Question

Find the general solution of the given system

$$\begin{gathered} \frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{5}\mathit{x}\mathbf{+}\mathit{y} \\ \frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\mathbf{2}\mathit{x}\mathbf{+}\mathbf{3}\mathit{y} \end{gathered}$$

Short answer

The general solution of the given derivative system is$X=c_1\left[\left(\begin{array}{c}\cos t\\ -\cos t-\sin t\end{array}\right)e^{4t}\right]+c_2\left[\left(\begin{array}{c}\sin t\\ \cos t-\sin t\end{array}\right)e^{4t}\right]$

Step-by-step solution

Finding of the eigenvalues of matrix of coefficient from the characteristics equation

The theorem of General Solution-Homogeneous System states that

Let be ${\mathit{\lambda }}_{\mathbf{1}}\mathbf{,}{\mathit{\lambda }}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{\lambda }}_{\mathbf{n}}$distinct real eigenvalues of the coefficient matrix A of the homogeneous system and let ${\mathit{K}}_{\mathbf{1}}\mathbf{,}{\mathit{K}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{K}}_{\mathbf{n}}$be the corresponding eigenvectors. Then the general solution of the homogeneous system on the interval$\left(\infty ,-\infty \right)$is given by,

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

Rewrite the given system in matrix form

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

Where

$A=\left(\begin{array}{cc}5& 1\\ -2& 3\end{array}\right)$

We need to find the eigenvalues and eigenvectors of the matrix of the coefficient.

Hence

$$\begin{gathered} det\left|A-\lambda \right|=0\to \left|\begin{array}{cc}5-\lambda & 1\\ -2& 3-\lambda \end{array}\right|=0 \\ \left(5-\lambda \right)\left(3-\lambda \right)-\left(-2\right)=0 \\ 15-5\lambda -3\lambda +{\lambda }^2+2=0 \\ {\lambda }^2-8\lambda +17=0 \end{gathered}$$a

So our eigenvalues are . Notice that these are ${\lambda }_1=4+i,{\lambda }_2=4-i$complex conjugates, so obtaining an eigenvector for one of them would be sufficient.

Finding of the corresponding eigenvectors and the general solution

For ${\lambda }_1=4+i$, we have

$$\begin{gathered} \left(A-\left(4+i\right)I|0\right)=\left(\begin{array}{ccc}5-\left(4+i\right)& 1& 0\\ -2& 3-\left(4+i\right)& 0\end{array}\right) \\ =\left(\begin{array}{ccc}1-i& 1& 0\\ -2& -1-i& 0\end{array}\right) \end{gathered}$$

From the first row, we have

$\left(1-i\right)k_1+k_2=0\to \left(1-i\right)k_1=-k_2$

Choosing $k_1=1$yields role="math" localid="1664197255596" $k_2=i-1$. This gives an eigenvector,

$K_1=\left(\begin{array}{c}1\\ i-1\end{array}\right)=\left(\begin{array}{c}1\\ -1\end{array}\right)+i\left(\begin{array}{c}0\\ 1\end{array}\right)$

So,

$B_1=\left(\begin{array}{c}1\\ -1\end{array}\right),B_2=\left(\begin{array}{c}0\\ 1\end{array}\right)$

Also, ${\lambda }_1=\alpha +\beta i=4+i$.

Where $\alpha =4\beta =1$

Therefore,

$$\begin{gathered} X_1=\left[B_1\cos \beta t-B_2\sin \beta t\right]e^{\alpha t} \\ =\left(\begin{array}{c}0\\ 1\end{array}\cos t-\begin{array}{c}1\\ -1\end{array}\sin t\right)e^{4t}=\left(\begin{array}{c}\cos t\\ -\cos t-\sin t\end{array}\right)e^{4t} \end{gathered}$$

And

role="math" localid="1664199068626" $$\begin{gathered} X_2=\left[B_2\cos \beta t-B_1\sin \beta t\right]e^{\alpha t} \\ =\left(\begin{array}{c}0\\ 1\end{array}\cos t+\begin{array}{c}1\\ -1\end{array}\sin t\right)e^{4t}=\left(\begin{array}{c}\sin t\\ \cos t-\sin t\end{array}\right)e^{4t} \end{gathered}$$

Thus the general solution that, we could finally conclude is,

$$\begin{gathered} X=c_1{X}_1+c_2{X}_2 \\ {c}_1\left(\begin{array}{c}\cos t\\ -\cos t-\sin t\end{array}\right)e^{4t}+c_2\left(\begin{array}{c}\sin t\\ \cos t-\sin t\end{array}\right)e^{4t} \end{gathered}$$