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Thread: Finding solution to a DFQ

  1. #1
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    Finding solution to a seperable ODE

    Hi!

    I have this DFQ;


    Im struggling on how to find the solution on this one. I think im starting wrong somehow.

    Could it be something like;

    $\displaystyle y = (e^-t + C)^2$

    How do I continue solving $\displaystyle y(0) = 1$

    Any help is appriciated!
    Last edited by jokke22; Nov 6th 2008 at 05:59 AM.
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  2. #2
    Senior Member Peritus's Avatar
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    it's a separable ODE:
    $\displaystyle
    \begin{gathered}
    \frac{{dy}}
    {{dt}} = \sqrt y e^{ - t} \hfill \\
    \frac{{dy}}
    {{\sqrt y }} = e^{ - t} dt \hfill \\
    \end{gathered} $

    integrate both sides:

    $\displaystyle
    2\sqrt y = C - e^{ - t} $

    apply initial conditions:


    $\displaystyle \begin{gathered}
    2 = C - 1 \Rightarrow C = 3 \hfill \\
    \sqrt y = \frac{{3 - e^{ - t} }}
    {2} \hfill \\
    \end{gathered}
    $
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  3. #3
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    I see! A little wrong calculations by me as I can see now. Thanks for your answere!

    Just to check if I've done this correctly... Could you take a look on this one;

    $\displaystyle \frac{{dy}} {{dt}} = y*sint$

    Solving this;

    $\displaystyle y = e^{-cost} + C$
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  4. #4
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    If anyone or you Peritus can comfirm/correct this one as well;

    $\displaystyle \frac{{dy}} {{dt}} = yk$

    Where;

    $\displaystyle \frac{{1}} {{y}}dy = kdt$

    Where;
    $\displaystyle ln|y| = k + C$

    Where;
    $\displaystyle e^{ln|y|} = e^{k + C}$

    Which is;

    $\displaystyle y = Ce^{k}$

    ???
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by jokke22 View Post
    I see! A little wrong calculations by me as I can see now. Thanks for your answere!

    Just to check if I've done this correctly... Could you take a look on this one;

    $\displaystyle \frac{{dy}} {{dt}} = y*sint$

    Solving this;

    $\displaystyle y = e^{-cost} + C$
    Correct!

    Quote Originally Posted by jokke22 View Post
    If anyone or you Peritus can comfirm/correct this one as well;

    $\displaystyle \frac{{dy}} {{dt}} = yk$

    Where;

    $\displaystyle \frac{{1}} {{y}}dy = kdt$

    Where;
    $\displaystyle ln|y| = k + C$

    Where;
    $\displaystyle e^{ln|y|} = e^{k + C}$

    Which is;

    $\displaystyle y = Ce^{k}$

    ???
    Correct!

    Now, if you have initial conditions, apply them to find C, and then write out the exact solution(s).

    --Chris
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  6. #6
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    Thanks Chris!

    One question, how can I show that this is periodic;
    $\displaystyle y = e^{-cost} + C$

    I know cos has a periodic value =
    $\displaystyle cos(x)2kpi$

    First initialise with a value e.g
    $\displaystyle y(0) = 1 $
    and then;
    $\displaystyle y(x+2pi) = y(x)$

    ?
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  7. #7
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    If I try to initialize;
    $\displaystyle y = e^{-cost} + C$
    $\displaystyle y(0) = 1$

    $\displaystyle y = e^{-cost} + C = 1$

    $\displaystyle = 0,37 + C = 1$

    $\displaystyle C = 0.63$

    Would this be correct?

    Thanks yet again for answers!
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by jokke22 View Post
    If I try to initialize;
    $\displaystyle y = e^{-cost} + C$
    $\displaystyle y(0) = 1$

    $\displaystyle y = e^{-cost} + C = 1$

    $\displaystyle = 0,37 + C = 1$

    $\displaystyle C = 0.63$

    Would this be correct?

    Thanks yet again for answers!
    Yes, that would be correct.

    However, I'm not quite sure how to show its periodic...

    --Chris
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  9. #9
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    Im not quite sure myself... Hehe!

    However, when I have your "attention" and appriciate all the help I can get,

    If you were to show values for k in;
    $\displaystyle y = Ce^{k}$
    and explain what happens in the cases of
    $\displaystyle k < 0$
    $\displaystyle k > 0$
    $\displaystyle k = 0$

    when

    $\displaystyle k \Rightarrow infinity$(cant remeber the code for that sign)

    How can you approach that...
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