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Initial Value Differential Equation Problem

  1. Dec 4, 2008 #1
    1. The problem statement, all variables and given/known data

    y1' = y1 + 4y2 - t2 +6t
    y2' = y1 + y2 - t2 + t -1

    y1(0) = 2
    y2(0) = -1

    Could someone give me a nudge as to how to properly complete this problem? We never really went over anything with multiple y values and I am a little confused as to how I should approach this problem.

    Thank you in advance.
     
    Last edited: Dec 4, 2008
  2. jcsd
  3. Dec 4, 2008 #2

    Defennder

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    Is there another expression for [tex]y_2 ' [/tex] which you didn't include?
     
  4. Dec 4, 2008 #3
    Yes Defender, I have added it to my original post.

    I am just not sure what the first step would be at this point.
     
  5. Dec 4, 2008 #4

    tiny-tim

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    Hi Wildcat04! :smile:

    hmm :rolleyes: … hint: eigenvectors! :wink:
     
  6. Dec 4, 2008 #5
    Ok, I should have thought of that...I think that I have found the homogen. part of the solution but I am struggling with the particular solution.

    Sorry about the matrices, I havent figured out how to make them look nice so I am using TI89 notation :shy:

    y' = [[1 4][1 1]] [y1 y2]

    det => [tex]\lambda = 3, -1[/tex]

    yhomo = c1 [[2][-1]] e3t + c2 [[2][-1]] e-t

    Now for the particular solution

    g1 = t2 + 6t
    g2 = -t2 + t - 1

    yp of the form:

    (at2 + bt + c)
    (dt2 + et + d)

    Am I heading the right direction?



     
  7. Dec 4, 2008 #6

    tiny-tim

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    Are they both [[2][-1]]?
    hmm … i'm a bit lost here …

    i expect there is a matrix way of doing it …

    but i was just going to separate into the two eigenvectors, and solve for each separately. :redface:
     
  8. Dec 4, 2008 #7

    HallsofIvy

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    Or equivalently, a little simpler in concept but harder calculations:

    Differentiate y1'= y1 + 4y2 - t2 +6t with respect to t to get y1"= y1'+ 4y2'- t2+ 6t. Since y2'= y1 + y2 - t2 + t -1, that is y1"= y1'+ 4y1- 4y2- 4t2+ 4t- 4- t2+ 6t= y1'+ 4y1- 4y2- 5t2+ 10t- 4.

    But from the first equation, 4y2= y1'- y1+ t2- 6t so
    y1"= y1'+ (y1'- y1+ t2[/sup- 6t)- 5t2+ 10t- 4 or

    y1"= 2y1'- y1- 4t2+ 4t+ 4 or

    y1"- 2y1'+ y1= -4t2+ 4t+ 4
    with y1(0)= 2 and, since 4y2(0)= -4= y1'(0)- y1(0)+ 02- 6(0)= y1'(0)-2,
    y'(0)= -2.

    Solve that equation for y1(x) and then use 4y2= y1'- y1+ t2- 6t to solve for y2(x).
     
  9. Dec 5, 2008 #8
    Going off of Halls of Ivy's equations,

    Please let me know if these are correct assumptions

    y1'(0) = -2 => y1'(t) = t - 2
    y1(0) = 2 => y1(t) = .5t2 - 2t +2

    4y2 = y1' + y1 + t2 - 6t
    4y2 = t - 2 - .5t2 + 2t - 2 + t2 - 6t
    4y2 = -.5t2 - 3t -4

    => y2(t) = -.125t2 - .75t -1
     
  10. Dec 5, 2008 #9
    Tiny Tim,

    For my own sake I also wanted to try and do it using eigenvectors and I think that I have reached a solution (or at least I think that I am close). If you get a chance could you please take a look and let me know if it is correct?

    [tex]\lambda = 3, -1[/tex]

    y(t)homogen = c1[[1][2]]e3t + c2[[1][-2]]e-t

    [tex]\varsigma[/tex] (1) = 1/(12 + 22).5 [[1][2]] = 1/(5).5 [[1][2]]

    [tex]\varsigma[/tex] (2) = 1/(5).5 [[1][-2]]

    T=(1/(5).5) [[1,1][2,-2]]

    T-1 = [[1.2][1,-2]]

    => =>

    y1' - 3y1 = -3/(5).5t2 + 8/(5).5t

    y1(t) = 1/(5).5t2 + 2.66/(5).5t + (c1e3t)/3 + (c2e-t)/3

    y2' + y2 = 1/(5).5t2 + 4/(5).5t

    y2(t) = 1/(5).5t2 + 4/(5).5t - c1e3t -c2e-t


    Am I completely off base or is this a reasonable solution?

    Thanks!
     
  11. Dec 5, 2008 #10

    HallsofIvy

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    Are you assuming that y1"(t)= 1 for all t? Why?
    As I said before, y1"- 2y1'+ y1= -4t2+ 4t+ 4

    You said before that the problem was "We never really went over anything with multiple y values" so I assume you can solve a d.e. with a single y1.

     
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