Question

In problems 21-30 find the general solution of the given system.

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

Short answer

The general solution of the given system is,

$X=c_1\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right)e^t+c_2\left[\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right)t{e}^t+\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)e^t\right]+c_3\left[\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\frac{t^2}{2}{e}^t+\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)t{e}^t+\left(\begin{array}{c}1/2\\ 1\\ 1\end{array}\right)e^t\right]$

Step-by-step solution

Definition of general solution-Homogeneous Systems

Let ${\mathit{\lambda }}_{\mathbf{1}}\mathbf{,}{\mathit{\lambda }}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{\lambda }}_{\mathbf{n}}$be ndistinct real eigen values of the coefficient matrix Aof the homogeneous system and let ${\mathit{K}}_{\mathbf{1}}\mathbf{,}{\mathit{K}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{K}}_{\mathbf{n}}$be the correspondingeigen vectors.

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

$\mathit{X}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{K}}_{\mathbf{1}}{\mathit{e}}^{{\mathbf{\lambda }}_{\mathbf{1}}\mathbf{t}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{K}}_{\mathbf{2}}{\mathit{e}}^{{\mathbf{\lambda }}_{\mathbf{2}}\mathbf{t}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{c}}_{\mathbf{n}}{\mathit{K}}_{\mathbf{n}}{\mathit{e}}^{{\mathbf{\lambda }}_{\mathbf{n}}\mathbf{t}}$

We are given,

$X\text{'}=\left(\begin{array}{ccc}1& 0& 0\\ 2& 2& -1\\ 0& 1& 0\end{array}\right)X$

where, $A=\left(\begin{array}{ccc}1& 0& 0\\ 2& 2& -1\\ 0& 1& 0\end{array}\right)$

Eigen values of the coefficient matrix 

Obtaining the eigen values of the coefficient matrix,

$$\begin{gathered} det\left(\mathrm{A}-\lambda \mathrm{I}\right)=0 \\ \left|\begin{array}{ccc}1-\lambda & 0& 0\\ 1& 2-\lambda & -1\\ 0& 1& 0-\lambda \end{array}\right|=0 \end{gathered}$$

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

So our eigen values are $\lambda =1$, multiplicity of three.

Eigen vector and its corresponding solution vector

For $\lambda =1$

localid="1664230494723" $$\begin{gathered} \left(\mathrm{A}-\lambda \mathrm{I}\right)\mathrm{K}=0 \\ \left(\begin{array}{ccc}1-1& 0& 0\\ 2& 2-1& -1\\ 0& 1& 0-1\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\\ k_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ \left(\begin{array}{ccc}0& 0& 0\\ 2& 1& -1\\ 0& 1& -1\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\\ k_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \end{gathered}$$

Apply row operation $R_2-R_3\to R_2$:

$\left(\begin{array}{ccc}0& 0& 0\\ 2& 0& 0\\ 0& 1& -1\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\\ k_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)$

So we have $k_1=0$and $k_2-k_3=0\to k_2=k_3$

Choosing $k_3=1$yields $k_2=1$.

This gives an eigen vector and a corresponding solution vector

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

Next,

localid="1664231529384" $$\begin{gathered} \left(A-\lambda I\right)P=K \\ \left(\begin{array}{ccc}1-1& 0& 0\\ 2& 2-1& -1\\ 0& 1& 0-1\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\\ p_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ \left(\begin{array}{ccc}0& 0& 0\\ 2& 1& -1\\ 0& 1& -1\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\\ p_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \end{gathered}$$

Apply row operation $R_2-R_3\to R_2$:

$\left(\begin{array}{ccc}0& 0& 0\\ 2& 0& 0\\ 0& 1& -1\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\\ p_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)$

So we have $p_1=0$and $k_2-k_3=1$

Choosing $k_3=0$yields $k_2=1$.

This gives an eigen vector and a corresponding solution vector

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

Next,

$$\begin{gathered} \left(A-\lambda I\right)Q=P \\ \left(\begin{array}{ccc}1-1& 0& 0\\ 2& 2-1& -1\\ 0& 1& 0-1\end{array}\right)\left(\begin{array}{c}{q}_1\\ q_2\\ q_3\end{array}\right)=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right) \\ \left(\begin{array}{ccc}0& 0& 0\\ 2& 1& -1\\ 0& 1& -1\end{array}\right)\left(\begin{array}{c}{q}_1\\ q_2\\ q_3\end{array}\right)=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right) \end{gathered}$$

Apply row operation $R_2-R_3\to R_2$:

$\left(\begin{array}{ccc}0& 0& 0\\ 2& 0& 0\\ 0& 1& -1\end{array}\right)\left(\begin{array}{c}{q}_1\\ q_2\\ q_3\end{array}\right)=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)$

So we have $2q_1=1 \to q_1=\frac{1}{2}$and $q_2-q_3=0 \to q_2=q_3$

Choosing $q_3=1$yields $q_2=1$.

This gives an eigen vector and a corresponding solution vector

$Q=\left(\begin{array}{c}1/2\\ 1\\ 1\end{array}\right),X_3=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\frac{t^2}{2}{e}^t+\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)t{e}^t+\left(\begin{array}{c}1/2\\ 1\\ 1\end{array}\right)e^t$

General solution of given system

Finally, the general solution is given by

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

$\Rightarrow c_1\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right)e^t+c_2[\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right)t{e}^t+\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)e^t]+c_3[\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\frac{t^2}{2}{e}^t+\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)t{e}^t+\left(\begin{array}{c}1/2\\ 1\\ 1\end{array}\right)e^t]$

Therefore, the general solution of the given system is

$X=c_1\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right)e^t+c_2[\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right)t{e}^t+\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)e^t]+c_3[\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\frac{t^2}{2}{e}^t+\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)t{e}^t+\left(\begin{array}{c}1/2\\ 1\\ 1\end{array}\right)e^t]$