### Author Topic: Last question from todays lecture  (Read 2376 times)

#### Boyu Zheng

• Jr. Member
•  • Posts: 12
• Karma: 8 ##### Last question from todays lecture
« on: September 13, 2018, 08:08:58 PM »
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 $a-b-2=0$?

« Last Edit: September 14, 2018, 01:04:40 AM by Victor Ivrii »

#### Wei Cui

• Full Member
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• Karma: 11 ##### Re: last question from todays lecture
« Reply #1 on: September 13, 2018, 10:57:41 PM »
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 ##### Re: last question from todays lecture
« Reply #2 on: September 14, 2018, 01:04:24 AM »
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

• Jr. Member
•  • Posts: 6
• Karma: 1 ##### Re: Last question from todays lecture
« Reply #3 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)?

#### Victor Ivrii ##### Re: Last question from todays lecture
« Reply #4 on: September 15, 2018, 04:12:30 PM »
Is it because there is a redundant constant $-2$, so we would like to change from the equation of $(x,y)$ to $(w,t)$?
Indeed