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Math Help - Second Order Differential Equation

  1. #1
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    Smile Second Order Differential Equation

    Hi, I need some help with a question involving differential equation.

    The first one is:
    Find the solution to the differential equation.

    y"-9y'+8y = 0, y(0) = 8, y'(0) = 8

    Express your answer as a function of the variable t.

    _________________________
    Here's what I did for the first question:
    I used y = e^(λt) and found the first and second derivatives for y.
    y' = λ*e^(λt)
    y" = (λ^2)*e^(λt)

    And I subbed them into y"-9y'+8y = 0, which is:
    (λ^2)*e^(λt) - 9λe^(λt) + 8e^(λt) = 0

    factored out the e^(λt) and got (λ^2 - 9λ +8) = 0.
    Which results - (λ - 8)(λ - 1) = 0.

    λ = 8 and λ= 1. I plugged the numbers into e^(λt) and got roots
    A*e^(1(t)) + B*e^(8t) where A and B are constants.

    After that I not sure what to do with it or how to plug in y(0) = 8 and
    y'(0) = 8 into the equation.

    I would be grateful for any advice offered .
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Kenny12345 View Post
    Hi, I need some help with a question involving differential equation.

    The first one is:
    Find the solution to the differential equation.

    y"-9y'+8y = 0, y(0) = 8, y'(0) = 8

    Express your answer as a function of the variable t.

    _________________________
    Here's what I did for the first question:
    I used y = e^(λt) and found the first and second derivatives for y.
    y' = λ*e^(λt)
    y" = (λ^2)*e^(λt)

    And I subbed it into y"-9y'+8y = 0, which is:
    (λ^2)*e^(λt) - 9λe^(λt) + 8e^(λt) = 0

    factored out the e^(λt) and got (λ^2 - 9λ +8) = 0.
    Which results - (λ - 8)(λ - 1) = 0.

    λ = 8 and λ= 1. I plugged the numbers into e^(λt) and got roots
    A*e^(1(t)) + B*e^(8t) where A and B are constants.

    After that I not sure what to do with it or how to plug in y(0) = 8 and
    y'(0) = 8 into the equation.

    I would be grateful for any advice offered .
    you have y(t) = Ae^t + Be^{8t}

    so that y'(t) = Ae^t + 8Be^{8t}

    Now, y(0) = 8 means that when you plug in t = 0 into y(t) the result is 8. y'(0) = 8 means something similar.

    can you finish?
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  3. #3
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    Unhappy

    Yea I did solve but I got stuck when I plugged y(0) = 8 and y'(0) = 8 into the equation y = A*e^(t) + B*e^(8t).

    For y(0) = 8, I got 8 = Ae^(0) + Be^(8(0)), which results in 8 = A + B
    and for y'(0) = 8 I got 8 = Ae^(0) + 8*Be^(8(0)), which is 8 = A + 8B.

    e^(0) = 1.

    I am stranded with what to do with the two equations; I tried to solve for one equation for A or B and subbed it in to the other equation.

    For example, solving for B in 8 = A + B. I got 8 - B = A and plugged it into 8 = A + 8B, which results in 8 = (8 - B) + 8B, which I got B = 0.

    I'm confused .
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Kenny12345 View Post
    Yea I did solve but I got stuck when I plugged y(0) = 8 and y'(0) = 8 into the equation y = A*e^(t) + B*e^(8t).

    For y(0) = 8, I got 8 = Ae^(0) + Be^(8(0)), which results in 8 = A + B
    and for y'(0) = 8 I got 8 = Ae^(0) + 8*Be^(8(0)), which is 8 = A + 8B.

    e^(0) = 1.

    I am stranded with what to do with the two equations; I tried to solve for one equation for A or B and subbed it in to the other equation.

    For example, solving for B in 8 = A + B. I got 8 - B = A and plugged it into 8 = A + 8B, which results in 8 = (8 - B) + 8B, which I got B = 0.

    I'm confused .
    that's a long way to solve it, but that's fine. given these initial conditions, A = 8 and B = 0

    don't see why you were confused...
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  5. #5
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    What's an easier way to solve it then? I would like to know .
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  6. #6
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    Smile

    Thanks for the advice. I found the solution while I was typing my post.
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