find the solution of the cauchy equation below
$4u_x+8u_y-u=1$
$u(x,3x)=cos(x)$
my attempt:
char. eq. is $\dfrac{dy}{dx}=\dfrac{8}{4}=2$
$dy=2~dx$, $y=2x+k$
we can take $\xi=x$ and $\eta=y-2x$
$y=\eta+2\xi$
$w(\xi,\eta)=u(x,y)$
$u_x=w_\xi-2w_\eta$ and $u_y=w_\eta$ so the equation is $4(w_\xi-2w_\eta)+8w_\eta-w=1$
so $w=-\frac{1}{4} e^{\frac{\xi}{4} }e^{-\xi}+g(\eta) $ then $u(x,y)=-\frac{1}{4} e^{\frac{x}{4} }e^{-x}+g(y-2x)$$\quad$$\quad$(*)
when we subsitute $u(x,3x)=cos(x)=-\frac{1}{4} e^{\frac{x}{4} }e^{-x}+g(x)$
$ g(x)=\frac{1}{4} e^{\frac{-3x}{4} }+cos(x)$ so the solution is $ u(x,y)=-\frac{1}{4} e^{\frac{-3x}{4} }+\frac{1}{4} e^{\frac{-3(y-2x)}{4} }+cos(y-2x)$ isnt this approach correct? edit2 : mistake was at (*) w(ξ,η) had to equal to $g(η)e^{ξ/4}−1. $
To solve $4 u_x + 8 u_y = u + 1$ with boundary condition $u(x,3x) = \cos x$, I solve the three ordinary differential equations $$\frac{dx_r}{ds}=4 \\ \frac{dy_r}{ds}=8 \\ \frac{dz_r}{ds}=z + 1$$ with initial conditions $$x_r(0) = r \\y_r(0) = 3r \\ z_r(0) = \cos r\,.$$ The solutions are $$x = 4s + r \\ y = 8s + 3r \\ z = (1+\cos r)e^s - 1\,.$$ Expressing $r$ and $s$ in terms of $x$ and $y$, we have $r = y - 2x$ and $s = \frac{3}{4}x - \frac{1}{4}y$, so that $$z = u(x,y) = (1 + \cos{(y-2x)})e^{\frac{3}{4}x - \frac{1}{4}y} - 1\,.$$
Some good notes on this solution method may be found online here.