Question

In problems 1-12 find the general solution of the given system.

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\frac{\mathbf{5}}{\mathbf{2}}\mathit{x}\mathbf{+}\mathbf{2}\mathit{y} \\ \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{4}}\mathit{x}\mathbf{-}\mathbf{2}\mathit{y} \end{gathered}$$

Short answer

The general solution of the given system is $X=c_1\left(\begin{array}{c}4\\ 3\end{array}\right)e^{-t}+c_2\left(\begin{array}{c}2\\ 1\end{array}\right)e^{-\frac{7}{2}t}$.

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 $\left(-\infty ,\infty \right)$is given by

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

We are given,

$$\begin{gathered} \left\{\frac{dx}{dt}=-\frac{5}{2}x+2y \\ \frac{dy}{dt}=\frac{3}{4}x-2y\right. \end{gathered}$$

Which can be written in the matrix form

localid="1663954771185" $X\text{'}=\left(\begin{array}{cc}-5/2& 2\\ 3/4& -2\end{array}\right)X$

Where,

$A=\left(\begin{array}{cc}-5/2& 2\\ 3/4& -2\end{array}\right)$

Eigen values of the coefficient matrix 

Obtaining the eigen values of the coefficient matrix,

$det\left(\mathrm{A}-\lambda \mathrm{I}\right)=0\to \left|\begin{array}{cc}-5/2-\lambda & 2\\ 3/4& -2-\lambda \end{array}\right|=0$

$$\begin{gathered} \left(-\frac{5}{2}-\lambda \right)\left(-2-\lambda \right)-\frac{3}{2}=0 \\ 5+\frac{5}{2}\lambda +2\lambda +{\lambda }^2-\frac{3}{2}=0 \\ {\lambda }^2+\frac{9}{2}\lambda +\frac{7}{2}=0 \\ \left(\lambda +1\right)\left(\lambda +\frac{7}{2}\right)=0 \end{gathered}$$

So our eigen values are ${\lambda }_1=-1$and${\lambda }_2=-\frac{7}{2}$

Eigen vector and its corresponding solution vector

For ${\lambda }_1=-1$:

localid="1663959391822" $$\begin{gathered} \left(\mathrm{A}+\mathrm{I}\left|0\right.\right)=\left(-5/2-(-1) \\ 3/4\right. \left.2 \\ -2-(-1)\right|\left. 0 \\ 0\right) \\ =\left(-3/2 \\ 3/4\right. \left.2 \\ -1\right|\left. 0 \\ 0\right) \end{gathered}$$

Apply row operation

localid="1663958143680" $$\begin{gathered} 2R_2+\frac{1}{2}{R}_1\to R_2:\Rightarrow \left(-3/2 \\ 3/4\right. \left.2 \\ -1\right|\left. 0 \\ 0\right) \end{gathered}$$

So we have $-\frac{3}{2}{k}_1+2k_2=0\to k_1=\frac{4}{3}{k}_2$

Choosing$k_2=3$yields$k_1=4$.

This gives an eigen vector and a corresponding solution vector

$K_1=\left(\begin{array}{c}4\\ 3\end{array}\right),X_1=\left(\begin{array}{c}4\\ 3\end{array}\right)e^{-t}$ ,

For ${\lambda }_2=-\frac{7}{2}$:

localid="1663958969990" $$\begin{gathered} \left(\mathrm{A}+\frac{7}{2}\mathrm{I}\left|0\right.\right)=\left(-5/2-(-7/2) \\ 3/4\right. \left.2 \\ -2-(-7/2)\right|\left. 0 \\ 0\right) \\ =\left(1 \\ 3/4\right. \left.2 \\ 3/2\right|\left. 0 \\ 0\right) \end{gathered}$$

Apply row operation

localid="1663959115448" $$\begin{gathered} R_2+\frac{3}{4}{R}_1\to R_2:\Rightarrow \left(1 \\ 0\right. \left.2 \\ 0\right|\left. 0 \\ 0\right) \end{gathered}$$

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

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

This gives an eigen vector and a corresponding solution vector

localid="1663959466583" $X_2=\left(\begin{array}{c}-2\\ 1\end{array}\right),X_2=\left(\begin{array}{c}-2\\ 1\end{array}\right)e^{-\frac{7}{2}t}$ ,

General solution of given system

Finally, the general solution is given by

$$\begin{gathered} X=c_1{X}_1+c_2{X}_2 \\ X=c_1\left(\begin{array}{c}4\\ 3\end{array}\right)e^{-t}+c_2\left(\begin{array}{c}-2\\ 1\end{array}\right)e^{-\frac{7}{2}t} \end{gathered}$$

Therefore, the general solution of the given system is $X=c_1\left(\begin{array}{c}4\\ 3\end{array}\right)e^{-t}+c_2\left(\begin{array}{c}-2\\ 1\end{array}\right)e^{-\frac{7}{2}t}$.