I am stuck on this PDE. The text says to use separation of variables, but I cannot figure out how to get the extra terms that show up in the solution.
$$\frac{\partial u}{\partial x}=2\frac{\partial u}{\partial y}+u\\ u(x,0) = 3e^{-5x}+2e^{-3x}$$
Solution:
$$u=3e^{-5x-3y}+2e^{-3x-2y}$$
I use separation of variables and get $u=Ce^{\frac{\lambda}{2}(2x+y)}$. Then, I try using $u=e^{(ax+by)}$, I get $u=e^{b(2x+y)}$. So, I keep getting that the exponential has the form $2x+y$, but I am stuck on how the solution is reached from th3 given boundary condition.
Making the coordinate transformation
$$ \cases{\eta = x\\ \xi = c x + y } $$
we have $v(\eta,\xi) = u(x,y)$
$$ v_{\eta}+(c-2)v_{\xi}=v $$
then choosing $ c = 2$ we have
$$ v_{\eta}=v $$
with solution
$$ v(\eta,\xi) = \phi(\xi) e^{\eta} $$
or
$$ v(x,\xi) = \phi(\xi) e^x $$
and the condition
$$ u(x,0) = v(x,2x) = \phi(2x)e^x = 3e^{-5x}+2e^{-3x}\Rightarrow \phi(x) = 3e^{-3x}+2e^{-2x} $$
gives
$$ u(x,y) = e^x\phi(2x+y) = 3e^{-5x-3y}+2e^{-3x-2y} $$