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Home Assignment 2 / Re: problem4 (20)

« on: January 19, 2019, 07:24:22 PM »

We can make the substitution $y = 3x - C$ because we are restricting $u(x, y)$ to the characteristic curves, so I believe we can treat $y$ as equal to $3x - C$ when finding the general solution. We do this because we need the $xydx$ totally in terms of $x$ or we will not be able to integrate both sides. After integrating, we have to get rid of $C$ by replacing it with $3x-y$ again because we want our final solution $u$ to be a function of $x$ and $y$, not of $C$. $C$ is just a constant but it is still in terms of $x, y$ by the characteristic curves.

$C$ is a constant only along integral curves. V.I.

2

Home Assignment 1 / Home Assignment 1

« on: January 14, 2019, 10:04:54 PM »

Problem 4 was to find the general solution to $uu_{xy} = u_xu_y$. I used the hint and got $\frac{\partial}{\partial y} \ln u_x = \frac{\partial}{\partial y} \ln u$, but I'm not sure where to go from here.