Question

Question: In Problems 1-12 find the general solution of the given system.
$$\begin{gathered} \mathrm{X}\text{'}=\left(10-5 \\ 8-12\right)\mathrm{X} \end{gathered}$$

Short answer

The general solution of the given system is$$\begin{gathered} \mathbf{X}={\mathrm{c}}_1\left(5 \\ 2\right){\mathrm{e}}^{8\mathrm{t}}+{\mathrm{c}}_2\left(1 \\ 4\right){\mathrm{e}}^{-10\mathrm{t}} \end{gathered}$$

Step-by-step solution

Definition of General Solution of Homogeneous Systems

Let${\mathrm{\lambda }}_1,{\mathrm{\lambda }}_2,.......,{\mathrm{\lambda }}_{\mathrm{n}}$be n distinct real eigenvalues of the coefficient matrix A

of the homogeneous system and let ${\mathrm{K}}_1,{\mathrm{K}}_2,.......,{\mathrm{K}}_{\mathrm{n}}$be the corresponding

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

$\mathrm{X}={\mathrm{c}}_1{\mathrm{K}}_1{\mathrm{e}}^{{\mathrm{\lambda }}_1\mathrm{t}}+{\mathrm{c}}_2{\mathrm{K}}_2{\mathrm{e}}^{{\mathrm{\lambda }}_2\mathrm{t}}+....+{\mathrm{c}}_{\mathrm{n}}{\mathrm{K}}_{\mathrm{n}}{\mathrm{e}}^{{\mathrm{\lambda }}_{\mathrm{n}}\mathrm{t}}$.

The given information is written as:
$$\begin{gathered} \mathrm{X}\text{'}=\left(10-5 \\ 8-12\right)\mathrm{X}\mathrm{where}\mathrm{A}=\left(10-5 \\ 8-12\right) \end{gathered}$$

Eigenvalues of the Coefficient Matrix

Obtaining the eigenvalues of the coefficient matrix as:


Simplify the equation as:
$$\begin{gathered} -120\lambda -10\lambda +12\lambda +{\lambda }^2+40=0 \\ {\lambda }^2+2\lambda -80=0 \\ \left(\lambda -8\right)\left(\lambda +10\right)=0 \end{gathered}$$
So our eigenvalues are ${\lambda }_1=8$and ${\lambda }_2=-10$.

Eigenvector and its Corresponding Solution Vector

For${\lambda }_1=8$, find its corresponding solution vector as:

Apply row operation$R_2-4R_1\to R_2$ as:

Apply row operation$\frac{1}{2}{R}_1\to R_1$ as:

So we have $k_1-\frac{5}{2}{k}_2=0\Rightarrow k_1=\frac{5}{2}{k}_2$

Choosing $k_1=5$yields$k_2=2$

This gives an eigenvector and a corresponding solution vector $$\begin{gathered} {\mathrm{K}}_1=\left(5 \\ 2\right),{\mathrm{X}}_1=\left(5 \\ 2\right){\mathrm{e}}^{8\mathrm{t}} \end{gathered}$$
For${\lambda }_2=-10$, find its corresponding solution vector as:

Apply row operation$\frac{1}{20}{R}_1\to R_1$ as:


Apply row operation$R_2-8R_1\to R_2$ as:


So we have$k_1-\frac{1}{4}{k}_2=0\Rightarrow k_1=\frac{1}{4}{k}_2$ .

Choosing $k_1=1$yields$k_2=4$ .
This gives an eigenvector and a corresponding solution vectorlocalid="1664176673595" $$\begin{gathered} {\mathrm{K}}_2=\left(1 \\ 4\right),{\mathrm{X}}_2=\left(1 \\ 4\right){\mathrm{e}}^{-10\mathrm{t}} \end{gathered}$$

General solution of the given system

Finally, the general solution is found using$$\begin{gathered} {\mathrm{X}}_1=\left(5 \\ 2\right){\mathrm{e}}^{8\mathrm{t}}\mathrm{and}{\mathrm{X}}_2=\left(1 \\ 4\right){\mathrm{e}}^{-10\mathrm{t}} \end{gathered}$$ as:
$$\begin{gathered} \mathbf{X}=c_1{\mathbf{K}}_1{e}^{{\lambda }_{1}t}+c_2{\mathbf{K}}_2{e}^{{\lambda }_{2}t} \\ \mathit{}\mathit{}\mathit{}\mathit{=}{c}_{\mathit{1}}\left(5 \\ 2\right)e^{\mathit{8}t\mathit{}}\mathit{+}{c}_{\mathit{2}}\left(1 \\ 4\right)e^{\mathit{-}\mathit{10}t} \end{gathered}$$

Therefore, the general solution of the given system is $$\begin{gathered} \mathrm{X}={\mathrm{c}}_1\left(5 \\ 2\right){\mathrm{e}}^{8\mathrm{t}}+{\mathrm{c}}_2\left(1 \\ 4\right){\mathrm{e}}^{-10\mathrm{t}} \end{gathered}$$