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Convert second order of diff. equations to first order

  1. Mar 17, 2017 #1
    1. The problem statement, all variables and given/known data
    I have this set of equation:
    My''+Cy'+Ky=0 but C=0
    M is a matrix consist of {(-m) (0)/( -1/12mb^2) (-1/12mb^3)}
    and K is a matrix of {(-K1-K2) (-K2b)/ ((K1b-K2b)/(2)) (-K2b^2/2)}
    and y is a coordinate system which is (x1,θ)
    Now i have to convert these two equations of second order to first order and i really got lost since its two equations and using matrices.
    We can stack y' and y into z
    so the final equation will be : z'=Az
    Can anybody guide me how to do it?
     
    Last edited: Mar 17, 2017
  2. jcsd
  3. Mar 17, 2017 #2
    If M and K were just constants and y were a single variable, would you recognise the equation?
     
  4. Mar 17, 2017 #3
    I didn't really get what u mean? what do you mean y is a single variable? its a coordinate system (x1,θ)
     
    Last edited by a moderator: Mar 17, 2017
  5. Mar 17, 2017 #4
    I know that; that's what "if . . . y were" implies. I'm asking if you recognise the corresponding scalar equation with C=0, which is often solved by reducing it to first order.
     
    Last edited: Mar 17, 2017
  6. Mar 17, 2017 #5

    haruspex

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    Assuming M-1K is invertible, you can diagonalise it as P-1DP.
    See if that gives you some clues.
    @Basem, do you need more hints?
     
    Last edited: Mar 18, 2017
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