dx1/ dt = x1 + x2

dx2/dt = 4x1 + x2

Note the first step, again, is to form a matrix from the coefficients, which we see are (1, 1) for the top, and (4, 1) for the bottom. Thus: A =

(1 .....1)

(4......1)

Then, it must be true from the properties of determinants that:

(A - l) D =

[(1 - l).......1] [k1]

[4 ..... .(1 -l)] [k2]

Again, we allow l to be subtracted from the first element in the upper left, and from the last element in the lower right). Cross-multiplying and using matrix properties we obtain the characteristic equation:

l

^{2}- 2l - 3 = 0

where l1 = 3 and l2 = -1 We need to first find a vector that solves the equation: (A - l1 I)D = 0

In the first instance, we substitute the first eigenvalue, l= 3, into the matrix, whence:

(A - 3I) D = 0 =

[-2.....1] [

**k1**]

[4 ...-2] [

**k2**]

Therefore:

-2

**k1**+

**k2**= 0, and 4

**k1**– 2

**k2**= 0, So

**k1**= ½ and

**k2**= 1

Then our first eigenvector is: K1 =

[½]

[1]

Therefore, the first linearly independent solution for the system is:

X1 = K1 exp (l1 t) = K1 exp (3t)

The second eigenvalue was l2 = -1 so we repeat the process again to obtain the equation to be solved:

(A - l2 I)D = (A - (-1)I) D = (A + I)D

Then, (A + I) D = 0 =

[2.....1] [

**k1**]

[4 ….2] [

**k2**]

Or: 2

**k1**+

**k2**= 0 and 4

**k1**+ 2

**k2**

**= 0**

So that:

**k1**= 1, then

**k2**= - 2

The second eigenvector is then: K2 =

[1]

[-2]

So another linearly independent solution is:

X2 = K2 exp (-t)

Then add the two solutions, to obtain:

X = X1 + X2 = K1 exp (3t) + K2 exp (-t)

With, of course, the 2 column vectors (as computed above) substituted in for K1, K2. This serves as a general approach for solving all such systems.

---------

*Problems for the Math Maven:*

*1) Obtain the general solution(s) for the system:*

dx/dt = 2x + y

dy/dt = 2x + 3y

2) Obtain the general solution(s) for the system:

dx1/ dt = 3x1 - 4x2

dx2/dt = x1 - x2

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