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System of two second order ODE's. Solution does not agree with Wolfram.

  1. Oct 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Solve the following system of differential equations:

    ##y''(x) = y'(x) + z'(x) - z(x)##
    ##z''(x) = -5*y'(x) - z'(x) -4*y(x) + z(x)##

    2. The attempt at a solution

    I converted the two second order equations to 4 first order equations by substituting:
    ##g(x) = y'(x)## and ##h(x) = z'(x)##

    So now I have the following system of equations:

    ##g' = g + h - z##
    ##h' = -5g - h - 4y + z##
    ##y' = g##
    ##z' = h##

    I wrote the system in matrix form, solved the characteristic equation for 4 eigenvalues and got
    λ1 = ##1##
    λ2 = ##-1##
    λ3 = ##2i##
    λ4 = ##-2i##

    So the eigenvectors are as follows:

    V1 = ##(-1,9,-1,9)##
    V2 = ##(-1,1,1,-1)##
    V3 = ##(1,2i,-i/2,1)##
    V4 = ##(1,-2i,i/2,1)##

    So then

    X1 = ##exp(x)##V1
    X2 = ##exp(-x)##V2
    X3 = ##(cos2x+i*sin2x)##V3
    X4 = ##(cos2x-i*sin2x)##V4

    Skipping some steps (separating complex and real parts of the matrices)

    I get that the general solution is (in matrix form):

    ϒ = C1*X1 + C2*X2 + C3*X3 + C4*X4

    Now, since I need the solution just to the last two functions ( ##y(x)## and ##z(x)## )

    I multiply out the last two rows of the matrices and get:

    ##y(x) = -C_1exp(x) + C_2exp(-x) + (C_3/2)sin2x-(C_4/2)cos2x + (C_5/2)sin2x + (C_6/2)cos2x##

    I am not too sure how to proceed further because if I group some arbitrary constants together (near sines and cosines), these constants will not match with the constants in the solution for function Z(x). Nevertheless, ##y(x)## appears to not agree with WolframAlpha:

    http://www.wolframalpha.com/input/?i=system of equations&a=*C.system of equations-_*Calculator.dflt-&a=FSelect_**SolveSystemOf2EquationsCalculator--&f3=y''(x)=y'(x)+z'(x)-z(x)&f=SolveSystemOf4EquationsCalculator.equation1_y''(x)=y'(x)+z'(x)-z(x)&f4= z''(x)=-5*y'(x)-z'(x)-4*y(x)+z(x)&f=SolveSystemOf4EquationsCalculator.equation2_ z''(x)=-5*y'(x)-z'(x)-4*y(x)+z(x)&f5=&f=SolveSystemOf4EquationsCalculator.equation3_&f6=&f=SolveSystemOf4EquationsCalculator.equation4_


    I can't seem to find a mistake. Can anyone help me?
     
  2. jcsd
  3. Oct 25, 2014 #2

    vela

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    Nice problem… good review for me. It sounds like your method is correct, so it's probably just algebra and sign mistakes. I'm not sure what your ##c_5## and ##c_6## are, however.

    One thing I learned is that if you calculate ##\vec{v}_3 e^{2it}##, the real and imaginary parts give you the two independent solutions for the complex roots. That saves doing a bit of algebra, and I think it'll help you avoid whatever error you made. This is the solution I ended up with (with ##x## replaced by ##t##):
    $$\begin{bmatrix} g \\ h \\ y \\ z\end{bmatrix} =
    c_1 \begin{bmatrix} -1 \\ 9 \\ -1 \\ 9 \end{bmatrix} e^t +
    c_2 \begin{bmatrix} -1 \\ 1 \\ 1 \\ -1 \end{bmatrix} e^{-t} +
    c_3 \begin{bmatrix} \cos 2t \\ -2 \sin 2t \\ \frac 12 \sin 2t \\ \cos 2t \end{bmatrix} +
    c_4 \begin{bmatrix} \sin 2t \\ 2 \cos 2t \\ -\frac 12 \cos 2t \\ \sin 2t \end{bmatrix}.$$ The results from Wolfram Alpha are kind of a pain to compare to because they're not simplified. I did verify the solution I got in Mathematica.
     
    Last edited: Oct 26, 2014
  4. Oct 26, 2014 #3
    Alright. So if I understand you correctly the real parts and the complex parts of

    X3 = ##(cos2x+i*sin2x)##V3
    X4 = ##(cos2x-i*sin2x)##V4

    are linearly dependant matrices so I can eliminate one real and one complex part? That does makes sense.
     
    Last edited: Oct 26, 2014
  5. Oct 26, 2014 #4

    vela

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    Yes, though I wouldn't say you're eliminating anything. You're just rearranging terms and combining constants. ##\vec{x}_3## and ##\vec{x}_4## are complex conjugates, so each component of ##c_3 \vec{x}_3 + c_4 \vec{x}_4## is of the form ##c_3 (p+iq) + c_4 (p-iq) = (c_3+c_4) p + i(c_3-c_4)q##.
     
  6. Oct 26, 2014 #5
    Thanks for clearing things up.
     
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