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Math Help - Help with variables seperable ODEs needed

  1. #1
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    Help with variables seperable ODEs needed

    Ok, so i just started a 2nd year engineering maths unit, but i got credit for the 1st year prerequisite from a subject i did about 4 yrs ago...
    need some help.

    eg. dy/dt + 2y = 1 , y(0)= 5/2

    i started with dy/(1-2y)= dt
    ln(1-2y) = t+c
    1-2y = e^(t+c)
    y= 1/2(1-e^(t+c))
    am i right so far?

    i then put y(0)= 1/2(1-e^c) = 5/2
    1-e^c = 5
    e^c = -4
    c = ln(-4) which is clearly wrong.
    where did i go wrong? any help greatly appreciated.. thanks
    Last edited by CaptainBlack; March 29th 2011 at 01:59 AM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by linalg123 View Post
    i started with dy/(1-2y)= dt ln(1-2y) = t+c

    It should be -(1/2)\ln |1-2y|=t+c .
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  3. #3
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    ok, so -(1/2)ln(1-2y) = t+c
    ln(1-2y) = -2t - 2c
    1-2y = e^(-2t-2c)
    y= 1/2( 1 - e(-2t-2c))
    is this right?
    because i am still getting a c=ln(negative) when i try to solve the i.v.p
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  4. #4
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by linalg123 View Post
    ok, so -(1/2)ln(1-2y) = t+c
    ln(1-2y) = -2t - 2c
    1-2y = e^(-2t-2c)
    y= 1/2( 1 - e(-2t-2c))
    This can reduce to \displaystyle y(t)=\frac{1}{2}(1-Ae^{-2t})

    where A=-2c.

    Now continue on and find A using your condition.
    Last edited by bugatti79; March 29th 2011 at 04:39 AM. Reason: changed y(x) to y(t)
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    can you explain the step how 1/2(1- e(-2t-2c)) = 1/2 (1- Ae^-2t) where A= -2c.. wouldn't A= e^-2c?
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  6. #6
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by linalg123 View Post
    can you explain the step how 1/2(1- e(-2t-2c)) = 1/2 (1- Ae^-2t) where A= -2c.. wouldn't A= e^-2c?
    Yes, you are correct. That is my typo. Can you continue on from there?
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    y(0)= 1/2(1-A)= 5/2
    1-A = 5
    A= -4

    y(t) = 1/2( 1 + 4e^(-2t))

    is this correct?
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  8. #8
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by linalg123 View Post
    y(0)= 1/2(1-A)= 5/2
    1-A = 5
    A= -4

    y(t) = 1/2( 1 + 4e^(-2t))

    is this correct?
    Looks good to me.
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