Question

Find the general solution of the given system

$\mathit{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{ccc}1& -1& 2\\ -1& 1& 0\\ -1& 0& 1\end{array}\right)\mathit{X}$

Short answer

The general solution of the given derivative system is

$X=c_1\left(\begin{array}{c}0\\ 2\\ 1\end{array}\right)e^t+c_2\left(\begin{array}{c}\cos t\\ -\sin t\\ -\sin t\end{array}\right)e^t+c_3\left(\begin{array}{c}\sin t\\ \cos t\\ \cos t\end{array}\right)e^t$

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,

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

Where

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

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

Hence

$$\begin{gathered} det\left|A-\lambda I\right|=0\to \left|\begin{array}{ccc}1-\lambda & -1& 2\\ -1& 1-\lambda & 0\\ -1& 0& 1-\lambda \end{array}\right|=0 \\ (1-\lambda )\left|\begin{array}{cc}1-\lambda & 0\\ 0& 1-\lambda \end{array}\right|+1\left|\begin{array}{cc}-1& 0\\ -1& 1-\lambda \end{array}\right|+2\left|\begin{array}{cc}1& 1-\lambda \\ -1& 0\end{array}\right| \end{gathered}$$

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

For ${\lambda }^2-2\lambda +2=0$, we have

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

So our eigenvalues are ${\lambda }_1=1,{\lambda }_2=1+i,{\lambda }_3=\overline{{\lambda }_2}=1-i$.

Finding of the corresponding eigenvectors and the general solution

For ${\lambda }_1=1$, we have

$$\begin{gathered} (A-I|0)=\left(\begin{array}{cccc}1-1& -1& 2& 0\\ -1& 1-1& 0|& 0\\ -1& 0& 1-1& 0\end{array}\right) \\ =\left(\begin{array}{cccc}0& -1& 2& 0\\ -1& 0& 0& 0\\ -1& 0& 0& 0\end{array}\right) \end{gathered}$$

From the second and third row, we have K1=0, and from the first row we have

$-k_2+2k_3=0\to k_2=3k_3$

Choosing K3=1 yields K2=2. This gives an eigenvector and corresponding solution vector,

$K_1=\begin{array}{c}0\\ 2\\ 1\end{array},X_1=\left(\begin{array}{c}0\\ 2\\ 1\end{array}\right)e^t$

For${\lambda }_2=1+i$

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

From the third row, we have,

$-k_1-i{k}_3=0\to k_1=-i{k}_3$

Choosing k3 = i yields k1 = 1. We can solve for k2 by substituting these values into the equation we obtain from the first row,

$$\begin{gathered} -i{k}_1-k_2+2k_3=0\to -i-k_2+2i=0 \\ \Rightarrow i-k_2=0 \\ \Rightarrow k_2=i \end{gathered}$$

This gives an eigenvector,

$K_2=\begin{array}{c}1\\ i\\ i\end{array}=\begin{array}{c}1\\ 0+i\\ 0\end{array}\begin{array}{c}0\\ 1\\ 1\end{array}$

And the corresponding column vector,

$B_1=\begin{array}{c}1\\ 0\\ 0,\end{array}{B}_2=\begin{array}{c}0\\ 1\\ 1\end{array}$

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

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

Therefore,

localid="1668012267698" $$\begin{gathered} X_2=\left[B_1\cos \beta t-B_2\sin \beta t\right]e^{\alpha t} \\ =\left[\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\cos t-\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right)\sin t\right]e^t=\left(\begin{array}{c}\cos t\\ -\sin t\\ -\sin t\end{array}\right)e^t \end{gathered}$$

And

$$\begin{gathered} X_3=\left[B_2\cos \beta t+B_1\sin \beta t\right]e^{\alpha t} \\ =\left[\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right)\cos t+\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\sin t\right]e^t=\left(\begin{array}{c}\sin t\\ \cos t\\ \cos t\end{array}\right)e^t \end{gathered}$$

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

localid="1668011055069" $$\begin{gathered} X=c_1{X}_1+c_2{X}_2 \\ X=c_1\left(\begin{array}{c}0\\ 2\\ 1\end{array}\right)e^t+c_2\left(\begin{array}{c}\cos t\\ -\sin t\\ -\sin t\end{array}\right)e^t+c_3\left(\begin{array}{c}\sin t\\ \cos t\\ \cos t\end{array}\right)e^t \end{gathered}$$