Solving PDE $u_x+y^2\cdot u_y=0$ I am having a hard time understanding the general solution to this question and require some assistance. The question is:

Let there be a solution $u(x,y) $ for the PDE:
  $$u_x+y^2\cdot u_y=0$$Which satisfies the conditions:
  $u(3,2)=7, u(4,2)=-6, u(8,1)=-2, u(6,-1)=3, u(6,-2)=0$.  Find :
  $u\left(\frac{5}{2}, \frac{1}{2}\right)$ and $u\left(8, -\frac{2}{5}\right)$. Does the solution for $u\left(\frac{9}{2}, 1\right)$ exist?

Now I found that $\frac{\partial y}{\partial x}=y^2$, and used separation of variables to get:$$c=-x-\frac{1}{y}$$
Now I think this leads to the fact that:
$$u(x,y) = f\left(-x-\frac{1}{y}\right)$$
Do I just plug in numbers from here? I'm kinda confused by how to proceed. 
 A: You are right, the general solution is :
$$u(x,y) = f\left(-x-\frac{1}{y}\right)$$
The function $f(X)$ is unknown, except for the given data, for example :
$$u(3,2)=7=f\left(-3-\frac{1}{2}\right)=f\left(-\frac{7}{2}\right)$$
Thus, for $X=-\frac{7}{2}$ we now know that $f(X)=7$
I let you find the four others values of $f(X)$.
So, now you have a list of five known values of $f(X)$.
Suppose that you were asked to find the value of $u\left(\frac{3}{2},\frac{1}{2}\right)$
$$u\left(\frac{3}{2},\frac{1}{2}\right)=  f\left(-\frac{3}{2}-\frac{1}{\frac{1}{2}}\right)=f\left(-\frac{7}{2}\right)$$
You look in the known list of $f(X)$ and see if there is (or not) the value $X=-\frac{7}{2}$


*

*If it doesn't exist, you cannot answer.

*If it exists, you can answer. Well, it is in the list : $f\left(-\frac{7}{2}\right)=7$. So the answer is $u\left(\frac{3}{2},\frac{1}{2}\right)= 7$.


You where not asked to find the value of $u\left(\frac{3}{2},\frac{1}{2}\right)$. This was an example to show you how to proceed.
I suppose that you can proceed on the same manner for the three cases that you are asked for.
