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Math Help - Finding the general solution from a given particular solution.

  1. #1
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    Finding the general solution from a given particular solution.

    for One solution of this differential equation is Using this information, what is the general solution of the DE?

    y(t) = ?
    Last edited by mr fantastic; October 7th 2009 at 12:40 AM. Reason: Re-titled
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  2. #2
    MHF Contributor chisigma's Avatar
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    The general solution of the 'incomplete equation' y^{'} + y =0 is y_{g}(t)= c\cdot e^{-t}. The particular solution of the 'complete equation' y^{'} + y = t You know is y_{p} (t) = e^{-t} + t -1, so that the general solution of the 'complete equation' is...

    y(t)=y_{g} (t) + y_{p} (t)= (1+c)\cdot e^{-t} + t -1

    Kind regards

    \chi \sigma
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  3. #3
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    Exclamation Straight-line solution?!?!?

    Quote Originally Posted by chisigma View Post
    The general solution of the 'incomplete equation' y^{'} + y =0 is y_{g}(t)= c\cdot e^{-t}. The particular solution of the 'complete equation' y^{'} + y = t You know is y_{p} (t) = e^{-t} + t -1, so that the general solution of the 'complete equation' is...

    y(t)=y_{g} (t) + y_{p} (t)= (1+c)\cdot e^{-t} + t -1

    Kind regards

    \chi \sigma
    sorri i accidently posted the wrong question lol i meant to post this one:
    for:
    What is the straight-line solution of this DE?
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  4. #4
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    Quote Originally Posted by bsanghera View Post
    sorri i accidently posted the wrong question lol i meant to post this one:
    for:
    What is the straight-line solution of this DE?
    The answer is already in this thread. Look for it.
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  5. #5
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    Cool

    E^(-t) +t -1 is what I think the answer is but I checked on the website where I am submitting it and it tells me my answer is incorrect that is why I asked again!!!
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  6. #6
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    Quote Originally Posted by bsanghera View Post
    E^(-t) +t -1 is what I think the answer is but I checked on the website where I am submitting it and it tells me my answer is incorrect that is why I asked again!!!
    y = e^{-t} + t - 1 is not linear! However, y = t - 1 is ....
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