he given system can be expressed as
X ′ = A X X'=AX X ′ = A X , where X ′ = [ d x d t d y d t d z d t ] , A = [ − 1 4 2 4 − 1 2 0 0 6 ] , X = [ x y z ] X' = \begin{bmatrix}
\frac{dx}{dt} \\[0.3em]
\frac{dy}{dt} \\[0.3em]
\frac{dz}{dt} \end{bmatrix},A = \begin{bmatrix}
-1 & 4 & 2 \\[0.3em]
4 & -1 & 2 \\[0.3em]
0 & 0 & 6
\end{bmatrix},X = \begin{bmatrix}
x \\[0.3em]
y \\[0.3em]
z \end{bmatrix} X ′ = ⎣ ⎡ d t d x d t d y d t d z ⎦ ⎤ , A = ⎣ ⎡ − 1 4 0 4 − 1 0 2 2 6 ⎦ ⎤ , X = ⎣ ⎡ x y z ⎦ ⎤
Find the eigenvalues and eigenvectors of the coefficient matrix A A A .
Characteristic equation of the matrix A A A is given by
d e t ( A − λ I ) = 0 det(A-\lambda I)=0 d e t ( A − λ I ) = 0
⇒ ∣ λ + 1 4 2 4 λ + 1 2 0 0 6 − λ ∣ = 0 \Rightarrow \left |
\begin{matrix}
\lambda +1 & 4 & 2 \\
4 & \lambda +1 & 2 \\
0 & 0 & 6-\lambda
\end{matrix}
\ \right |=0 ⇒ ∣ ∣ λ + 1 4 0 4 λ + 1 0 2 2 6 − λ ∣ ∣ = 0
⇒ − λ 3 + 4 λ 2 + 27 λ − 90 = 0 \Rightarrow -\lambda ^3+4\lambda ^2+27\lambda -90=0 ⇒ − λ 3 + 4 λ 2 + 27 λ − 90 = 0
⇒ ( λ + 5 ) ( λ − 3 ) ( λ − 6 ) = 0 \Rightarrow (\lambda +5)(\lambda -3)(\lambda -6)=0 ⇒ ( λ + 5 ) ( λ − 3 ) ( λ − 6 ) = 0
⇒ λ = − 5 , 3 , 6 \Rightarrow \lambda =-5,3,6 ⇒ λ = − 5 , 3 , 6
So, the eigenvalues of the matrix A A A are: ⇒ λ = − 5 , 3 , 6 \Rightarrow \lambda =-5,3,6 ⇒ λ = − 5 , 3 , 6
Now let us find the eigenvectors corresponding to the eigenvalues.
Eigenvector corresponding to the eigenvalue λ = − 5 \lambda =-5 λ = − 5 :
Let v 1 v_{1} v 1 be the eigenvector corresponding to the eigenvalue λ = − 5 \lambda=-5 λ = − 5
Then, ( A + 5 ∗ I ) v = 0 (A+5*I)v=0 ( A + 5 ∗ I ) v = 0
⇒ [ 4 4 2 4 4 2 0 0 11 ] v = 0 \Rightarrow \begin{bmatrix}
4 & 4 & 2 \\[0.3em]
4 & 4 & 2 \\[0.3em]
0 & 0 & 11
\end{bmatrix}v=0 ⇒ ⎣ ⎡ 4 4 0 4 4 0 2 2 11 ⎦ ⎤ v = 0
Solve the homogeneous system by Gaussian elimination.
⇒ [ 2 2 1 0 0 0 0 0 1 ] v = 0 \Rightarrow \begin{bmatrix}
2 & 2 & 1 \\[0.3em]
0 & 0 & 0 \\[0.3em]
0 & 0 & 1
\end{bmatrix}v=0 ⇒ ⎣ ⎡ 2 0 0 2 0 0 1 0 1 ⎦ ⎤ v = 0
⇒ x 1 + x 2 = 0 , x 3 = 0 \Rightarrow x_{1}+x_{2}=0,x_{3}=0 ⇒ x 1 + x 2 = 0 , x 3 = 0
⇒ v 1 = [ − 1 1 0 ] \Rightarrow v_{1}=\begin{bmatrix}
-1 \\[0.3em]
1 \\[0.3em]
0
\end{bmatrix} ⇒ v 1 = ⎣ ⎡ − 1 1 0 ⎦ ⎤
Similarly, the eigenvectors corresponding to the eigenvalues λ = 3 , 6 \lambda=3,6 λ = 3 , 6 are
v 2 = [ 1 1 0 ] , v 3 = [ 2 2 3 ] v_{2}=\begin{bmatrix}
1 \\[0.3em]
1 \\[0.3em]
0
\end{bmatrix},v_{3}=\begin{bmatrix}
2 \\[0.3em]
2 \\[0.3em]
3
\end{bmatrix} v 2 = ⎣ ⎡ 1 1 0 ⎦ ⎤ , v 3 = ⎣ ⎡ 2 2 3 ⎦ ⎤
Therefore, the general solution of the given system is
X ( t ) = c 1 e − 5 t [ − 1 1 0 ] + c 2 e 3 t [ 1 1 0 ] + c 3 e 6 t [ 2 2 3 ] X(t)= c_{1}e^{-5t}\begin{bmatrix}
-1 \\[0.3em]
1 \\[0.3em]
0
\end{bmatrix}+c_{2}e^{3t}\begin{bmatrix}
1 \\[0.3em]
1 \\[0.3em]
0
\end{bmatrix}+c_{3}e^{6t}\begin{bmatrix}
2 \\[0.3em]
2 \\[0.3em]
3
\end{bmatrix} X ( t ) = c 1 e − 5 t ⎣ ⎡ − 1 1 0 ⎦ ⎤ + c 2 e 3 t ⎣ ⎡ 1 1 0 ⎦ ⎤ + c 3 e 6 t ⎣ ⎡ 2 2 3 ⎦ ⎤
That is, [ x ( t ) y ( t ) z ( t ) ] = [ − c 1 e − 5 t + c 2 e 3 t + 2 c 3 e 6 t c 1 e − 5 t + c 2 e 3 t + 2 c 3 e 6 t 3 c 3 e 6 t ] \begin{bmatrix}
x(t) \\[0.3em]
y(t) \\[0.3em]
z(t)
\end{bmatrix}= \begin{bmatrix}
- c_{1}e^{-5t} +c_{2} e^{3t}+2c_{3}e^{6t} \\[0.3em]
c_{1}e^{-5t} +c_{2} e^{3t}+2c_{3}e^{6t} \\[0.3em]
3c_{3}e^{6t}
\end{bmatrix} ⎣ ⎡ x ( t ) y ( t ) z ( t ) ⎦ ⎤ = ⎣ ⎡ − c 1 e − 5 t + c 2 e 3 t + 2 c 3 e 6 t c 1 e − 5 t + c 2 e 3 t + 2 c 3 e 6 t 3 c 3 e 6 t ⎦ ⎤
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