Toronto Math Forum
MAT244-2013F => MAT244 Math--Tests => MidTerm => Topic started by: Victor Ivrii on October 09, 2013, 07:20:21 PM
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Solve the following initial value problem:
\begin{equation*}
y'(t) = 1- y^2(t),\qquad y(0)=0.
\end{equation*}
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answer to p1
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1
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For Xiaozeng Yu,
Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?
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For Xiaozeng Yu,
Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?
no mistake. $d(\ln(1-y))/dt = -1/(1-y)$.
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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
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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 $\int \frac{dy}{1+y^2}=\arctan(y)$).
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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...
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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
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t is defined everywhere...
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$t$ is defined everywhere...
You think right but write incorrectly: Solution $y(t)$ is defined everywhere.
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thank u prof.