1) Obtain the general solution(s) for the system:
dx/dt = 2x + y
dy/dt = 2x + 3y
Solution:
In this case, the alert math maven will note the x, y derivatives are with respect to t, but the equivalence is to x, y forms with no t variable present. We reconcile our terms and allow the solution to proceed via the following substitutions:
x = A exp(lt) and y = B exp(lt)
With these substitutions we than re-write our system in a consistent form:
l A exp(lt) = 2 A exp(lt) + B exp(lt)
lB exp(lt) = 2 A exp(lt) + 3B exp(lt)
(Recall: dx/dt = d (A exp(lt))/dt = l A exp(lt) - Analogous for dy/dt!)
Now, collect like terms:
2 A exp(lt) - l A exp(lt) + B exp(lt) = 0
2 A exp(lt) + 3B exp(lt) -lB exp(lt) = 0
Factor out the exp(lt) terms top and bottom to get:
(2 - l) A + B = 0
2A + (3 - l) B = 0
Set up the determinant as per prior problems: (A - l) D = 0 =
(2 - l…………1)
(2………...3 - l )
Find the characteristic equation, viz.
(2 - l) (3 - l) – 2 = 0 = 6 - 3 l - 2 l + l2 – 2 =
l2 – 5l + 4 = 0
Or: (l - 4) (l - 1) = 0
So we have eigenvalues : l1 = 4 and l2 = 1
As before with previous problems, substitute the eigenvalue l1 = 4 into the matrix to get:
[-2 ……..1] [A]
[2……..-1] [B]Whence: -2 A + B = 0 and 2A – B = 0
So, A = 1, B = 2, so c1 =
[1]
[2]
The solution is then: X1 = c1 exp (4t)
Now, substitute the eigenvalue l2 = 1 into the matrix to get:
[1 ……..1] [A][2……..2] [B]
Whence: A + B = 0 and 2A + 2B = 0 so: A = 1 and B = -1
so c2 =
[1]
[-1]
The solution is then: X2 = c2 exp (t)
The full general solution for the system is:
X = X1 + X2 = c1 exp (4t) + c2 exp (t)
2)Obtain the general solution(s) for the system:
dx1/ dt = 3x1 - 4x2
dx2/dt = x1 - x2
When roots are real and equal, then assume soln. of form:
x1 = A exp(lt) and x2 = B exp(lt)
If Det (A - lI) D = 0 only has one soln. l then (A - lI) D = 0 has at least one independent solution. A first solution is: X1 = D exp (lt)
Now take: X = D exp (lt) + Bt exp (lt)
dX/ dt = l D exp (lt) + B exp(lt) + l Bt exp(lt)
= AX + AD exp (lt) + ABt exp (lt)
For all t we need to satisfy:
B = (A - lI) D + t (A - lI) B
Take (A - lI) B = 0
(A - lI) D = B
The independent solution is then:
X2 = D exp (lt) + Bt exp (lt)
Applied to the given linear system we get:
The characteristic equation is then:
l2 – 2l + 1 = 0
Whence:
2d1 – 4d2 = 0
d1 - 2d2 = 0
=> 3d1 - 6d2 = 0
Or: d1 - 2d2 = 0
Then:
d1 = 2d2 so: d1 = 2 and d2 = 1
Eigenvector =
(2)
(1)
Finally:
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