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Math Help - Diff Equation

  1. #1
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    Diff Equation

    Hello,

    Who can you solve this y^{'}+1=4e^{-y}\sin x the problem is that I don't know who to bring in a standaard form

    Who can help? Thanks.
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  2. #2
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    Quote Originally Posted by Bert
    Hello,

    Who can you solve this y^{'}+1=4e^{-y}\sin x the problem is that I don't know who to bring in a standaard form

    Who can help? Thanks.
    Try this:

    let z(x)=y(x)+x, then the equation becomes:

    <br />
z'=4e^{-z+x}\sin(x)=4 e^{-z}\ e^x \sin(x)<br />
,

    which is of variables seperable type.

    RonL
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  3. #3
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    Yes then z^{'}(x)=y^{'}+1 is the same as z^{'}(x)=y^{'} ??

    Greets.
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  4. #4
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    Quote Originally Posted by Bert
    Yes then z^{'}(x)=y^{'}+1 is the same as z^{'}(x)=y^{'} ??

    Greets.
    No z'=y'+1, so the LHS of the DE is z'

    RonL
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  5. #5
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    what do you mean with LHS ? Greets.
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  6. #6
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    After using CaptainBlack's suggestion you get that \frac{dz}{dx}=4e^{-z}e^x\sin{(x)}. And as he also said this is separable. Can you take it from here?
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  7. #7
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    Quote Originally Posted by CaptainBlack
    Try this:

    let z(x)=y(x)+x, then the equation becomes:

    <br />
z'=4e^{-z+x}\sin(x)=4 e^{-z}\ e^x \sin(x)<br />
,

    which is of variables seperable type.

    RonL
    I am not am expert in differencial equation (I do not even know how to spell the word) how did you know how to do that? Which class of linear-equations it falls into? Or it that something you can up with yourself?
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  8. #8
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    Quote Originally Posted by ThePerfectHacker
    I am not am expert in differencial equation (I do not even know how to spell the word) how did you know how to do that?
    I didn't. I looked at the problem for a while trying to see a way forward
    short of brute force (that is getting out my copy of Piaggio or Kreysig).

    After about 20 seconds it became obvious that the LHS (left hand side)
    could be converted to a derivative by that change of variable, and "Bob
    was my uncle".

    I expect that I have seen this all before, and it is just subconscious memories
    surfacing.

    Which class of linear-equations it falls into? Or it that something you can up with yourself?
    Variables separable is one of the simplest classes of ODE, I always check
    if something can be converted to one of these first (Its the only class,
    other than linear constant coefficients, I can solve without having to
    remind myself what to do next ).



    RonL
    Last edited by CaptainBlack; May 9th 2006 at 12:34 AM.
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by Bert
    what do you mean with LHS ? Greets.
    LHS - Left Hand Side
    RHS - Right Hand Side

    RonL
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  10. #10
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    thanks I lot but I' can't see were you stay wit the +1 on the RHS can you drop him of?

    Greets.
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  11. #11
    Grand Panjandrum
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    Quote Originally Posted by Bert
    thanks I lot but I' can't see were you stay wit the +1 on the RHS can you drop him of?

    Greets.
    The change of variables gives a new DE for z as a function of x.

    Solve this, then change back to y, using y(x)=z(x)-x.

    RonL
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  12. #12
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    Of course I see the equation y^{'}+1=4e^{-y}sin(x)

    And not y'=4e^{-y}sin(x) I try to do your substitution on this equation and of course it can't

    So thank you verry well you solove the problem Greets.
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  13. #13
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    I have calculated and I get 0.5 \ cos(x)e^{-x}+0.5 \ sin(x)e^{-x}-e^{x}=e^y

    Is this the same as e^y=2(sin(x)-cos(x))+ce^{-x}

    Thanks.
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  14. #14
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    Quote Originally Posted by Bert
    I have calculated and I get 0.5 \ cos(x)e^{-x}+0.5 \ sin(x)e^{-x}-e^{x}=e^y

    Is this the same as e^y=2(sin(x)-cos(x))+ce^{-x}

    Thanks.
    First thing to note is that you have lost the arbitary constant somewhere.
    (that is the c in the second of the equations in your post).

    RonL
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  15. #15
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    it must so <br />
0.5 \ cos(x)e^{-x}+0.5 \ sin(x)e^{-x}-ce^{x}=e^y<br />

    is this oké?
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