Solving PDE $u_x+y^2\cdot u_y=0$

Question

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.

Answer 1

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.