Solving PDE by variable coefficient equation method I'm taking my first course on PDEs and the text used by the professor has next to 0 examples. I was just wondering if I am approaching this question correctly. 

$$ e^{x^2}u_x + xu_y = 0 $$

The characteristic curve satisfies the following ODE:
    $$ \frac{dy}{dx} = \frac{x}{e^{x^2}}$$
    Solving this ODE gives:
$$ y= -\frac{1}{2e^{x^2}} + C$$
Then we can isolate for $C$ to find the general solution.
$$ C = y + \frac{1}{2e^{x^2}} $$
Thus the general solution is:
$$u(x,y) = f( y + \frac{1}{2e^{x^2}})$$


Consider the initial value problem:
$$ u_x + xu_y = 0, u(0,y) = sin(y) $$

So using the steps above I get the general solution to be:
$$ u(x,y) = f(y-\frac{x^2}{2})$$
Then using the initial value:
$$u(0,y) = siny = f(y)$$
I am not sure where to go from this step
Any guidance would be greatly appreciated.
 A: I'm studying the same material from Partial Differential Equations: Methods and Applications by McOwen. Another way to do the second problem is shown below. 

$$ u_x + xu_y = 0 $$
$$ u(0,y) = sin(y)$$

One could first write the initial data curve as $\Gamma: (0,s,\sin(s))$. Then, the method of characteristics would produce the following characteristic equations
$$\frac{dx}{dt}=1, ~\frac{dy}{dt}=x, ~\frac{dz}{dt}=0$$
with the initial data
$$x(s,0)=0, ~y(s,0)=s, ~z(s,0)=\sin(s)$$
For the first equation,
$$\frac{dx}{dt}=1 ~\Rightarrow dx = dt ~\Rightarrow x = t + C$$
Then, the initial data of $x(s,0)=0$ forms $x = 0 + C = 0$. Therefore, 
$$x=t$$
For the third equation,
$$\frac{dz}{dt}=0 ~\Rightarrow dz = 0 ~\Rightarrow z = C$$
Then, the initial data of $z(s,0)=\sin(s)$ forms $z = C = \sin(s)$. Therefore, 
$$z=\sin(s)$$
For the last equation,
$$\frac{dy}{dt}=x ~\Rightarrow dy = {t}dt ~\Rightarrow y = \frac{t^2}{2} + C$$
Then, the initial data of $y(s,0)=s$ forms $y = 0 + C = s$. Therefore, 
$$y=\frac{t^2}{2} + s$$
Putting all of our equations together,
$$x=t$$
$$y=\frac{t^2}{2} + s ~\Rightarrow s = y - \frac{t^2}{2} = y - \frac{x^2}{2}$$
$$z=\sin(s) = \sin(y - \frac{x^2}{2})$$
Hence,
$$z=u(x,y)=\sin(y - \frac{x^2}{2})$$
It might be easier to find a constant function $\phi(x,y,z)$ first. Once you find that $\phi(x,y,z)=c=constant$, you could then find a different function $\psi(x,y,z)$ that is constant and is independent of $\phi$. Letting $\phi=f(\psi)$ for an arbitrary $f\in{C^1}$ would produce a solution.
