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General solution for system of differential equations

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the general solution to the following differential equations
    y'1 = -12y1 + 13y2 +10y3
    y'2 = 4y1 - 3y2 - 4y3
    y'3 = -21y1 +21y2 +19y3


    3. The attempt at a solution
    I'm a little unsure about what to do at the end, or what form to put it in.
    The eigenvalues are
    λ1 = 5, λ2 = -2, λ3 = 1
    and the eigenvectors are
    v1 = [1, -1, 3], v2 = [1, 0, 1], v3 = [1, 1, 0] (respectively)
    so once I have that, what do I do? I need to put them into an exponential form like
    a*eλ2*t*v1 + b*eλ2*t*v2 + c*eλ3*t*v3 ???
    I don't understand why, or what this means.
     
  2. jcsd
  3. Nov 1, 2011 #2

    Mark44

    Staff: Mentor

    Your system of diff. equations is coupled, meaning that the three derivative components all involve the three y components. It's much easier to solve an uncoupled system such as
    z1' = a1z1
    z2' = a2z2
    z3' = a3z3

    and this is the primary motivation for diagonalizing a system of diff. equations.

    Your system can be represented as a matrix equation as:
    Y' = AY

    If we can find a diagonal matrix D that is similar to A, then we can find a related system of equations that is uncoupled, so easier to solve.

    Since you have already found that there are three distinct eigenvalues, that means your matrix A is diagonalizable. You have also found three eigenvectors for the three eigenvalues.

    Form a matrix P whose columns are the eigenvectors. The order doesn't matter, but the order will determine which values appear on the diagonal of matrix D (the diagonal matrix).

    Let Z = P-1Y, or equivalently, Y = PZ.

    Then Z' = P-1Y' = P-1AY = P-1APZ = DZ, where D is the diagonal matrix. As already noted, this system is easy to solve, and its solution is given by:

    [tex]Z = Z(t) = \begin{bmatrix}e^{\lambda_1 t} & 0 & 0 \\ 0 & e^{\lambda_2 t} & 0 \\ 0&0&e^{\lambda_3 t} \end{bmatrix} \begin{bmatrix}c_1 \\ c_2 \\ c_3 \end{bmatrix}[/tex]

    We're not quite done. We're interested in Y = Y(t), not Z(t).

    Since Y = PZ, then our solution is
    [tex]Y = Y(t) = PZ = P\begin{bmatrix}e^{\lambda_1 t} & 0 & 0 \\ 0 & e^{\lambda_2 t} & 0 \\ 0 & 0 & e^{\lambda_3 t}\end{bmatrix}\begin{bmatrix}c_1 \\ c_2 \\ c_3 \end{bmatrix}[/tex]
     
  4. Nov 1, 2011 #3
    OK thank you, that makes sense! But how come the diagonal matrix's entries are in the form e? And the matrix Z is given here by [c1, c2, c3]?
     
  5. Nov 1, 2011 #4

    Mark44

    Staff: Mentor

    What's the general solution of z1' = λ1z1? Once you answer that, you might see that the matrix equation summarizes the solution for the vector Z.
     
  6. Nov 2, 2011 #5
    Of course! Thank you, you are beautiful.
     
  7. Nov 2, 2011 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If you think of a solution of the form y1 = a*exp(r*x), y2 = b*exp(r*x), y3 = c*exp(r*x) [same r in all three] you will get a cubic equation in r --- essentially, the characteristic equation in the matrix representation. This cubic has 3 roots (r1 = -2, r2 = 1, r3 = 5 in this case), and so your solution will have the form y1 = a1*exp(r1*x) + a2*exp(r2*x) + a3*exp(r3*x), etc. You get equations for the ai, bi and ci by substituting these expressions into the DE; the solutions are, essentially, constants times eigenvectors of the matrix. Finally, initial conditions serve to identify the constants.

    RGV
     
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