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Differential equation is hard

  1. Apr 11, 2008 #1
    !!!For The Love of GOD!!! Diff Eq

    1. The problem statement, all variables and given/known data

    I have done this problem 3 times. I am getting a sign error somewhere and I cannot find it.

    When I solve this DE and then plug it back into the original, it is not checking out!!! Can anybody see my error? I would really appreciate a second set of eyes here. Clearly I am at that point where I just keep seeing what I think I am supposed to see!



    [​IMG]



    Re writing the solution to the DE with the new constants is

    [tex]x=-3/2\cos4t+3\sin4t+7/2\cos4t+1/2\sin4t=2\cos4t+19/4\sin4t[/tex]
    [tex]y=3\cos4t-7/2\sin4t[/tex]

    But when I differentiate and plug back into either if the originals, I am coming up with two sign errors...
     
    Last edited: Apr 12, 2008
  2. jcsd
  3. Apr 12, 2008 #2

    HallsofIvy

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    Staff Emeritus
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    At one point in your solution you have:
    [tex]4 k_1+ (-2- 4i)k_2= 0[/tex]
    and then
    [tex]k_1= \frac{-2- 4i}{4}k_2[/tex]
    It should be
    [tex]k_1= \frac{2+ 4i}{4}k_2[/tex]
     
  4. Apr 12, 2008 #3
    I do not understand why it should be positive? From the matrix I have [itex]4k_1+(-2-\lambda)k_2[/itex] where lambda=4i......unless I am interpreting matrices wrong??
     
  5. Apr 12, 2008 #4

    Vid

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    Basic Algebra.

    4K1 - (2+4i)k2 = 0
    K1 = (2+4i)k2/4
     
  6. Apr 12, 2008 #5
    Okay. Obviously I am misinterpreting how a matrix operates. Not how to do basic algebra.






























    I was just kidding, maybe that was inappropriate...my bad.
     
    Last edited: Apr 12, 2008
  7. Apr 12, 2008 #6
    If the original equation was
    x'=2x-5y
    y'=4x-2y

    To find the charectaristic equation I create this matrix thing...I don't know what it is called

    [tex]
    \left[\begin{array}{cc}2-\lambda & -5 \\ 4 & -2-\lambda\end{array}\right]=0[/tex]

    Now if I plug in [itex]\lambda= 4i[/itex] to the 2nd row, how do you suppose I will get anything positive?
     
  8. Apr 12, 2008 #7
    Can anybody see what I am doing wrong in this procedure? What am I misunderstanding?

    If I plug [itex]\lambda=4i[/itex] into [itex]-2-\lambda[/itex] why would I get 2+4i ?

    I have five more problems similar to this and cannot move on until I clear this up.

    Any help would be great!
     
  9. Apr 12, 2008 #8

    Kind of desperate here.....
     
  10. Apr 12, 2008 #9

    Vid

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    Do you still not see the glaring algebra error in the bottom left...?
     
  11. Apr 12, 2008 #10
    Thank you. That's all I needed....
     
  12. Apr 12, 2008 #11
    Your K1 value is where the sign error is. Change it to what Hallsofivy did and solve again from there.
     
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