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Posted by on Thursday, March 17, 2011 at 8:49am.

Find the general solution of the following system of equation?
Please help, i don't know what to do...
dx/dt = 3x +2y +2z
dy/dt = -5x - 4y - 2z
dz/dt = 5x + 5y +3z.

  • Differential equation- Math - , Friday, March 18, 2011 at 11:47pm

    I assume you covered the eigenvalues method in school for solution of system of differential equations:

    Given:
    x'=3x+2y+2z
    y'=-5x-4y-2z
    z'=5x+5y+3z

    Using the eigenvalue method, and using uppercase symbols to represent matrices and/vectors, we have

    X'=AX
    [ 3 2 2 ]
    [-5 -4 -2] X
    [ 5 5 3 ]
    X'=[x',y'z'] and
    X=[x,y,z]

    From which the eigenvalues of A can be found to be
    λ=[3,-2,1]
    where λ1=3, λ2=-2, λ3=1.

    There will be three independent solutions to the homogeneous equation, each associated with an arbitrary integration constant named C1, C2 and C3 respectively.

    Each solution will be associated with a vector K whose elements apply to each of the dependent variable x,y and z, such that:

    X=CKe^(λt)
    where
    C=[C1,C2,C3], arbitrary constants
    and
    K=[k1,k2,k3] solution vector for x,y & z respectively.

    For λ=3,
    substitute λ=3 in the matrix A to get A-λ1I
    [ 3-3 2 2 ]
    [-5 -4-3 -2]
    [ 5 5 3-3]
    from which we get
    [ 0 2 2 ]
    [-5 -7 -2] K1=0
    [ 5 5 0]
    From the above matrix, we solve for k1 and k3 in terms of k2 to get
    K1=[-1 1 -1]

    Similarly, for λ=-2, we get
    K2=[0,-1,1]
    and
    for λ=1, we get
    K3=[1,-1,0]

    Therefore the final solution of
    X=CKe^(λt) is

    [x,y,z]=
    C1*K1*e^(3t) + C2*K2*e^(-2t) + C3*K3*e^t

    Separating the equation into three scalar equations, we get by extracting the individual values from K1, K2 and K3,

    x(t)=-C1*e^(3t)+C3*e^t
    y(t)=2C1*e^(3t)-C2*e^(-2t)-C3*e^t
    z(t)=-C1*e^(3t)+C2*e^(-2t)

    I suggest you substitute these solutions into the original equations to make sure they equal to the left hand side, x'(t), y'(t), etc.

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