Question

Find the general solution of the given system

$\mathit{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{cc}1& -8\\ 1& -3\end{array}\right)\mathit{X}$

Short answer

The general solution of the given derivative system is

$X=c_1\left[\left(\begin{array}{c}2\cos 2t-2\sin 2t\\ \cos 2t\end{array}\right)e^{-t}\right]+c_2\left[\left(\begin{array}{c}2\cos 2t+2\sin 2t\\ \sin 2t\end{array}\right)e^{-t}\right]$

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 ${\lambda }_1,{\lambda }_2,....,{\lambda }_n$be n distinct real eigenvalues of the coefficient matrix of A the homogeneous system and let $K_1,K_2,...,K_n$be the corresponding eigenvectors. Then the general solution of the homogeneous system 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 matrix form is,

$X\text{'}=\left(\begin{array}{cc}1& -8\\ 1& -3\end{array}\right)X$$X\text{'}=\left(\begin{array}{cc}1& -8\\ 1& -3\end{array}\right)X$

Where

$A=\left(\begin{array}{cc}1& -8\\ 1& -3\end{array}\right)$

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

Hence

$det\left|A-\lambda I\right|=0\to \left|\begin{array}{cc}1-\lambda & -8\\ 1& -3-\lambda \end{array}\right|$

$$\begin{gathered} \left(1-\lambda \right)\left(-3-\lambda \right)-\left(-8\right)=0 \\ -3-\lambda +3\lambda +{\lambda }^2+8=0 \\ {\lambda }^2+2\lambda +5=0 \end{gathered}$$

Where

$$\begin{gathered} \lambda =\frac{-2\pm \sqrt{4-4\left(5\right)}}{2} \\ =\frac{-2\pm \sqrt{-16}}{2} \\ =\frac{-2\pm 4i}{2} \\ =-1\pm 2i \end{gathered}$$

So our eigenvalues are ${\lambda }_1=-1+2i,{\lambda }_2=\overline{{\lambda }_1}=-1-2i$.

Finding of the corresponding eigenvectors and the general solution

For ${\lambda }_1=-1+2i$, we have

$$\begin{gathered} (A+(-1+2i\left)I\right|0)=\left(\begin{array}{ccc}1-(-1+2i)& -8& 0\\ 1& -3-(-1+2i)& 0\end{array}\right) \\ =\left(\begin{array}{ccc}2-2i& -8& 0\\ 1& -2-2i& 0\end{array}\right) \end{gathered}$$

Apply row operation,$\frac{1}{2-2i}{R}_1\to R_1$

$=\left(\begin{array}{ccc}1& -8/(2-2i)& 0\\ 1& -2-2i& 0\end{array}\right)$

Multiply -8/(2-2i) by the conjugate, which will yield -2-2i=-(2+2i),

$=\left(\begin{array}{ccc}1& -2-2i& 0\\ 1& -2-2i& 0\end{array}\right)$

Apply row operation$R_2-R_1\to R_2$

=$=\left(\begin{array}{ccc}1& -(2+2i)& 0\\ 0& 0& 0\end{array}\right)$

So now we have,

$k_1-\left(2+2i\right)k_2=0\to k_1=\left(2+2i\right)k_2$

Choosing $k_2=1$yields $k_1=2+2i$. This gives an eigenvector,

$K_1=\left(\begin{array}{c}2+2i\\ 1\end{array}\right)=\left(\begin{array}{c}2\\ 1\end{array}\right)+i\left(\begin{array}{c}2\\ 0\end{array}\right)$

So,

$B_1=\left(\begin{array}{c}2\\ 1\end{array}\right),B_2=\left(\begin{array}{c}2\\ 0\end{array}\right)$

Also,${\lambda }_1=\alpha +\beta i=-1+2i$ .

Where$\alpha =-1,\beta =2$

Therefore,

$$\begin{gathered} X_1=\left[B_1\cos \beta t-B_2\sin \beta t\right]e^{\alpha t} \\ =\left(\begin{array}{c}2\\ 1\end{array}\cos 2t-\begin{array}{c}2\\ 0\end{array}\sin 2t\right)e^{-t}=\left(\begin{array}{c}2\cos 2t-2\sin 2t\\ \cos 2t\end{array}\right)e^{-t} \end{gathered}$$

And

$$\begin{gathered} X_2=\left[B_2\cos \beta t+B_1\sin \beta t\right]e^{\alpha t} \\ =\left(\begin{array}{c}2\\ 0\end{array}\cos 2t+\begin{array}{c}2\\ 1\end{array}\sin 2t\right)e^{-t}=\left(\begin{array}{c}2\cos 2t+2\sin 2t\\ \sin 2t\end{array}\right)e^{-t} \end{gathered}$$

$$\begin{gathered} X=c_1{X}_1+c_2{X}_2 \\ =c_1\left(\begin{array}{c}2\cos 2t-2\sin 2t\\ \cos 2t\end{array}\right)+c_2\left(\begin{array}{c}2\cos 2t+2\sin 2t\\ \sin 2t\end{array}\right) \end{gathered}$$Thus the general solution that, we could finally conclude is,