## Friday, January 7, 2011

### Solution to Linear System DE Problem

Problem: We were to obtain the general solution(s) for the system:

dx/dt = 2x + y

dy/dt = 2x + 3

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

A =

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

Then, it must be true from the properties of determinants (And recall again that I is the identity matrix) that:

(A - LI) D =

[(2 - L)......1] [d1]
[4 .....(3 -L)] [d2]

For any determinant D such that D =

[a ……b]
[c……..d]

We compute according to (a*d – b*c) in order to find the characteristic equation.

We then obtain for this:

L^2 - 5L +6 = 0, for which:

L1 = -2 and L2 = -3

We need to find a vector that solves the equation:

(A - LI)D = 0

Now, in the first instance, we substitute the first eigenvalue, L= -2, into the matrix for L,

whence:

(A +2 I) D = 0 =

[4.....1] [d1]
[4 ...5] [d2]

Therefore, combining equations in d1, d2:

8 d1 + 6 d2 = 0, or d1 = (-3/4) d2

Let d2 = 4, then d1 = -3/4 (4) = -3

Then our first eigenvector is: D1 =

[-3]
[4]

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

X1 = D1 exp ( L1t) = D1 exp (-2t)

since L1 = -2

The second eigenvalue was L2 = -3 so we repeat the process again to obtain the equation to be solved:

(A - L2 I)D = (A - (-3)I) D = (A + 3I)D

Then, (A + 3 I) D = 0 =

[5.....1] [d1]
[4 ...6] [d2]

Or, combining equations:

- d1 – 5 d2 = 0, so d2 = -d1/5

Now, let: d1 = 5, then d2 = - (5)/5 = -1

The second eigenvector is then: D2 =

[5 ]
[-1]

And another linearly independent solution is:

X2 = D2 exp (-3t)

Where D2 =

[5]
[-1]

The full general solution is:

X = D1 exp (-2t) + D2 exp (-3t)