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Question

...... [ 0 6 0 ]
X' = [ 1 0 1 ] X
...... [ 1 1 0 ]
on the interval (-∞, ∞) is
............. [ 6 ] ...................... [ -3 ] ..................... [ 2 ]
X = c_1 [ -1 ] e^(-t) + c_2 [ 1 ] e^(-2t) + c_3 [ 1 ] e^(3t)
.............. [ -5 ] .................... [ 1 ] ...................... [ 1 ]

N.B. : Please ignore the "......"s, plus "_" means subscript, and the three " [ "s (which are to the right of c_1, c_2 and c_3) should actually be one big " [ " (for each term).

Thanks!

Answer 1

Let A be the coefficient matrix of the system of differential equations.
A=
| 0 6 0|
|1 0 1|
|1 1 0|
then
X' = A·X

Assume X = V·e^(λ·t) as solution (V is a vector ans λ a scalar)
Insert this to the DE system and divide by e^(λ·t):
λ·V·e^(λ·t) = A·V·e^(λ·t)
=>
λ·V = A·V
<=>
(A - λ·I ) · V = 0 (i is the identity matrix)

which is the description of the eigenvalue problem of the matrix A, where λ is a eigenvalue of A and V a corresponding eigenvector.

The eigenvalues of A are found from
det| A -λ·I | = 0
<=>
-λ³ + 6 + 0 + λ + 6λ = 0
<=>
λ³ - 7λ - 6 = 0
with solutions
λ1 = -1, λ2 = -2 and λ3 = 3.

You obtain the eigenvectors by solving the equation system f
(A - λ·I )·X = 0
or each value of λ. Here you will find no unique solution. Thus fix one variable to a value and calculate the other two variables:
e.g. for λ1=-1
| 1 6 0|-0
|1 1 1|-0
|1 1 1|-0
<=>
x + 6y = 0
x + y + z = 0
<=>
x = 6y = 0
z = -x - y
set y = -1 → x = 6 and z = -5
Therefore
V1 =
| 6 |
| -1|
| -5|
is a eigenvector to λ1.

By similar calculation you find, that the vectors
V2 =
| -3 |
| 1|
| 1|
and
V3 =
| 2 |
| 1 |
| 1 |
are eigenvectors to λ2 and λ3 respectively.
Of course any multiple of these vectors is a eigenvector too.
Hence you've got three solutions
X_i = c_i·V_i·e^(λ_i·t)

Because each of these three functions a solution of the DE system, any linear combination of them is a solution too. Hence the general solution is:

X = c1·V1·e^(λ1·t) + c2·V2·e^(λ2·t) + c3·V3·e^(λ3·t)

Answer 2

Let

|0 6 0|
|1 0 1| = A
|1 1 0|

and (a, b, c)^T denote the column vector
|a|
|b|
|c|.

First note that A*(6,-1,-5)^T = (-6,1,5)^T; denote this by B. Also A*(-3,1,1)^T = (6,-2,-2)^T, and call this C. Finally, A*(2,1,1)^T = (6,3,3)^T, and call this D.

Now, differentiate X and find X' = -c_1*(6,-1,-5)^T*e^(-t) - 2*c_2*(-3,1,1)^T*e^(-2t) + 3*c_3*(2,1,1)*e^(3t) = c_1*B*e^(-t) + c_2*C*e^(-2t) + c_3*D*e^(3t).

Next, look at A*X and you will get the same result as in the previous paragraph.