## Friday, July 12, 2013

### Solutions to DE Problems

1) Set l = 2 - 3i in the algebraic system to get:

(1 + 3i) A  + 2B = 0

-5A  +    - (1 + 3i) B = 0

A simple, nontrivial solution of this system (using the method to solve for A, B we did before) yields: A = 2 and B = -1 - 3i. Using these one obtains the complex solutions:

x = 2 exp (2 - 3i)t

y =  (-1 +3i) exp (2 - 3i)t

Which can alternatively be written :

x = exp (2t) [(2 cost 3t) -  i(2 sin 3t)]
y = exp(2t)[(-cos 3t – 3 sin 3t) -  i (3 cos 3t – sin 3t)]

We can write the general solution of the system as:
x = 2exp (2t) (c1 cost 3t -  c2 sin 3t)

y = exp(2t)[c1(-cos 3t – 3 sin 3t) -  c2 (3 cos 3t – sin 3t)]
Solve, giving the complex and general solutions:
dx/dt = 5x – 2y
dy/dt = 4x – y

Let: x = A exp(lt) and y = B exp(lt)

Substitute the preceding into the system to get:

l A exp(lt)   =  5 A exp(lt)    -  2B exp(lt)

lB exp(lt)   =    4A exp(lt)     -  B exp(lt)

And the algebraic system is:
(5 - l) A  -  2B = 0
4A  -    (1 - l) B = 0

Yielding the standard determinant form:
[(5 - l)  …..-  2]
[4 ……..-(1-  l)]
Expand to obtain the characteristic equation:
l2 – 4l  + 3  = 0
Or: (l -3) (l - 1) = 0
The roots are: l1 = 3 and l2 = 1.  The roots are REAL so there are NO complex solutions!
In the first instance, we substitute the first eigenvalue, l1= 3, into the matrix, whence:
(A - lI)D = 0 =

[2.....-2] [A]
[4  ...2 ] [B]

Then:   2 A – 2B= 0   so A = 1, B = 1
The first eigenvector is then: K1 =
[1]
[1]
So: X1 =  K1 exp (t)
The second eigenvalue was l2 = 1 so we repeat the process again to obtain the equation to be solved:(A - l2 I)D =
[4.....-2] [A]
[4 ...-.0] [B]

Then:  4A  - 2B = 0  or A = 1,  B = 2

Then: K2 =
[1]
[2]

So: X2 =  K2 exp (3t)

Yielding  the general solutions:
x = c1 exp (t) + c2 exp (3t)
y = 2c1 exp (t) + c2 exp (3t)