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Math Help - find the general solution

  1. #1
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    find the general solution

    Find the general solution using the method of elimination
    x'=x+2y+z
    y'=6x-y
    z'=-x-2y-z

    I used a matrix and found
    D^3 + D^2 - 12D=0
    and then I think
    x(t)= A + Be^(-4t) +Ce^(3t)

    But then I don't know where to go, or how to solve it,

    Please Help

    Thanks
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  2. #2
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    Quote Originally Posted by mandy123 View Post
    Find the general solution using the method of elimination
    x'=x+2y+z
    y'=6x-y
    z'=-x-2y-z

    I used a matrix and found
    D^3 + D^2 - 12D=0
    and then I think
    x(t)= A + Be^(-4t) +Ce^(3t)

    But then I don't know where to go, or how to solve it,

    Please Help

    Thanks
    Let's do it a little different. First solve the second equation for x so x = \frac{y'+y}{6}\;\;(1). Substitute this into the first equation. This gives
    \frac{y''+y'}{6} = \frac{y'+y}{6} + 2y + z. Solving for z gives z = \frac{y'' - 13y}{6}\;\;(2). Now substitute this into the first equation giving (after simplying gives

    y''' + y'' - 12y' = 0 (same equation as for x)

    The solution as you said is y = c_1 + c_2 e^{-4t} + c_3e^{3t} (I've chosen different constants not to be confusing with your x(t) solution). Now substitute this into (1) and (2) above to get x\; \text{and}\; z.
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  3. #3
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    wow, this makes a lot more sense than what the book was telling me to do, thank you so much!!!!
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