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Math Help - 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;

    y = (e^-t + C)^2

    How do I continue solving y(0) = 1

    Any help is appriciated!
    Last edited by jokke22; November 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:
    <br />
\begin{gathered}<br />
  \frac{{dy}}<br />
{{dt}} = \sqrt y e^{ - t}  \hfill \\<br />
  \frac{{dy}}<br />
{{\sqrt y }} = e^{ - t} dt \hfill \\ <br />
\end{gathered}

    integrate both sides:

    <br />
2\sqrt y  = C - e^{ - t}

    apply initial conditions:


    \begin{gathered}<br />
  2 = C - 1 \Rightarrow C = 3 \hfill \\<br />
  \sqrt y  = \frac{{3 - e^{ - t} }}<br />
{2} \hfill \\ <br />
\end{gathered} <br />
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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;

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

    Solving this;

    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;

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

    Where;

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

    Where;
    ln|y| = k + C

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

    Which is;

    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;

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

    Solving this;

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

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

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

    Where;

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

    Where;
    ln|y| = k + C

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

    Which is;

    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;
    y = e^{-cost} + C

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

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

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

    y = e^{-cost} + C = 1

    = 0,37 + C = 1

    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;
    y = e^{-cost} + C
    y(0) = 1

    y = e^{-cost} + C = 1

    = 0,37 + C = 1

    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;
    y = Ce^{k}
    and explain what happens in the cases of
    k < 0
    k > 0
    k = 0

    when

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

    How can you approach that...
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