Particular solution of PDE I cannot understand how to find particular solution $u(x,0)$ of the below problem:
$$
\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=u, \quad u(x, 0)=2 e^{x}+3 e^{2 x}
$$
By using the general technique $u(x,t) = X(x)T(t)$, I come to the solution:
$$
\left\{\begin{array}{l}
X(x)=A e^{\lambda x} \\
Y(y)=B e^{(1-\lambda) y}
\end{array}\right.
$$
with some arbitrary $\lambda$. But then to obtain $u(x, 0)=2 e^{x}+3 e^{2 x}$, according to the solution, I need to set in two different values of $\lambda$, $\lambda=1$ and $\lambda=2$ in the same solution?
I do not understand in other words how to get to this particular solution. I would be grateful for any guidance.
 A: $$\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=u$$
Another method (Method of characteristics) :
The Charpit-Lagrange characteristic ODEs are :
$$\frac{dx}{1}=\frac{dy}{1}=\frac{du}{u}$$
A first characteristic equation comes from solving $\frac{dx}{1}=\frac{dy}{1}$ :
$$y-x=c_1$$
A second characteristic equation comes from solving $\frac{dx}{1}=\frac{du}{u}$ :
$$e^{-x}u=c_2$$
The general solution of the PDE expressed on implicit form $c_2=F(c_1)$ is :
$$e^{-x}u=F(y-x)$$
$$\boxed{u(x,y)=e^xF(y-x)}$$
$F$ is an arbitrary function, to be determined according to the boundary condition.
Without condition the PDE has infinity of solutions since they are an infinity of functions $F$.
Condition :
$$u(x,0)=2e^x+3e^{2x}=e^xF(0-x)$$
$$F(-x)=e^{-x}\left(2e^x+3e^{2x}\right)\quad\implies\quad F(-x)=2+3e^x$$
Let $x=-X$
$$F(X)=2+3e^{-X}$$
Now the function $F(X)$ is known. We put it into the above general solution where $X=y-x$ :
$$u(x,y)=e^x\left(2+3e^{-(y-x)}\right)$$
$$\boxed{u(x,y)=2e^x+3e^{2x-y}}$$
A: Let $u_1$ be the solution to $u_x + u_y - u = 0$, $u(x, 0) = 2e^x$ and let $u_2$ be the solution to $u_x + u_y - u =0$, $u(x, 0) = 3e^{2x}$. Since the equation is linear,
$u = u_1 + u_2$ solves the original equation. Solving each equation separately and using undetermined coefficients shows that $\lambda = 1$ for the first equation and  $\lambda = 2$ for the second.
