Last question from todays lecture

[Boyu Zheng]

Hi,
I have a question from today's lecture. The initial condition equation is $y'= (x-y)/(x+y-2)$ and after a couple steps we get
$$dw/dt = (t-w+a-b)/(t+w+a-b-2) = t-w/t+w.$$
And I am confused about why you say that $a-b=0$ and $a-b-2=0$?

[Wei Cui]

First of all, I wrote $a-b=0$ and $a+b-2=0$ in my notes (not $a-b-2=0$). And, I also wrote $x=t+a$, $y=w+b$, $x=t+1$, $y=w+1$, which may be the initial condition. In this case, $a=1$, and $b=1$, then the equation will make sense.

Well, actually I am confused about it too...

[Victor Ivrii]

We changed $x=t+a$, $y=w+b$ with constants $a,b$ to be chosen later and got
$$
\frac{dw}{dt}=\frac{t-w+a-b}{t+w+a-b-2}
\tag{*}
$$
We want to have a homogeneous equation in $(t,w)$ and it happens if $a-b=0$, $a-b-2=0$. So, we choose $a,b$ to satisfy these equations.

[Weina Zhu]

Is it because there is a redundant constant -2, so we would like to change from the equation of (x,y) to (w,t)?

[Victor Ivrii]

Is it because there is a redundant constant $-2$, so we would like to change from the equation of $(x,y)$ to $(w,t)$?

Indeed