Question

Exercise 3.3.2: Verify that the system x® 0 1 3 3 1 x® has the two solutions 1 1 e 4t and 1 −1 e −2t a) . b) Write down the general solution. Write down the general solution in the form x1 ?, x2 ? (i.e. write down a formula for each element of the solution).

Answer 1

Answer:

The general solution of the system is

$X(t)=A\left(\begin{array}{c}1 &-1 \end{array}\right)e^{-2t} + B\left(\begin{array}{c}1 &1 \end{array}\right)e^{4t}$

Step-by-step explanation:

To verify that the system $X^{'} =\left[\begin{array}{cc}1&3\\3&1\end{array}\right] X$ has the solutions given,

First, we determine the Eigen Values β of the matrix of the form $X^{'} =A X$Using |A-βI|=0, where I is the Identity Matrix, we have:

$|A-\beta I|=0$

$|\left(\begin{array}{cc}1&3\\3&1\end{array}\right )-\left(\begin{array}{cc}\beta &0\\0&\beta \end{array}\right )|=0$

Subtracting matrices

$\left|\begin{array}{cc}1-\beta &3\\3&1-\beta \end{array}\right |=0$

Taking the determinant

$(1-\beta)(1-\beta)-(3X3)=0\\1-\beta-\beta+\beta^{2}-9=0\\\beta^{2}-2\beta-8=0\\\beta^{2}-4\beta+2\beta-8=0\\\beta(\beta-4)+2(\beta-4)=0\\(\beta-4)((\beta+2)=0\\\beta=4 or -2$

We determine the eigen vector using$\left(\begin{array}{cc}1-\beta &3\\3&1-\beta \end{array}\right)\left(\begin{array}{c}c_{11} &c_{12} \end{array}\right)=0$

When $\beta$ = -2,

$\left(\begin{array}{cc}3 &3\\3&3 \end{array}\right)\left(\begin{array}{c}c_{11} &c_{12} \end{array}\right)=0$ which implies that $3c_{11}+3c_{12}=0$ and thus  

$If c_{11}=1, c_{12}=-1$

When $\beta$ = 4,

$\left(\begin{array}{cc}-3 &3\\3&-3 \end{array}\right)\left(\begin{array}{c}c_{21} &c_{22} \end{array}\right)=0$ which implies that $3c_{21}-3c_{22}=0$ and thus  

$If c_{21}=1, c_{22}=1$

A general solution of the system is given as

$X(t)=Ac_{1}e^{\beta _{1}} + Bc_{2}e^{\beta _{2}t$ where the c's are the eigen vectors.

Thus the general solution is

$X(t)=A\left(\begin{array}{c}1 &-1 \end{array}\right)e^{-2t} + B\left(\begin{array}{c}1 &1 \end{array}\right)e^{4t}$