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.
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.
1 answer
1 answer