Toronto Math Forum

APM346--2019 => APM346--Lectures & Home Assignments => Home Assignment 2 => Topic started by: Zhiman Tang on January 18, 2019, 11:53:36 PM

Title: problem4 (20)
Post by: Zhiman Tang 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?
Title: Re: problem4 (20)
Post by: Victor Ivrii on January 19, 2019, 07:03:34 AM
Please learn how to post math properly (http://forum.math.toronto.edu/index.php?topic=610.0)  Also, asking for help, copy the problem.
Title: Re: problem4 (20)
Post by: Brittany Palandra 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.
Title: Re: problem4 (20)
Post by: MikeMorris 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.