Question

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

Short answer

The general solution of the given system is $$\begin{gathered} \mathrm{X}={\mathrm{c}}_1\left(1 \\ 3\right)+{\mathrm{c}}_2\left(2 \\ 1\right){\mathrm{e}}^{5\mathrm{t}} \end{gathered}$$
.

Step-by-step solution

Definition of General Solution of Homogeneous Systems

Let${\mathbf{\lambda }}_{\mathbf{1}}\mathbf{,}{\mathbf{\lambda }}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathbf{\lambda }}_{\mathbf{n}}$be ndistinct real eigenvalues of the coefficient matrix A

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:
$$\begin{gathered} \mathrm{X}\text{'}=\left(-62 \\ -31\right)\mathrm{X}\mathrm{where}\mathrm{A}=\left(-62 \\ -31\right) \end{gathered}$$

Eigenvalues of the Coefficient Matrix

Obtaining the eigenvalues of the coefficient matrix as:

Simplify the equation as:

So our eigenvalues are${\lambda }_1=0$ and ${\lambda }_2=-5$ .

Eigenvector and its Corresponding Solution Vector

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

Apply row operation $-\frac{1}{6}{R}_1:$

Apply row operation $R_2+3R_1:$

So we have $k_1-\frac{1}{3}{k}_2=0\Rightarrow k_1=\frac{1}{3}{k}_2$
Choosing $k_1=1$ yields k₂ =3
This gives an eigenvector and a corresponding solution vector$$\begin{gathered} {\mathrm{K}}_1=\left(1 \\ 3\right),{\mathrm{X}}_1=\left(1 \\ 3\right) \end{gathered}$$ ,
For${\lambda }_2=-5$ , find its corresponding solution vector as:

Apply row operation

So we have$-k_1+2k_2=0\Rightarrow -k_1=-2k_2$ .

Choosing $k_1=2$yields$k_2=1$.

This gives an eigenvector and a corresponding solution vector$$\begin{gathered} {\mathrm{K}}_2=\left(2 \\ 1\right),{\mathrm{X}}_2=\left(2 \\ 1\right){\mathrm{e}}^{-5\mathrm{t}} \end{gathered}$$

General solution of the given system

Finally, the general solution is found using$$\begin{gathered} {\mathrm{X}}_1=\left(1 \\ 3\right)\mathrm{and}{\mathrm{X}}_2=\left(2 \\ 1\right){\mathrm{e}}^{-5\mathrm{t}} \end{gathered}$$ as:
$$\begin{gathered} \mathbf{X}=c_1{\mathbf{K}}_1{e}^{{\lambda }_{1}t}+c_2{\mathbf{K}}_2{e}^{{\lambda }_{2}t} \\ \mathit{}\mathit{}\mathit{}\mathit{=}{c}_{\mathit{1}}\left(1 \\ 3\right)\mathit{+}{c}_{\mathit{2}}\left(2 \\ 1\right)e^{\mathit{-}\mathit{5}t} \end{gathered}$$
Therefore, the general solution of the given system is $$\begin{gathered} \mathrm{X}={\mathrm{c}}_1\left(1 \\ 3\right)+{\mathrm{c}}_2\left(2 \\ 1\right){\mathrm{e}}^{-5\mathrm{t}} \end{gathered}$$.