I assume you covered the eigenvalues method in school for solution of system of differential equations:
Using the eigenvalue method, and using uppercase symbols to represent matrices and/vectors, we have
[ 3 2 2 ]
[-5 -4 -2] X
[ 5 5 3 ]
From which the eigenvalues of A can be found to be
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:
C=[C1,C2,C3], arbitrary constants
K=[k1,k2,k3] solution vector for x,y & z respectively.
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
for λ=1, we get
Therefore the final solution of
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,
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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