Author Topic: MT, P1  (Read 3502 times)

Victor Ivrii

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MT, P1
« on: October 09, 2013, 07:20:21 PM »
Solve the following initial value problem:
\begin{equation*}
y'(t) = 1- y^2(t),\qquad y(0)=0.
\end{equation*}

Yangming Cai

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Re: MT, P1
« Reply #1 on: October 09, 2013, 10:00:17 PM »
answer to p1

Xiaozeng Yu

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Re: MT, P1
« Reply #2 on: October 09, 2013, 10:18:28 PM »
1
« Last Edit: October 09, 2013, 10:38:57 PM by Xiaozeng Yu »

Xuewen Yang

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Re: MT, P1
« Reply #3 on: October 09, 2013, 10:47:45 PM »
For Xiaozeng Yu,

Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?

Xiaozeng Yu

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Re: MT, P1
« Reply #4 on: October 09, 2013, 10:52:39 PM »
For Xiaozeng Yu,

Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?

no mistake. $d(\ln(1-y))/dt = -1/(1-y)$.
« Last Edit: October 10, 2013, 05:54:53 AM by Victor Ivrii »

Xuewen Yang

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Re: MT, P1
« Reply #5 on: October 09, 2013, 10:57:34 PM »
For Xiaozeng Yu,

Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?

no mistake. d(ln(1-y))/dt = -1/1-y


You are right! My bad  :P

Victor Ivrii

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Re: MT, P1
« Reply #6 on: October 10, 2013, 06:03:59 AM »
And where this solution is defined? $\DeclareMathOperator{\arctanh}{arctanh}$

BTW, when collecting MT I have seen this problem and in many papers integration was atrocious.

As long as $|y|<1$ we have $\int \frac{dy}{1-y^2}=\arctanh (y)$ (inverse hyperbolic function $\tanh$) and then
$y=\tanh (t)=\frac{{e^t}-e^{-t}}{e^{t}+e^{-t}}$ (similar to $\int \frac{dy}{1+y^2}=\arctan(y)$).



« Last Edit: October 10, 2013, 06:07:28 AM by Victor Ivrii »

Xiaozeng Yu

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Re: MT, P1
« Reply #7 on: October 10, 2013, 09:53:07 AM »
And where this solution is defined? $\DeclareMathOperator{\arctanh}{arctanh}$

BTW, when collecting MT I have seen this problem and in many papers integration was atrocious.

As long as $|y|<1$ we have $\int \frac{dy}{1-y^2}=\arctanh (y)$ (inverse hyperbolic function $\tanh$) and then
$y=\tanh (t)=\frac{{e^t}-e^{-t}}{e^{t}+e^{-t}}$ (similar to $\int \frac{dy}{1+y^2}=\arctan(y)$).

hmm...yea...-1<y<1  T.T i feel i'm screwed...
« Last Edit: October 10, 2013, 09:54:50 AM by Xiaozeng Yu »

Victor Ivrii

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Re: MT, P1
« Reply #8 on: October 10, 2013, 10:51:37 AM »
And where this solution is defined?
hmm...yea...-1<y<1 

This is a range of $y$ (what values it takes), I asked about domain--for which $t$ it is defined.

So, my question about domain is pending
« Last Edit: October 11, 2013, 12:58:13 PM by Victor Ivrii »

Xiaozeng Yu

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Re: MT, P1
« Reply #9 on: October 10, 2013, 10:51:42 PM »
t is defined everywhere...

Victor Ivrii

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Re: MT, P1
« Reply #10 on: October 11, 2013, 12:19:32 AM »
$t$ is defined everywhere...

You think right but write incorrectly: Solution $y(t)$ is defined everywhere.


Xiaozeng Yu

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Re: MT, P1
« Reply #11 on: October 11, 2013, 10:05:54 AM »
thank u prof.