## Wednesday, July 3, 2013

### Solution to Problem (2) Of June 30 DE Systems Set. (Part A)

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

dx1/ dt = 3x1 - 4x2

dx2/dt = x1 - x2

Let:

x1 = A exp(lt) and y1 = B exp(lt)

With these substitutions we than re-write our system in a consistent form:

l A exp(lt)   =  3 A exp(lt)    -  4B exp(lt)

lB exp(lt)   =    A exp(lt)      -  B exp(lt)

Now, collect like terms:

3 A exp(lt) - l A exp(lt)     - 4 B exp(lt) =  0

A exp(lt)   -  B exp(lt)   -lB exp(lt)   = 0

Factor out the exp(lt) terms top and bottom to get:

(3 - l) A  - 4B = 0

A  +    (-1 - l) B = 0

Set up the determinant as per prior problems: (A - l) D = 0 =

(3 - l………-4)

(1………-1 - l )

Find the characteristic equation, viz.

(3 - l) (-1 - l) + 4 = 0  or:   l2 – 2l  + 1 = 0

=   (l - 1) (l - 1) = 0

When roots are real and equal we must seek additional solutions of the form shown below, given we have a case of repeated roots  l1 = 1  = l2.

x1 = (A1t + A2) exp (lt)

x2 = (B1t + B2) exp (lt)

As before with previous problems, substitute the eigenvalue l1 = 1   into the matrix to get:

[2 ……..-4] [A]

[1……..-2] [B]