Respuesta :
Answer:
The general solution of the system is
[tex]X(t)=A\left(\begin{array}{c}1 &-1 \end{array}\right)e^{-2t} + B\left(\begin{array}{c}1 &1 \end{array}\right)e^{4t}\\[/tex]
Step-by-step explanation:
To verify that the system [tex]X^{'} =\left[\begin{array}{cc}1&3\\3&1\end{array}\right] X[/tex] has the solutions given,
First, we determine the Eigen Values β of the matrix of the form [tex]X^{'} =A X[/tex]Using |A-βI|=0, where I is the Identity Matrix, we have:
[tex]|A-\beta I|=0[/tex]
[tex]|\left(\begin{array}{cc}1&3\\3&1\end{array}\right )-\left(\begin{array}{cc}\beta &0\\0&\beta \end{array}\right )|=0[/tex]
Subtracting matrices
[tex]\left|\begin{array}{cc}1-\beta &3\\3&1-\beta \end{array}\right |=0[/tex]
Taking the determinant
[tex](1-\beta)(1-\beta)-(3X3)=0\\1-\beta-\beta+\beta^{2}-9=0\\\beta^{2}-2\beta-8=0\\\beta^{2}-4\beta+2\beta-8=0\\\beta(\beta-4)+2(\beta-4)=0\\(\beta-4)((\beta+2)=0\\\beta=4 or -2[/tex]
We determine the eigen vector using[tex]\left(\begin{array}{cc}1-\beta &3\\3&1-\beta \end{array}\right)\left(\begin{array}{c}c_{11} &c_{12} \end{array}\right)=0[/tex]
When [tex]\beta[/tex] = -2,
[tex]\left(\begin{array}{cc}3 &3\\3&3 \end{array}\right)\left(\begin{array}{c}c_{11} &c_{12} \end{array}\right)=0[/tex] which implies that [tex]3c_{11}+3c_{12}=0[/tex] and thus Â
[tex]If c_{11}=1, c_{12}=-1[/tex]
When [tex]\beta[/tex] = 4,
[tex]\left(\begin{array}{cc}-3 &3\\3&-3 \end{array}\right)\left(\begin{array}{c}c_{21} &c_{22} \end{array}\right)=0[/tex] which implies that [tex]3c_{21}-3c_{22}=0[/tex] and thus Â
[tex]If c_{21}=1, c_{22}=1[/tex]
A general solution of the system is given as
[tex]X(t)=Ac_{1}e^{\beta _{1}} + Bc_{2}e^{\beta _{2}t[/tex] where the c's are the eigen vectors.
Thus the general solution is
[tex]X(t)=A\left(\begin{array}{c}1 &-1 \end{array}\right)e^{-2t} + B\left(\begin{array}{c}1 &1 \end{array}\right)e^{4t}\\[/tex]
