Question

Question: In Problems 1-12 find the general solution of the given system.
$$\begin{gathered} \frac{dx}{dt}=-4x+2y \\ \frac{dy}{dt}=-\frac{5}{2}x+2y \end{gathered}$$

Short answer

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

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

$X=c_1{K}_1{e}^{{\lambda }_{1}t}+c_2{K}_2{e}^{{\lambda }_{2}t}+....+c_n{K}_n{e}^{{\lambda }_{n}t}$

The given information is written as:
$$\begin{gathered} \frac{dx}{dt}=-4x+2y \\ \frac{dy}{dt}=-\frac{5}{2}x+2y \end{gathered}$$
And this can be written in the matrix form as:
$$\begin{gathered} \mathrm{X}\text{'}=\left(-42 \\ -\frac{5}{2}2\right)\mathrm{X}\mathrm{where}\mathrm{A}=\left(-42 \\ -\frac{5}{2}2\right) \end{gathered}$$

Eigenvalues of the Coefficient Matrix

Obtaining the eigenvalues of the coefficient matrix as:

Simplify the equation as:
$\left(\lambda +3\right)\left(\lambda -1\right)=0$

So our eigenvalues are ${\lambda }_1=1$and${\lambda }_2=-3$ .

Eigenvector and its Corresponding Solution Vector

For${\lambda }_1=1$, find its corresponding solution vector as:



And this yields two equations as:
$-5x+2y=0$ ………… (1)
$-\frac{5}{2}x+y=0$ …………. (2)
These two equations are the same if we multiply equation (2) by 2 as:
$$\begin{gathered} -5x+2y=0 \\ 2y=5x \end{gathered}$$

Let y=5 .
Then, x=2 .

Therefore, the first eigenvector is $$\begin{gathered} {\mathrm{K}}_1=\left(2 \\ 5\right) \end{gathered}$$.

This gives an eigenvector and a corresponding solution vector $$\begin{gathered} {\mathrm{K}}_1=\left(2 \\ 5\right),{\mathrm{X}}_1=\left(2 \\ 5\right){\mathrm{e}}^{\mathrm{t}} \end{gathered}$$
For ${\lambda }_2=-3$, find its corresponding solution vector as:

And this yields two equations,
$-x+2y=0$ ………… (3)
$-\frac{5}{2}x+5y=0$ …………. (4)
These two equations are the same if we multiply equation (2) by$\frac{2}{5}$ as:
$$\begin{gathered} -x+2y=0 \\ 2y=x \end{gathered}$$
Let $y=1$
Then,$x=2$ .

Therefore, the second eigenvector is$$\begin{gathered} {\mathrm{K}}_2=\left(2 \\ 1\right) \end{gathered}$$ .

This gives an eigenvector and a corresponding solution vector$$\begin{gathered} {\mathrm{K}}_2=\left(2 \\ 1\right),{\mathrm{X}}_2=\left(2 \\ 1\right){\mathrm{e}}^{-3\mathrm{t}} \end{gathered}$$.

General solution of the given system

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

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