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Math Help - Differential first order equations

  1. #1
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    Differential first order equations

    Hello i need help with this differential equations system!
    x,u and l are functions of t... so u(t), x(t) and l(t)
    tf is free.
    Could you find the solutions?

    <br /> <br />
\frac{dx}{dt}=-x+u<br />

    <br />
\frac{dl}{dt}=l<br />

    <br />
 l -2u-2=0<br /> <br /> <br />

    boundary conditions
    x(0)=1
    x(tf)=0
    l(tf)=2
    Last edited by magodiafano; May 17th 2010 at 09:47 AM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    First step is to 'attack' ...

    \frac{dl}{dt} = l (1)

    ... that contains only the 'unknown function' l. Its solution is...

    l(t)=2 c_{1} e^{t} (2)

    Now from the third equation we obtain...

    u(t) = c_{1} e^{t} - 1 (3)

    ... and if we insert (3) in the first equation we obtain...

    \frac{dx}{dt} = -x + c_{1} e^{t} - 1 (4)

    ... the solution of which is...

    x(t) = c_{2} e^{-t} + \frac{c_{1}}{2} e^{t} - 1 (5)

    The values of c_{1} and c_{2} are obtained from the 'initial conditions'...

    Kind regards

    \chi \sigma
    Last edited by chisigma; May 17th 2010 at 05:03 AM.
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  3. #3
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    uhm.. there is something that is not clear (sorry...):

    1) why did you use
    2 c_{1}
    ibstead of c_{1} ?? perhaps in order to simplify the calculations after?

    2)
    x(t) = c_{2} e^{-t} + c_{1}  e^{2 t} - e^{t}

    it's not clear the second and the third term of the result...
    why is it not:
    x(t) = c_{2} e^{-t} + c_{1}  e^{t} - t

    ???

    besides, since it is a definite integral between t0 and tf, what's the final result using this considerations?
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  4. #4
    MHF Contributor chisigma's Avatar
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    I apologize for some mistakes caused by my hurry ...

    The DE ...

     \frac{dx}{dt} = - x + c_{1} e^{t} - 1 (1)

    ... has solution...

    x (t) = e^{\int a(t) dt} \cdot \{b(t)\cdot e^{-\int a(t) dt} dt + c_{2}\} (2)

    ... where...

     a(t) = -1

    b(t) = c_{1} e^{t} - 1 (3)

    Today my 'computation capability' is not excellent... sorry again ...

    Kind regards

    \chi \sigma
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  5. #5
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    uhm i don't understand why do you do the multiplication and not the sum of the terms?

    <br />
\frac{dx}{dt} = - x + c_{1} e^{t} - 1<br />

    i think that result should be:

    <br />
x (t) = e^{-t} + c_{1} e^{-t} - t + c2<br />




    i think that i can consider each term per time and then do the sum of the results... is it right?
    Last edited by magodiafano; May 17th 2010 at 09:41 AM.
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  6. #6
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    is there somebody could help me?
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  7. #7
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by magodiafano View Post
    uhm i don't understand why do you do the multiplication and not the sum of the terms?

    <br />
\frac{dx}{dt} = - x + c_{1} e^{t} - 1<br />

    i think that result should be:

    <br />
x (t) = e^{-t} + c_{1} e^{-t} - t + c2<br />

    I think that i can consider each term per time and then do the sum of the results... is it right?
    For the DE...

    \frac{dx}{dt} = - x + c_{1} e^{t} -1 (1)

    ... it is easy to verify that the solution is...

    x(t) = c_{2} e^{-t} + \frac{c_{1}}{2} e ^{t} -1 (2)

    In fact is...

    \frac{dx}{dt}= - c_{2} e^{-t} + \frac{c_{1}}{2} e^{t}

    - x + c_{1} e^{t} -1 = -c_{2} e^{-t} + \frac{c_{1}}{2} e^{t} (3)

    Kind regards

    \chi \sigma
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  8. #8
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    Now it's clear...
    but i've another question for you!
    I have three boundary conditions.. one about l(t), with whom i can easily find C1... and the other two about x(t)! the problem is if i use the boundary condition for t=0 (so x(0) ), i obtain that the condition on x(1) is not satisfied unless i fix tf (final istant)... is should be right?
    Last edited by magodiafano; May 18th 2010 at 08:47 AM.
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