How to solve the non-homogeneous PDEs: $u_x + u_y=2u,\ u(x,0)=h(x)$?

I have this first-order non-homogeneous partial differential equation with initial condition:
$u_x + u_y=2u,\ u(x,0)=h(x)$
The following was what I tried:
By the method of characteristic curves, we have:
$dx/dt=1;\ dy/dt=1;\ du/dt=2u$
Then, we have $dy/dt=1$, and if we integrate both sides, we have $x-y=C_1$, meaning that the characteristic lines are $y-x=C_1$. And, we also have $du/dx=2u$. We will have $u=C_2 e^{2x}$, which is $u/e^{2x}=C_2$. So the question is where should I go next? My teacher did not give examples like these non-homogeneous equations. Any hints, please.

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Just a correction to your question, the mentioned equation is in fact Homogeneous. – user128180 Feb 12 '14 at 8:15
Sorry, what I meant the PDE itself is Homogeneous, however, the initial conditions is not. – user128180 Feb 12 '14 at 8:17

Consider the change of variables $(x,y)$ to $(a,v)=(x,x-y)$.

Then $u_x=u_a+u_v$ and $u_y=-u_v$. Your PDE then becomes $u_a=2u$. The solution to the PDE is $u(a,v)=C(v)e^{2a}$, where $C(v)$ is some function independent of $a$. Transforming back to the original variables gives $$u(x,y)=C(x-y)e^{2x}.$$ The initial condition then gives $u(x,0)=h(x)=C(x)e^{2x}\Rightarrow C(x)=h(x)e^{-2x}$. The solution of the PDE is then $$u(x,y)=h(x-y)e^{2y}.$$

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(-1): The OP wants us to solve this question by the method of characteristics rather than by the change of variables. – doraemonpaul Feb 5 '13 at 7:30
@doraemonpaul Clearly the OP was happy with this solution method, as it was accepted. Furthermore, my understanding of a variable change, in this case at least, is that it acts like a characteristic solution. – Daryl Feb 5 '13 at 9:17

$\dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$

$\dfrac{dx}{dt}=1$ , letting $x(0)=x_0$ , we have $x=t+x_0=y+x_0$

$\dfrac{du}{dt}=2u$ , letting $u(0)=f(x_0)$ , we have $u(x,y)=f(x_0)e^{2t}=f(x-y)e^{2y}$

$u(x,0)=h(x)$ :

$f(x)=h(x)$

$\therefore u(x,y)=h(x-y)e^{2y}$

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By method of characteristics: $u_x +u_y=2u$, and we know $du=u_xdx+u_ydy$, so $\frac {du}{dx}=u_x+\frac{dy}{dx} u_y$ and $\frac {dy}{dx}=1$ as shown by your analysis. So $u_x+u_y=\frac {du}{dx}$, substitute this to original equation we get:$\frac {du}{dx}=2u$, the. You can do the integration to get $ln(u)=2x+C$, which $C$ is a constant based on your initial condition. So $u(x,y)=e^{2x+C}$.

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