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March 1, 2015

March 1, 2015

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

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 -
**MathMate**, Friday, March 18, 2011 at 11:47pmI 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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