Question

Question: In Problems 1-12 find the general solution of the given system.

$$\begin{gathered} X^{\text{'}}=\left(101 \\ 010 \\ 101\right)X \end{gathered}$$

Short answer

The general solution of the given system is $$\begin{gathered} \mathbf{X}=c_{1 }\left(1 \\ 0 \\ -1\right)+ c_2\left(0 \\ 1 \\ 0\right)e^t+c_{3 }\left(1 \\ 0 \\ 1\right)e^{2t} \end{gathered}$$

Step-by-step solution

Definition of General Solution of 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 eigenvalues of the coefficient matrixA

of 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 corresponding

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

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

The given information is written as:$$\begin{gathered} X^{\text{'}}=\left(101 \\ 010 \\ 101\right)\mathrm{X} \end{gathered}$$, where$$\begin{gathered} A=\left(101 \\ 010 \\ 101\right) \end{gathered}$$

Eigenvalues of the Coefficient Matrix

Obtaining the eigenvalues of the coefficient matrix as:


Simplify the equation as:


Further simplify the equation as:
$\lambda \left(\lambda -1\right)\left(\lambda -2\right)=0$

So, our eigenvalues are ${\lambda }_1=0, {\lambda }_2=1$and ${\lambda }_3=2$.

Find Solution Vector for λ1=0

For ${\lambda }_1=0,$find its corresponding solution vector as:

Here, we have $k_2=0$and $k_1+ k_3=0\Rightarrow k_1=-k_{3 }$

Choosing $k_3=-1$yields $k_1=1$.

This gives an eigenvector and corresponding solution vector

Find Solution Vector for λ2=1

For${\lambda }_2=1$, find its corresponding solution vector as:


Here, we have $k_1=0$and$k_3=0$.
However, k₂ here is undetermined; we can choose an arbitrary value for it, say k₂ = 1 .

This gives an eigenvector and corresponding solution vector

Find Solution Vector for λ3=2

For${\lambda }_3=2$, find its corresponding solution vector as:

Here, we have $-k_1+k_{3 }=0\Rightarrow k_1=k_{3 }$and $-k_2=0\Rightarrow k_2=0$.

Choosing $k_1=1$yields$k_3=1$.

This gives an eigenvector and corresponding solution vector

General solution of a given system

Finally, the general solution is found using

as:

Therefore, the general solution of the given system is