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Converting an nth order equation to a system of first order equations

  1. Nov 2, 2009 #1
    1. The problem statement, all variables and given/known data
    convert y'' +x^2y'+12y=0 to a system of first order equations with initial conditions y(0)=0 y'(0)=7.


    2. Relevant equations



    3. The attempt at a solution
    first i isolate highest derivative y'' = -x^2y'-12y
    then i let u_1=y u_2=y'

    then (u_1)' = u_2 and (u_2)= y''

    then (u_2)' = (-12u_1)-(x^2u_2)

    i then write these as

    (u_1)' = 0*u_1 + 1u_2
    (u_2)' = (-12u_1) + (-x^2u_2)


    so then in matrix form i have

    matrix [u_1 u_2] = [top -0 1 bottom -12 -x^2] *[u_1 u_2] + [0 0]

    i think im close but i don't know how to get vector c but putting vec u(0) ....pls help and sorry for the poor notation...i can rewrite but i can't find link to use the symbols and such
     
  2. jcsd
  3. Nov 2, 2009 #2
    bump...pleaseee help
     
  4. Nov 2, 2009 #3

    Dick

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    I think you've basically got it. [u1,u2]'=[[0,1],[-12,-x^2]]*[u1,u2]. You can add [0,0] to that but there's no need to. The initial condition is then [u1(0),u2(0)]=[0,7], right? If you want some texing clues check out https://www.physicsforums.com/showthread.php?t=8997
     
  5. Nov 2, 2009 #4
    im sorry but if u can, please help me with that matrix algebra...my prof just copied chicken scratch notes and didn't explain anything.

    edit :this is what i get

    [u1,u2]'= [top 0 u1 bottom -12u2 -x^2u2]

    how to i get my final answer
     
    Last edited: Nov 2, 2009
  6. Nov 2, 2009 #5

    Dick

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    There's not much to explain. The problem was to convert the problem to a system of first order ode's. You have already done that, I think. It didn't say you should solve it, right? It just said convert.
     
  7. Nov 2, 2009 #6
    your right, it says to convert, but my prof also wants us to find vector c with the initial conditions (forgot to mention). thats how im not sure you get [0,7] which is the correct answer
     
  8. Nov 2, 2009 #7

    Dick

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    [u1(0),u2(0)]=[y(0),y'(0)], that was your definition of u1 and u2, right?
     
  9. Nov 2, 2009 #8
  10. Nov 2, 2009 #9

    Dick

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    Ok, so I'm guessing you also see why that's [0,7].
     
  11. Nov 2, 2009 #10
    ok i think i get it... the y values from the initial conditions are the ones put in the vector c matrix
     
  12. Nov 2, 2009 #11
    got it thanks!
     
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