I am trying to solve this PDE and not sure if I'm on the right track. I first found (using a change of variables) the general solution in the form $$u(x,y)=f(x-y)+g(x+\frac{y}{2})$$
Then, the initial conditions give me the following equations: $$u(x,0) = f(x) + g(x) =\phi(x) \tag{1}$$ $$u_y(x,0) = - f'(x) +\frac{1}{2}g'(x) =\psi(x) \tag{2}$$
From there I solved the system of equations of (1) and (2) to get $$\frac{3}{2} g'(x)=\phi'(x) +\psi(x)$$ $$\Rightarrow g'(x) = \frac{2}{3}(\phi'(x)+\psi(x))$$
Then from (1): $$ f'(x) = \phi'(x)-g'(x) =\frac{1}{3}\phi'(x) +\psi(x)$$
Integrating both sides, I got $$\int f' dx= f(x) = \frac{1}{3}\phi(x)+\int_{0}^x \psi(s) ds $$ $$\Rightarrow g(x) = \phi(x)-f(x) = \frac{2}{3}\phi(x) -\int_{0}^x \psi(s) ds$$ So then,
$$u(x,y) = f(x-y)+g(x+\frac{y}{2}) $$
$$ = \frac{1}{3}\phi(x-y)+\frac{2}{3}\phi(x+\frac{y}{2})+\int_0^{x-y}\psi(s)ds-\int_0^{x+\frac{y}{2}}\psi(s)ds$$
$$ = \frac{1}{3}\phi(x-y)+\frac{2}{3}\phi(x+\frac{y}{2})-\int_{x-y}^{x+\frac{y}{2}} \psi(s)ds $$
Im not sure if this is the right approach--this is my first time solving a problem of this type, so I was just following along with the derivation of the general solution of the wave equation. Any help would be appreciated!