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Last question from todays lecture

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**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**:

--- Quote from: Weina Zhu on September 15, 2018, 03:40:28 PM ---Is it because there is a redundant constant $-2$, so we would like to change from the equation of $(x,y)$ to $(w,t)$?

--- End quote ---

Indeed

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