Method of characteristic for PDE $xu_x + yu_y = u$ I'm studying method of characteristic. This is a homework. I'm very new to this topic so hope anyone can give a direction.

Given a PDE : $xu_x + yu_y = u$.
a) Solve it, where the initial curve is $u=1$ along $y=x^2, x>0$.
b) Find classes of function $f(t)$ that satisfy the initial curve satisfying $u=-f(t)$ along the arc parameterized by $x=t, y=t, t>0$ such that the problem has infinitely many solutions.

For a), I write $dx/dt=x$, and so on for the characteristic equation, obtaining $x=Ae^t, y=Be^t, u=Ce^t$. I also write $x(0,s)=s, y(0,s)=s^2, u(0,s)=1$ for the initial curve. I can see that the Jacobian is nonzero. Thus, unique solution is expected. Solving the given data, I have $u=x^2/y$.
Is this a right solution?
For b), I use the same method. Writing exactly the same as a) for the characteristic equation, obtaining the same functions in $t$. The initial curve is $x(0,s)=s, y(0,s)=s, u(0,s)=-f(s)$ (where I just switched the variable to $s$ instead of $t$ from the problem. And indeed, the Jacobian is zero, so it'll have infinitely many solutions or no solution. Now, I don't know where to go. I somehow realize I don't understand problem b). Should I just choose one $f(t)$ and add constant in it..or..how?
Any help is appreciated. Thanks a lot.
 A: The answer was mainly given in comments. This is another approach which is too long to be posted in comments.
The characteristics equation is $\quad \frac{dx}{x}= \frac{dy}{y}= \frac{du}{u}\quad$ from which we obtain : 
$\begin{cases}
\frac{y}{x}=c_1 \\
\frac{u}{x}=c_2 \\
\end{cases}$
The general solution expressed on implicit form is : $\quad \Phi \left( \frac{y}{x} \:,\: \frac{u}{x} \right) = 0\quad$ where $\Phi$ is any derivable function of two variables.
It can be solved for $u$ leading to the equivalent explicit form :
$$u(x,y)=x\:F\left(\frac{y}{x}\right)$$
where $F$ is any derivable function.
a) Condition $u=1$ along $y=x^2$
$u=1=x\:F\left(\frac{x^2}{x}\right)=x\:F(x)\quad \implies \quad F(x)=\frac{1}{x}\quad$ on the curve characteristic.
Elsewhere  $\quad u=x\:\left(\frac{1}{\frac{y}{x}}\right)$
$$u(x,y)= \frac{x^2}{y} $$
b) Condition  $u=-f(t)$ along $y=x=t$
$u=-f(t)=t\:F\left(\frac{t}{t}\right)=t\:F(1)= c\:t\quad$ where $c=F(1)=$constant.
If $f(t)=-c\:t\quad \implies u(x,y)=x\:F\left(\frac{y}{x}\right)\quad$ An infinity of functions $F$ such $F(1)=c$ are convenient, so an infinity of solutions.
If $f(t)\neq -c\:t\quad $ the non-linear function $f(t)$ cannot be equal to a linear function $t\:F(1)\quad$ : no solution.
