=-e^{x+y} + e^{4x}h(2x-y)$ is the general solution of original equation. 

*Since $u(x,-x)=x$, we have $x=-1+e^{4x}h(3x)$, hence $h(3x)=e^{-4x}(x+1)$. The function $h$ is $h(t) =  e^{-4t/3}(t/3+1)$.

*Plug known $h$ into $u$ to get the final answer: $$u(x,y)=-e^{x+y} + e^{4x}e^{-4(2x-y)/3}((2x-y)/3+1)$$
If you want to verify the result of 6, simplify it before taking derivatives.
A: I don't know if anyone will read this after 8 years, but I think there is another mistake higher up concerning the following two lines:
$2V_w+2(-V_w+V_z) -4(v) = e^{\frac{w+z}{2}+z}$
$2V_w-2V_w+V_z -4(v) = e^{\frac{w+z}{2}+z}$
In the second line it should instead read:
$2V_w-2V_w+2V_z -4(v) = e^{\frac{w+z}{2}+z}$
You forgot to also multiply the $V_z$ by 2, thus invalidating the rest of the calculation.
