Question

In problems 31 and 32 solve the given initial-value problem.

$\mathit{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right)\mathit{X}\mathbf{,}\mathit{X}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\left(\begin{array}{c}1\\ 2\\ 5\end{array}\right)$,

Short answer

The general solution of the given system using the given initial-value $X\left(0\right)=\left(\begin{array}{c}1\\ 2\\ 5\end{array}\right)$ is found to be $X=2\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)e^{-t}+3\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)e^t+2\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)e^t$.

Step-by-step solution

Definition of general solution-Homogeneous Systems

Let ${\lambda }_1,{\lambda }_2,.......,{\lambda }_n$be ndistinct real eigen values of the coefficient matrix A

of the homogeneous system and let $K_1,K_2,.......,K_n$be the corresponding

eigen vectors.

Then the general solution on the interval $\left(-\infty ,\infty \right)$is given by

$X=c_1{K}_1{e}^{{\lambda }_{1}t}+c_2{K}_2{e}^{{\lambda }_{2}t}+....+c_n{K}_n{e}^{{\lambda }_{n}t}$

We are given,

$X\text{'}=\left(\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right)X$

where, $A=\left(\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right)$

Eigen values of the coefficient matrix 

Obtaining the eigen values of the coefficient matrix,

$det\left(A-\lambda I\right)=0\to \left|\begin{array}{ccc}0-\lambda & 0& 1\\ 0& 1-\lambda & 0\\ 1& 0& 0-\lambda \end{array}\right|=0$

$$\begin{gathered} \left(0-\lambda \right)\left[\left(1-\lambda \right)\left(0-\lambda \right)\right]-\left(1-\lambda \right)=0 \\ {\lambda }^2-{\lambda }^3-1+\lambda =0 \\ {\left(\lambda -1\right)}^2\left(\lambda +1\right)=0 \end{gathered}$$

So our eigen values are ${\lambda }_1=-1$and role="math" localid="1667970171332" ${\lambda }_2={\lambda }_3=1$.

Eigen vector and its corresponding solution vector

For ${\lambda }_1=-1$:

$$\begin{gathered} \left(A+I\right)K=0\to \left(\begin{array}{ccc}0-(-1)& 0& 1\\ 0& 1-(-1)& 0\\ 1& 0& 0-(-1)\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\\ k_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ \Rightarrow \left(\begin{array}{ccc}1& 0& 1\\ 0& 2& 0\\ 1& 0& 1\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\\ k_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \end{gathered}$$

So we have k₂=0 and k₁+k₃=0, k₁ =-k₃

Choosing k₃ = 1 yields k₁ =-1.

This gives an eigen vector and a corresponding solution vector

$K_1=\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right),X_1=\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)e^{-t}$ ,

For , ${\lambda }_2={\lambda }_3=1$:

$$\begin{gathered} \left(A-I\right)K=0\to \left(\begin{array}{ccc}0-1)& 0& 1\\ 0& 1-1)& 0\\ 1& 0& 0-1)\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\\ k_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ \Rightarrow \left(\begin{array}{ccc}-1& 0& 1\\ 0& 0& 0\\ 1& 0& -1\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\\ k_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \end{gathered}$$

So we have $-k_1+k_3=0 \to k_1=k_3$

Let, k₂= 0 and choosing k₃= 1 yields k₁ = 1.

This gives an eigen vector and a corresponding solution vector

$K_2=\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right),X_2=\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)e^t$,

Let, k₂=1 and choosing k₃= 0 yields k₁=1 .

This gives an eigen vector and a corresponding solution vector

$K_3=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right),X_3=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)e^t$

General solution of given system

Finally, the general solution is given by

$\mathbf{X}=c_1{\mathbf{X}}_1+c_2{\mathbf{X}}_2+c_3{\mathbf{X}}_3$

Substituting the values, we get,

$X=c_1\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)e^{-t}+c_2\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)e^t+c_3\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)e^t$ ………….. (A)

Applying initial-value condition

Now, Apply the given initial-value condition $X\left(0\right)=\left(\begin{array}{c}1\\ 2\\ 5\end{array}\right)$ in (A),

$\left(\begin{array}{c}1\\ 2\\ 5\end{array}\right)=c_1\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)+c_2\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)+c_3\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)$

So we have, $1=-c_1+c_2$ ……… (1)

$2=c_3$ ……… (2)

$5=c_1+c_2$ ……… (3)

Adding (3) to (1), we obtain $1+5=c_2+c_2 \to 2c_2=6 \Rightarrow c_2=3$

Now, substitute c₂ in (1), $1=-c_1+3 \to c_1=2$

Thus, the solution is given by

$X=2\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)e^{-t}+3\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)e^t+2\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)e^t$

Therefore, the general solution of the given system using the given initial-value $X\left(0\right)=\left(\begin{array}{c}1\\ 2\\ 5\end{array}\right)$ is found to be$X=2\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)e^{-t}+3\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)e^t+2\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)e^t$