what is general solution for
dw/dt = 7w+6y-10z
dx/dt= 3w + x +3y-5z
dy/dt= 5w+2x+11y-14z
dz/dt=5w+2x+8y-11z
Represent the coupled differential equations in matrix form.
X = Av
Where X is the column vector [w’ x’ y’ z’] and x’ = dx/dt, etc.
And A is the component matrix,
|7 0 6 -10|
|3 1 3 -5|
|5 2 11 -14|
|5 2 8 -11|
And v is the column vector [w x y z]
To find the solutions to the original coupled differential equations we need to solve Av = λv for eigenvalues and eigenvectors.
i.e. det(A –λI) = 0
or,
|7-λ 0 6 -10| = 0
| 3 1-λ 3 -5|
| 5 2 11-λ -14|
| 5 2 8 -11-λ|
(7-λ){(1-λ)[(11-λ)(-11-λ) – (-14)(8)] – 3[2(-11-λ) – (-14)(2)] + (-5)[2*8 – (11-λ)*2]} – 0 + 6{3[2(-11-λ) – (-14)*2] – (1-λ)[5*(-11-λ) – (-14)*5] + (-5)[5*2 – 2*5]} – (-10){3[2*8 – (11-λ)2] – (1-λ)[5*8 – (11-λ)*5] + 3[5*2 – 2*5]} = 0
You could evaluate and solve the above expression for λ manually, or use an algebra software package such as Maple, Matlab or Mathematica.
The solutions for λ are,
λ = 1, λ = 1, λ = 3, λ = 3
i.e. four solutions with two roots repeated twice.
The eigenvectors are got by solving Av = λv, for λ = 1, 3 i.e.
|7 0 6 -10 ||w| = |w| and |7 0 6 -10 ||w| = |3w|
|3 1 3 -5 || x| |x| |3 1 3 -5 || x| |3x|
|5 2 11 -14|| y| | y| |5 2 11 -14|| y| | 3y|
|5 2 8 -11|| z| | z| |5 2 8 -11|| z| | 3z|
Giving,
7w + 6y – 10z = w and 7w + 6y – 10z = 3w
3w + x + 3y – 5z = x 3w + x + 3y – 5z = 3x
5w + 2x + 11y – 14z = y 5w + 2x + 11y – 14z = 3y
5w + 2x + 8y – 11z = z 5w + 2x + 8y – 11z = 3z
The solution of which is: The solution of which is:
w = 2, x = 1, y = 3, z = 3 w = 2, x = 1, y = 2, z = 2
i.e. v1 = [2, 1, 3, 3] i.e. v2 = [2, 1, 2, 2]
Independent solutions of the coupled equations then are,
x1 = v1.e^t, x3 = v2.e^(3t)
Repeated roots
λ = 1, and one solution is x1.
A 2nd solution is x2 = t.x1 + p.e^t
Where p is solved from (A – λI)p = v1
| 6 0 6 -10||p1| = |2|
| 3 0 3 -5||p2| |1|
| 5 2 10 -14||p3| |3|
| 5 2 8 -12||p4| |3|
Giving, p = [p1, p2, p3, p4] = [k, k, k, k]
Taking k = 1, p = [1, 1, 1, 1]
Then x2 = t.v1.e^t + p.e^t
λ = 3, and one solution is x3.
A 2nd solution is x4 = t.x3 + q.e^(3t)
Where q is solved from (A – λI)q = v2
| 4 0 6 -10||q1| = |2|
| 3 -2 3 -5||q2| |1|
| 5 2 8 -14||q3| |2|
| 5 2 8 -14||q4| |2|
Giving, q = [q1, q2, q3, q4] = [2k, k, 2k+ 2, 2k + 1]
Taking k = 0, q = [0, 0, 2, 1]
Then x4 = t.v2.e^(3t) + q.e^(3t)
Our general solution then is: x = c1.x1 + c2.x2 + c3.x3 + c4.x4
x = c1.v1.e^t + c2{ t.v1.e^t + p.e^t } + c3.v2.e^(3t) + c4{ t.v2.e^(3t) + q.e^(3t) }