Question

Find the general solution of the given system

$\mathit{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{ccc}4& 0& 1\\ 0& 6& 0\\ -4& 0& 4\end{array}\right)\mathit{X}$

Short answer

The general solution of the given derivative system is

$X=c_1\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)e^{6t}+c_2\left(\begin{array}{c}\sin 2t\\ 0\\ 2\cos 2t\end{array}\right)e^{4t}+c_3\left(\begin{array}{c}-\cos 2t\\ 0\\ 2\sin 2t\end{array}\right)e^{4t}$

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,

$X\text{'}=\left(\begin{array}{ccc}4& 0& 1\\ 0& 6& 0\\ -4& 0& 4\end{array}\right)X$

Where

$A=\left(\begin{array}{ccc}4& 0& 1\\ 0& 6& 0\\ -4& 0& 4\end{array}\right)$

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

Hence

$$\begin{gathered} det\left|A-\lambda I\right|=0\to \left|\begin{array}{ccc}4-\lambda & 0& 1\\ 0& 6-\lambda & 0\\ -4& 0& 4-\lambda \end{array}\right|=0 \\ (4-\lambda )\left|\begin{array}{cc}6-\lambda & 0\\ 0& 4-\lambda \end{array}\right|+1\left|\begin{array}{cc}0& 6-\lambda \\ -4& 0\end{array}\right|=0 \end{gathered}$$

$$\begin{gathered} {\left(4-\lambda \right)}^2\left(6-\lambda \right)+4\left(6-\lambda \right)=0 \\ \left(6-\lambda \right){\left(4-\lambda \right)}^2+4=0 \end{gathered}$$

role="math" localid="1668052303978" $$\begin{gathered} \left(6-\lambda \right)16-8\lambda +{\lambda }^2+4=0 \\ \left(6-\lambda \right){\lambda }^2-8\lambda +20=0 \end{gathered}$$

For ${\lambda }^2-8\lambda +20=0$, we have

$$\begin{gathered} \lambda =\frac{8\pm \sqrt{64-4\left(1\right)\left(20\right)}}{2} \\ =\frac{8\pm 4i}{2} \\ =4\pm 2i \end{gathered}$$

So our eigenvalues are ${\lambda }_1=6,{\lambda }_2=4+2i,{\lambda }_3=\overline{{\lambda }_2}=4-2i$.

Finding of the corresponding eigenvectors and the general solution

For ${\lambda }_1=6$, we have

$$\begin{gathered} (A-6I|0)=\left(\begin{array}{ccc}4-6& 0& 1\\ 0& 6-6& 0\\ -4& 0& 4-6\end{array}\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ =\left(\begin{array}{cccc}-2& 0& 1& 0\\ 0& 0& 0& 0\\ -4& 0& -2& 0\end{array}\right) \end{gathered}$$

Apply row operation,$R_3-2R_1\to R_3:$

$=\left(\begin{array}{cccc}-2& 0& 1& 0\\ 0& 0& 0& 0\\ 0& 0& -4& 0\end{array}\right)$

Apply row operation,$2R_1\to R_{1}and\frac{1}{2}{R}_3\to R_3:$

$=\left(\begin{array}{cccc}-4& 0& 2& 0\\ 0& 0& 0& 0\\ 0& 0& -2& 0\end{array}\right)$

Apply row operation,$R_1+R_3\to R_1:$

$=\left(\begin{array}{cccc}-4& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& -2& 0\end{array}\right)$

Apply row operation,$-\frac{1}{4}{R}_1\to R_{1}and-\frac{1}{2}{R}_3\to R_3:$

$=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 1& 0\end{array}\right)$

We see that k₁ = 0 and k₃=1, as for k₂, we will choose an arbitrary value, say k₂=1. This gives an eigenvector and a corresponding solution vector,

$K_1=\begin{array}{c}0\\ 1\\ 0\end{array},X_1=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)e^{6t}$

For ${\lambda }_2=4+2i$

$$\begin{gathered} (A-(4+2i\left)\right|0)=\left(\begin{array}{ccc}4-(4+2i)& 0& 1\\ 0& 6-(4+2i)& 0\\ -4& 0& 4-(4+2i)\end{array}\begin{array}{c}0\\ 0\\ 0\end{array}\right) \\ =\left(\begin{array}{cccc}-2i& 0& 1& 0\\ 0& 2-2i& 0& 0\\ -4& 0& -2i& 0\end{array}\right) \end{gathered}$$

Apply row operation,$-\frac{1}{2}i{R}_3+R_1\to R_3:$

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

From the second row, we have k₂=0, and from the first row we have,

$-2i{k}_1+k_3=0\to k_1=-\frac{1}{2}i{k}_3$

Choosing k₃=2 yields k₁=-i.

This gives an eigenvector,

$K_2=\begin{array}{c}-i\\ 0\\ 2\end{array}=\begin{array}{c}0\\ 0\\ 2\end{array}+i\begin{array}{c}-1\\ 0\\ 0\end{array}$

And the corresponding column vector,

$B_1=\begin{array}{c}0\\ 0\\ 2\end{array},B_2=\begin{array}{c}-1\\ 0\\ 0\end{array}$

Also,${\lambda }_1=\alpha +\beta i\to \lambda =4+2i$ .

Where$\alpha =4,\beta =2$

Therefore,

$$\begin{gathered} X_2=\left[B_1\cos \beta t-B_2\sin \beta t\right]e^{\alpha t} \\ =\left[\left(\begin{array}{c}0\\ 0\\ 2\end{array}\right)\cos 2t-\left(\begin{array}{c}-1\\ 0\\ 0\end{array}\right)\sin 2t\right]e^{4t}=\left(\begin{array}{c}\sin 2t\\ 0\\ 2\cos 2t\end{array}\right)e^{4t} \end{gathered}$$

And

$$\begin{gathered} X_3=\left[B_2\cos \beta t+B_1\sin \beta t\right]e^{\alpha t} \\ =\left[\left(\begin{array}{c}-1\\ 0\\ 0\end{array}\right)\cos 2t+\left(\begin{array}{c}0\\ 0\\ 2\end{array}\right)\sin 2t\right]e^{4t}=\left(\begin{array}{c}-\cos 2t\\ 0\\ 2\sin 2t\end{array}\right)e^{4t} \end{gathered}$$

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

$X=c_1\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)e^{6t}+c_2\left(\begin{array}{c}\sin 2t\\ 0\\ 2\cos 2t\end{array}\right)e^{4t}+c_3\left(\begin{array}{c}-\cos 2t\\ 0\\ 2\sin 2t\end{array}\right)e^{4t}$