Question

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

$\mathit{X}\mathbf{\text{'}}\mathbf{=}\mathbf{\left(}\begin{array}{ccc}\mathbf{4}& \mathbf{1}& \mathbf{0}\\ \mathbf{0}& \mathbf{4}& \mathbf{1}\\ \mathbf{0}& \mathbf{0}& \mathbf{4}\end{array}\mathbf{\right)}\mathit{X}$

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$(-\infty ,\infty )$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,

$$\begin{gathered} X\text{'}=\left(\begin{array}{ccc}4& 1& 0\\ 0& 4& 1\\ 0& 0& 4\end{array}\right)X \\ A=\left(\begin{array}{ccc}4& 1& 0\\ 0& 4& 1\\ 0& 0& 4\end{array}\right) \end{gathered}$$

Eigen values of the coefficient matrix 

Obtaining the eigen values of the coefficient matrix,

$$\begin{gathered} det\left(\mathrm{A}-\lambda \mathrm{I}\right)=0 \\ \left|\begin{array}{ccc}4-\lambda & 1& 0\\ 0& 4-\lambda & 1\\ 0& 0& 4-\lambda \end{array}\right|=0 \\ \left(4-\lambda \right)\left|\begin{array}{cc}4-\lambda & 1\\ 0& 4-\lambda \end{array}\right|=0 \\ {\left(4-\lambda \right)}^3=0 \end{gathered}$$

So our eigen values are $\lambda =4$, multiplicity of three.

Eigen vector and its corresponding solution vector

For $\lambda =4$:

$$\begin{gathered} \left(\mathrm{A}-\lambda \mathrm{I}\right)\mathrm{K}=0 \\ \left(\begin{array}{ccc}4-4& 1& 0\\ 0& 4-4& 0\\ 0& 0& 4-4\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\\ k_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ \left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\\ k_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \end{gathered}$$

So we have $k_2=0$and $k_3=0$

Choosing an arbitrary value, $k_1=1$

This gives an eigen vector and a corresponding solution vector

$K=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),X_1=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)e^{4t}$

Next,

localid="1664241639009" $$\begin{gathered} \left(A-\lambda I\right)P=K \\ \left(\begin{array}{ccc}4-4& 1& 0\\ 0& 4-4& 0\\ 0& 0& 4-4\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\\ p_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ \left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\\ p_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \end{gathered}$$

So we have $p_2=1$and$p_3=0$

Choosing an arbitrary value, $p_1=1$

This gives an eigen vector and a corresponding solution vector

$P=\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right),X_2=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)t{e}^{4t}+\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)e^{4t}$

Next,

$$\begin{gathered} (A-\lambda I)Q=P \\ \left(\begin{array}{ccc}4-4& 1& 0\\ 0& 4-4& 0\\ 0& 0& 4-4\end{array}\right)\left(\begin{array}{c}{q}_1\\ q_2\\ q_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ \left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)\left(\begin{array}{c}{q}_1\\ q_2\\ q_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right) \end{gathered}$$

So we have $q_2=1$and $q_3=1$

Choosing an arbitrary value, $q_1=1$.

This gives an eigen vector and a corresponding solution vector

$Q=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right),X_3=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\frac{t^2}{2}{e}^{4t}+\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)t{e}^{4t}+\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)e^{4t}$

General solution of given system

Finally, the general solution is given by

$$\begin{gathered} X=c_1{X}_1+c_2{X}_2+c_3{X}_3 \\ \Rightarrow c_1\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)e^{4t}+c_2[\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)t+\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)]e^{4t}+c_3[\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\frac{t^2}{2}+\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)t+\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)]e^{4t} \end{gathered}$$

Therefore, the general solution of the given system is

$X=c_1\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)e^{4t}+c_2[\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)t+\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)]e^{4t}+c_3[\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\frac{t^2}{2}+\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)t+\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)]e^{4t}$