# Toronto Math Forum

## APM346-2022S => APM346--Lectures & Home Assignments => Chapter 2 => Topic started by: Zicheng Ding on January 26, 2022, 09:22:59 PM

Title: Chapter 2.2 problem 2
Post by: Zicheng Ding on January 26, 2022, 09:22:59 PM
Chapter 2.2 problem 2 equation (9): $u_t + yu_x + xu_y = x$
and we get $\frac{dt}{1} = \frac{dx}{y} = \frac{dy}{x} = \frac{du}{x}$
By solving $\frac{dx}{y} = \frac{dy}{x}$ first we get $C = x^2 - y^2$
But we solving for either $\frac{dt}{1} = \frac{dx}{y}$ or $\frac{dt}{1} = \frac{dy}{x}$ we need to substitute $x$ for $y$ or the other direction, so in this case we will have a square root function like $x = \pm \sqrt{y^2 + C}$
I am not entirely sure whether or not we should proceed with the $\pm$ sign here or is there another easier approach to this question?
Title: Re: Chapter 2.2 problem 2
Post by: Victor Ivrii on January 27, 2022, 08:04:10 AM
Both signs. What are curves $x^2-y^2=C$?
Title: Re: Chapter 2.2 problem 2
Post by: Zicheng Ding on January 27, 2022, 08:19:58 AM
$x^2 - y^2 = C$ is a hyperbolic curve so it has two parts for x. I see now, thank you professor.