This can be rewritten in matrix form as y' = Ay, where
A = [0 0 1]
......[1 -1 2]
......[1 -1 1], and y(0) = (0, 0, -1)^T.
First, we find the eigenvalues of A by solving |A - λI| = 0.
|-λ....0.....1|
|1...-1-λ...2| = 0.
|1...-1...1-λ|
==> -λ [(λ^2 - 1) + 2] - 0 + 1[-1 + (1+λ)] = 0, expanding on the top row
==> -λ(λ^2 + 1) + λ = 0
==> -λ [(λ^2 + 1) - 1] = 0
==> λ^3 = 0
==> λ = 0 (eigenvalue with multiplicity three).
----
Next, we find its corresponding eigenvectors by solving (A - 0I)v = 0
[0 0 1|0]
[1 -1 2|0]
[1 -1 1|0], which row reduces to
[1 -1 0|0]
[0 0 0|0], yielding an 1-dim. eigenspace spanned by v = (1, 1, 0)^T.
--
Since we are 'missing' 3 - 1 = 2 eigenvectors, we now compute generalized eigenvectors.
First, we solve (A - 0I)w = v (where v is given as above):
[0 0 1|1]
[1 -1 2|1]
[1 -1 0|-1]
[0 0 0|0], having general solution w = (-1,0,1)^T + k(1, 1, 0)^T.
Letting k = 0 (as the second term is v), we find that one generalized eigenvector is w = (-1, 0, 1)^T.
To find another (linearly independent) generalized eigenvector, we next solve (A - 0I)x = w (where w is given as above):
[0 0 1|-1]
[1 -1 1|1], which row reduces to
[1 -1 0|2]
[0 0 0|0]; this has general solution x = (2, 0, -1)^T + k(1, 1, 0)^T.
Letting k = 0 yields x = (2, 0, -1)^T as the second generalized eigenvector.
Now, we can finally find the Jordan canonical form for A.
Since we have a triple eigenvalue of 0, we let J equal
[0 1 0]
[0 0 1]
[0 0 0].
Then, P is the matrix whose columns are the eigenvector and generalized eigenvectors:
[1 -1 2]
[1 0 0]
[0 1 -1].
Hence, A = PJP⁻¹.
Onto solving the given system of DEs.
A general solution is given by y = e^(tA) * c, where c is a constant vector.
Since y(0) = (0, 0, -1)^T, letting t = 0 yields
(0, 0, -1)^T = e^(0 * A) * c = I * c = c.
Thus, c = (0, 0, -1)^T.
Finally, we compute e^(tA).
e^(tA) = e^(tPJP⁻¹)
..........= e^(P(tJ)P⁻¹)
..........= P e^(tJ) P⁻¹.
To compute e^(tJ), we use its series definition
I + (tJ) + (1/2!)(tJ)² + ...
However, J² = JJ equals
[0 0 2]
[0 0 0]
[0 0 0], and thus J³ = 0 (as well as higher powers of J).
So, e^(tJ) equals
[1 t t²]
[0 1 t]
[0 0 1].
In summary, y = (Pe^(tJ) P⁻¹) * c,
where c = (0, 0, -1)^T and e^(tJ) is as computed above.
I'll leave the final computation of y (whose rows are y₁, y₂, y₃) to you.
-------
I hope this helps!
i dont know and i dont ******' care, you worthless pieace of ****, among with a dick of a ****
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Answers & Comments
This can be rewritten in matrix form as y' = Ay, where
A = [0 0 1]
......[1 -1 2]
......[1 -1 1], and y(0) = (0, 0, -1)^T.
First, we find the eigenvalues of A by solving |A - λI| = 0.
|-λ....0.....1|
|1...-1-λ...2| = 0.
|1...-1...1-λ|
==> -λ [(λ^2 - 1) + 2] - 0 + 1[-1 + (1+λ)] = 0, expanding on the top row
==> -λ(λ^2 + 1) + λ = 0
==> -λ [(λ^2 + 1) - 1] = 0
==> λ^3 = 0
==> λ = 0 (eigenvalue with multiplicity three).
----
Next, we find its corresponding eigenvectors by solving (A - 0I)v = 0
[0 0 1|0]
[1 -1 2|0]
[1 -1 1|0], which row reduces to
[1 -1 0|0]
[0 0 1|0]
[0 0 0|0], yielding an 1-dim. eigenspace spanned by v = (1, 1, 0)^T.
--
Since we are 'missing' 3 - 1 = 2 eigenvectors, we now compute generalized eigenvectors.
First, we solve (A - 0I)w = v (where v is given as above):
[0 0 1|1]
[1 -1 2|1]
[1 -1 1|0], which row reduces to
[1 -1 0|-1]
[0 0 1|1]
[0 0 0|0], having general solution w = (-1,0,1)^T + k(1, 1, 0)^T.
Letting k = 0 (as the second term is v), we find that one generalized eigenvector is w = (-1, 0, 1)^T.
To find another (linearly independent) generalized eigenvector, we next solve (A - 0I)x = w (where w is given as above):
[0 0 1|-1]
[1 -1 2|0]
[1 -1 1|1], which row reduces to
[1 -1 0|2]
[0 0 1|-1]
[0 0 0|0]; this has general solution x = (2, 0, -1)^T + k(1, 1, 0)^T.
Letting k = 0 yields x = (2, 0, -1)^T as the second generalized eigenvector.
----
Now, we can finally find the Jordan canonical form for A.
Since we have a triple eigenvalue of 0, we let J equal
[0 1 0]
[0 0 1]
[0 0 0].
Then, P is the matrix whose columns are the eigenvector and generalized eigenvectors:
[1 -1 2]
[1 0 0]
[0 1 -1].
Hence, A = PJP⁻¹.
----
Onto solving the given system of DEs.
A general solution is given by y = e^(tA) * c, where c is a constant vector.
Since y(0) = (0, 0, -1)^T, letting t = 0 yields
(0, 0, -1)^T = e^(0 * A) * c = I * c = c.
Thus, c = (0, 0, -1)^T.
Finally, we compute e^(tA).
e^(tA) = e^(tPJP⁻¹)
..........= e^(P(tJ)P⁻¹)
..........= P e^(tJ) P⁻¹.
To compute e^(tJ), we use its series definition
I + (tJ) + (1/2!)(tJ)² + ...
However, J² = JJ equals
[0 0 2]
[0 0 0]
[0 0 0], and thus J³ = 0 (as well as higher powers of J).
So, e^(tJ) equals
[1 t t²]
[0 1 t]
[0 0 1].
In summary, y = (Pe^(tJ) P⁻¹) * c,
where c = (0, 0, -1)^T and e^(tJ) is as computed above.
I'll leave the final computation of y (whose rows are y₁, y₂, y₃) to you.
-------
I hope this helps!
i dont know and i dont ******' care, you worthless pieace of ****, among with a dick of a ****