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Finding y(t) from ODE (HELP NEEDED QUICK)

  1. Mar 17, 2010 #1
    Find y(t) from the following first order ODE



    2. Relevant equations

    y' = 20SQRT10 x (1-e^(t/SQRT10))/(1+e(t/SQRT10))

    3. The attempt at a solution
    I know I must integrate to gain the solution but am having problems getting started please any help greatly appreciated!
     
  2. jcsd
  3. Mar 17, 2010 #2

    rock.freak667

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    It is a separation of variables type question. So try that. Then a substitution to make integration easier.
     
  4. Mar 17, 2010 #3
    So far I have done the following:
    dy/dt = 20SQRT10 x (1-e^(t/SQRT10))/(1+e(t/SQRT10))
    dy . 1/20SQRT10 = (1-e^(t/SQRT10))/(1+e(t/SQRT10)) dt

    I can then integrate the LHS to give,
    y/20SQRT 10 WHich I think is correct?

    How could I tackle the RHS? Tried numerous methods!!!
     
  5. Mar 17, 2010 #4

    rock.freak667

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    [tex]u=e^{\frac{t}{\sqrt{10}}}[/tex]


    Then split into partial fractions.
     
  6. Mar 17, 2010 #5
    so if u=e^t/SQRt10 the RHS becomes:

    ∫ (1- u)÷(1+u) dt

    From there you say split into partial fractions like the following ths is the part that is confusing me. How would you go about doing that?
     
  7. Mar 17, 2010 #6
    Okay from integrating I have got:

    t-te^(t/SQRT10)÷ e^(t/SQRT10) + 1

    is this correct?
     
  8. Mar 17, 2010 #7

    rock.freak667

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    I think you forgot a 'u' in the denominator. You should get

    [tex]\int \frac{1-u}{u(1+u)}du[/tex]

    Just put

    [tex]\frac{1-u}{u(1+u)} = \frac{A}{u}+\frac{B}{1+u}[/tex]

    solve for A and B (the expression is true for all values of u)
     
  9. Mar 17, 2010 #8
    Thanks using your method I have gotten,

    A=1 and B = -2 which gives

    1/(u) - 2/(1+u)

    1/e^(t/SQRT10) - 2/(1+e^(t/SQRT10))

    correct?
    Giving me a final answer of:

    Y/20SQRT(10) = 1/e^(t/SQRT10) - 2/(1+e^(t/SQRT10))
     
  10. Mar 17, 2010 #9

    rock.freak667

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    But you do understand how there is an extra 'u' right and how partial fractions work?

    Also, you are to integrate it, not just immediately replace u.
     
  11. Mar 17, 2010 #10
    Okay Ive completly forgotton to integrate!

    ∫1/(u) - 2/(1+u) = ln(u) - 2ln(1+u)

    Then subs in value for u:
    ln(e^t/SQRT10) - 2ln(1+e^(t/SQRT10)

    As ln and e cancel each other out this becomes:

    t/SQRT10 - 2ln1 - 2ln(e^(t/SQRT10)

    which becomes:

    t/SQRT10 - 2(t/SQRT10)

    is that correct?

    To answer other question Im not clear where you get the extra u from!
     
  12. Mar 17, 2010 #11

    rock.freak667

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    From here ln(a+b)≠ lna + lnb, you will need to leave it as is

    Say you had u=ekx, then du=kekx dx, and if u=ekx, then wouldn't du=ku dx and thus dx= du/(ku) ?
     
  13. Mar 17, 2010 #12
    Ah yes I see now where the extra u comes from so just to finish, the final answer I have taking everything into account is:

    Y/20SQRT(10) = t/SQRT(10)-2ln(1+e^(t/SQRT10)

    And from there I can simply solve to get equation in terms of Y as originally asked at the start.

    Y = (t/SQRT(10)-2ln(1+e^(t/SQRT10) ) x 20SQRT(10)

    This I know will simplify
     
  14. Mar 17, 2010 #13

    rock.freak667

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    You've already found y(t) so all should be well now.
     
  15. Mar 17, 2010 #14
    Thanks for help :)
     
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