## Thursday, January 6, 2011

### Homogeneous Linear DE Systems: Breaking down the solutions

We return once more to homogeneous linear systems of differential equations, to see how the solutions can be broken down and what we need to do. This time we will look at this example (and I invite readers who go through the steps here, to also apply them to the previous problem - shown in the last blog). We consider:

dx1/ dt = x1 + x2

dx2/dt = 4x1 + x2

Not the first step, again, is to form a determinant (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 - LI) D =

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

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:

L^2 - 2L - 3 = 0

where L1 = 3 and L2 = -1

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= 3, into the matrix for L, whence:

(A - 3I) D = 0 =

[-2.....1] [d1]
[4 ...-2] [d2]

Therefore:

-2d1 + d2 = 0, so d2 = 2d1

Now let d1 = ½ and d2 = 2d1 = 2( ½)= 1

Then our first eigenvector is: D1 =

[½]
[1]

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

X1 = D1 exp ( L1t) = D1 exp (3t) since L1 = 3

(Thus, the coefficient of the exponent's power function is none other than the eigenvalue, and this is L1 or L2 depending on which solution is sought first)

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] [d1]
[4 ...2] [d2]

Or: 2d1 + d2 = 0 so d2 = - 2d1

Now, let: d1 = 1, then d2 = - 2d1 = -2(1) = -2

The second eigenvector is then: D2 =

[1]
[-2]

And another linearly independent solution is:

X2 = D2 exp (-t)

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

X = X1 + X2 = D1 exp (3t) + D2 exp (-t)

With, of course, the row vectors as computed above substituted in for D1, D2.

This serves as a general approach for solving all such systems.

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

dx/dt = 2x + y

dy/dt = 2x + 3