### Author Topic: problem4 (20)  (Read 1855 times)

#### Zhiman Tang

• Jr. Member
•  • Posts: 6
• Karma: 0 ##### problem4 (20)
« on: January 18, 2019, 11:53:36 PM »
the expression on the right hand contains both x and y,
the characteristic line is 3x-y=c
then, we evaluate du = xydx. in this step, we have to replace y by 3x-c before integrating both sides.
My question is, why we have to do so?
And after integrte both sides, why we have to replace c back by 3x-y?

#### Victor Ivrii ##### Re: problem4 (20)
« Reply #1 on: January 19, 2019, 07:03:34 AM »
Please learn how to post math properly  Also, asking for help, copy the problem.

#### Brittany Palandra ##### Re: problem4 (20)
« Reply #2 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.
« Last Edit: January 19, 2019, 09:18:34 PM by Victor Ivrii »

#### MikeMorris

• Jr. Member
•  • Posts: 6
• Karma: 0 ##### Re: problem4 (20)
« Reply #3 on: January 21, 2019, 05:42:29 PM »
Would we be incorrect to define $C = y-3x$ since saying $y = 3x-C$ is essentially equivalent to $y = 3x+C$ for arbitrary $C$? This is what I did in my solution.