Let $c$ be a constant and $f,g$ are continuous functions. Then im asked to solve $f(y)u_x+u_y = cu$ where $u(x,0) = g(x)$ ?
I end up with parametrization $\frac{dx}{f(y)} = \frac{dy}{1} = \frac{du}{cu}$.
Then i solve $\frac{dx}{f(y)} = \frac{dy}{1}$ which i write $f(y)\frac{dy}{dx} = 1$ and by the chain rule i obtain $\frac{d}{dx}(F(y)) = 1$ and then i integrate both sides $F(y) = x+c_1$.
Then i solve $\frac{du}{ds} = cu$ which have a solution $e^{cy}\phi(c_1)$ so the general solution should be $e^{cy}\phi(F(y)-x)$. Then i try to apply the initial condition to the general solution, but i get stuck here $u(x,0) = \phi(F(0)-x) = g(x)$.
So my question: if this is at all correct how to proceed? If not suggest another way.