## Thursday, June 27, 2013

### Solution of Linear, Homogeneous Differential Equation Problem

Note the first step, again, is to form a determinant (matrix) from the coefficients, which we see are (2, 3) for the top, and (2, 1) for the bottom. Thus:  A =

(2 .....3)
(2......1)

Then, it must be true from the properties of determinants that: (A - lI) D =

[(2 - l)......3] [k1]

[2 .......(1 -l)]  [k2]

Note how we allow l ('lambda') 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:

(l2 -3l +2) – 6 = 0  or    l2 -3l -4 = 0

So that:  (l - 4) (l + 1) = 0,    where l1 = -1 and l2 = 4

We need to find a vector that solves the equation: (A -  lI) D = 0

In the first instance, we substitute the first eigenvalue, l1= -1, into the matrix, whence:

(A - (-1))D = 0 =

[3.....3] [k1]

[2  ...2] [k2]

Therefore:
3k1 + 3 k2 = 0,  and 2k1 + 2 k2 = 0

So: k1 =1  and    k2   = -1

The first eigenvector is then: K1 =

[1]

[-1]

So: X1 =  K1 exp (-t)

The second eigenvalue was l2 = 4 so we repeat the process again to obtain the equation to be solved:

(A - l2 I)D = (A - (4)I) D = (A + I)D = 0 =

[-2.....3] [k1]

[2 ...-.3] [k2]

Or: -2k1 + 3k2 = 0 and 2 k1  - 3k2 = 0

So that: k1 = 3  and k2 = 2

The second eigenvector is then: K2 =

[3]
[2]

And another linearly independent solution is:  X2 =  K2 exp (4t)

Because of the principle of linear superposition, one can also add the two solutions, to obtain:

X = X1 + X2 = K1 exp (-t) + K2 exp (4t)