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Homework Help: Solve the following intial value problem

  1. Aug 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Solve the following inital value problem

    y1' = y2
    y2' = -5y1-4y2

    with inital conditions y1(0) = 1, y2(0) = 0

    2. Relevant equations

    (A-[itex]\lambda[/itex]I)x = 0

    3. The attempt at a solution
    So I first started by setting out like this

    y2' = -5y1 - 4y2
    which then suggest 0 = -[itex]\lambda[/itex](-4-[itex]\lambda[/itex]) + 5
    0 = [itex]\lambda[/itex]2+4[itex]\lambda[/itex] + 5
    But this is going to give me complex roots.. so I was just wondering have I made a mistake in my characteristic equation?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 3, 2011 #2

    SammyS

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    y1' = y2 → y1'' = y2'

    So, the second equation becomes: y1'' = -5y1 - 4y1' .

    D2y1 + 4Dy1 + 5y1 = 0

    Can you take it from there?
     
  4. Aug 4, 2011 #3
    No sorry I don't understand why you have a D in there..
     
  5. Aug 4, 2011 #4
    D is the differential operator, basically another way of writing d/dx. D2y = d2y/dx2 = y''
     
  6. Aug 4, 2011 #5

    SammyS

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    It's an 'operator' which just represents d/dx .

    I get the same characteristic equation as you did, so not much help.
     
  7. Aug 4, 2011 #6
    \begin{equation}y_1^{\prime} = y_2 \end{equation} implies that \begin{equation}y_1^{\prime \prime} = y_2^{\prime} \end{equation}
    The second equation you gave can thus be rewritten in terms of just y_1 and its derivatives,
    \begin{equation}y_2^{\prime} = y_1^{\prime \prime}= -5y_1 - 4y_1^{\prime} \rightarrow y_1^{\prime \prime} + 4y_1^{\prime}+5y_1 = 0\end{equation}
    You just have to solve the equation. Think of the physical parts of the equation. This should be decaying exponentially, but oscillating as well. Try guessing something like
    \begin{equation}
    y_1(t) = e^{-2t}(c_1 \sin(t) + c_2 \cos(t))
    \end{equation}
    Use your initial conditions to determine the constants.
     
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