### Author Topic: LEC 0201 Question  (Read 864 times)

#### RunboZhang

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##### LEC 0201 Question
« on: September 18, 2020, 02:20:45 PM »
Hi, we just had a lecture on homogenous ODE this morning, and when I went through the slides afterwards, I have found a confusion in the slide P6(pic is also attached below). I am wondering why the domain is strictly greater than minus root two? Why can't it be smaller than minus root two? Is it because the initial condition of this question is greater than 0 so that our answer needs to be greater than 0? (But why?)

#### Victor Ivrii

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##### Re: LEC 0201 Question
« Reply #1 on: September 18, 2020, 07:51:52 PM »
We have initial condition at point $xt_0=0$, and solution blows up as $t=-\l(2)$. While we can consider solution, given by this formula as $t<-ln(2)$ it is disconnected from intial condition. Our solution blew up as $t=-\ln (2)$.

#### kavinkandiah

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##### Re: LEC 0201 Question
« Reply #2 on: September 19, 2020, 12:43:26 PM »
I'm sorry, I still don't understand why we can ignore t<-ln2 just because the initial condition is y(0)=1

#### RunboZhang

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##### Re: LEC 0201 Question
« Reply #3 on: September 19, 2020, 02:59:05 PM »
I think the idea behind this is that we need to make our range consistent to our initial condition and that is why we need to abandon the part t<-ln2. Otherwise it would be self-contradicted.

#### kavinkandiah

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##### Re: LEC 0201 Question
« Reply #4 on: September 19, 2020, 04:16:00 PM »
Still don't get it . If we let this range be all real numbers t excluding t=-ln2, the initial condition would still be included in that. Is just the fact that there's an asymptote at -ln2 and therefore everything left of t=-ln2 isn't "connected" to the part of the solution that has the point (0,-1)? In other words, we're including the only continuous portion?

#### kavinkandiah

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##### Re: LEC 0201 Question
« Reply #5 on: September 19, 2020, 04:21:20 PM »
Ah, I just read 2.4 in the textbook and the Existence and Uniqueness Theorem. I get it now, thank you for trying to make me understand it