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Math Help - Separation of variables

  1. #1
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    Separation of variables

    Hi,

    Need to solve the following equation, but i can't seem to separate the variables correctly so i can integrate through.

    dx/dt = t - x

    cheers

    Oli
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  2. #2
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    colby2152's Avatar
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    Quote Originally Posted by olip View Post
    Hi,

    Need to solve the following equation, but i can't seem to separate the variables correctly so i can integrate through.

    dx/dt = t - x

    cheers

    Oli
    This is a non-homogeneous differential equation.

    x'+x-t=0

    You have to guess an initial solution for x. Check this out: Method of undetermined coefficients - Wikipedia, the free encyclopedia
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  3. #3
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    Quote Originally Posted by olip View Post
    dx/dt = t - x
    As colby2152 wrote x'+x=t, multiply by e^t,

    x' + x = t \implies e^t x' + e^t x = te^t \,\therefore \,\left( {xe^t } \right)' = te^t .

    By integrating both sides

    xe^t  = \int {te^t \,dt}, where the last integral is fairly straightforward via integration by parts.
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  4. #4
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    Quote Originally Posted by olip View Post
    Hi,

    Need to solve the following equation, but i can't seem to separate the variables correctly so i can integrate through.

    dx/dt = t - x

    cheers

    Oli
    dx/dt = t - x

    => dx/dt + x = t

    is solved using the integrating factor method (as has been implied by Krizalid).
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  5. #5
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    Hi,

    thanks for all help so far, that working makes sense to me Krizalid, but on integrating \int {te^t \,dt} i get e^t(t-1) i.e. xe^t = e^t(t-1), or x = t-1, which doesnt seem right to me.

    I should give more information. This was just part of some dynamics book and i couldn't figure out how they got the solution. Their solution was

    x(t) = t-1 + exp(-t) (x_0 + 1)

    where x_0 is the solution at t=0

    thanks again,

    Oli
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  6. #6
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    Hi again,

    realised i made a lazy mistake there, forgot to add in the constant of integration, so the solution actually becomes  t-1 + Cexp(-t)

    When i substitute this back in to the original equation  x' + x = t it works, which is good, however it is different than the solution given in the book:
     t-1 + exp(-t)(C + 1) which doesn't give the right answer when substituted back into the equation. Is the one in the book wrong?? I am confused!

    cheers,

    Oli
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  7. #7
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    Quote Originally Posted by olip View Post
    [snip]
    the solution actually becomes  t-1 + Cexp(-t)
    [snip]
    however it is different than the solution given in the book:
     t-1 + exp(-t)(C + 1) which doesn't give the right answer when substituted back into the equation. Is the one in the book wrong?? I am confused!

    cheers,

    Oli
    The books solution does give the correct answer and is in fact equivalent to your solution. Since C is an arbitrary constant, C + 1 is also arbitrary ..... Re=brand C + 1 as K ...... x = t-1 + K exp(-t).
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