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

[2sin54]

Homework Statement

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?

 

[vela]

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.

 

[2sin54]

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.

 

[vela]

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##.

 

[2sin54]

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##.

Thanks for clearing things up.