Question

Find the general solution of the given system

$$\begin{gathered} \frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{2}\mathit{x}\mathbf{+}\mathit{y}\mathbf{+}\mathbf{2}\mathit{z} \\ \frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{3}\mathit{x}\mathbf{+}\mathbf{6}\mathit{z} \\ \frac{\mathbf{d}\mathbf{z}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\mathbf{4}\mathit{x}\mathbf{-}\mathbf{3}\mathit{z} \end{gathered}$$

Short answer

The general solution of the given derivative system is

$X=c_1\left(\begin{array}{c}0\\ -2\\ 1\end{array}\right)e^{-3t}+c_2\left(\begin{array}{c}-2\cos 2t+\sin 2t\\ 3\sin 2t\\ 2\cos 2t\end{array}\right)e^t+c_3\left(\begin{array}{c}-\cos 2t-2\sin 2t\\ -3\cos 2t\\ 2\sin 2t\end{array}\right)e^t$

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,

$X=c_1{K}_1{e}^{{\lambda }_{1}t}+c_2{K}_2{e}^{{\lambda }_{2}t}+...+c_n{K}_n{e}^{{\lambda }_{n}t}$

Rewrite the given system in matrix form

$X\text{'}=\left(\begin{array}{ccc}2& 1& 2\\ 3& 0& 6\\ -4& 0& -3\end{array}\right)X$

Where

$A=\left(\begin{array}{ccc}2& 1& 2\\ 3& 0& 6\\ -4& 0& -3\end{array}\right)$

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

Hence

$$\begin{gathered} det\left|A-\lambda I\right|=0\to \left|\begin{array}{ccc}2-\lambda & 1& 2\\ 3& 0-\lambda & 6\\ -4& 0& -3-\lambda \end{array}\right|=0 \\ (2-\lambda )\left|\begin{array}{cc}-\lambda & 6\\ 0& -3-\lambda \end{array}\right|-1\left|\begin{array}{cc}3& 6\\ -4& -3-\lambda \end{array}\right|+2\left|\begin{array}{cc}3& -\lambda \\ -4& 0\end{array}\right|=0 \end{gathered}$$

$$\begin{gathered} -\lambda \left(2-\lambda \right)\left(-3-\lambda \right)-3\left(-3-\lambda \right)-24-8\lambda =0 \\ 6\lambda -3{\lambda }^2+2{\lambda }^2-{\lambda }^3+9+3\lambda -24-8\lambda =0 \\ -{\lambda }^3-{\lambda }^2+\lambda -15=0 \\ -\left(\lambda +3\right){\lambda }^2-2\lambda +5=0 \end{gathered}$$

For ${\lambda }^2-2\lambda +5=0$, we have

$$\begin{gathered} \lambda =\frac{2\pm \sqrt{4-4\left(1\right)\left(5\right)}}{2} \\ =\frac{2\pm 4i}{2} \\ =1\pm 2 \\ {\lambda }_1=-3,{\lambda }_2=1+2i,{\lambda }_3=\overline{{\lambda }_2}=1-2i \end{gathered}$$

Finding of the corresponding eigenvectors and the general solution

For ${\lambda }_1=-3$, we have

$$\begin{gathered} \left(A+3I|0\right)=\left(\begin{array}{ccc}2-(-3)& 1& 2\\ 3& 0-(-3)& 6\\ -4& 0& -3-(-3)\end{array}\right)\begin{array}{c}0\\ 0\\ 0\end{array} \\ =\left(\begin{array}{cccc}5& 1& 2& 0\\ 3& 3& 6& 0\\ -4& 0& 0& 0\end{array}\right) \end{gathered}$$

Apply row operation$R_2-3R_1\to R_2$

$=\left(\begin{array}{cccc}5& 1& 2& 0\\ -12& 0& 0& 0\\ -4& 0& 0& 0\end{array}\right)$

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

$5k_1+k_2+2k_3=0\to k_2=-2k_3$

Choosing K3=1 yields K2=-2. This gives an eigenvector,

$K_1=\left(\begin{array}{c}0\\ -2\\ 1\end{array}\right),X_1=\left(\begin{array}{c}0\\ -2\\ 1\end{array}\right)e^{-3t}$

For ${\lambda }_2=1+2i$, we have

$$\begin{gathered} (A-(1+2i\left)I0\right)=\begin{array}{cccc}2-(1+2i)& 1& 2& 0\\ 3& 0-(1+2i)& 6& 0\\ -4& 0& -3-(1+2i)& 0\end{array} \\ =\left(\begin{array}{cccc}1-2i& 1& 2& 0\\ 3& -(1+2i)& 6& 0\\ -4& 0& -4-2i& 0\end{array}\right) \end{gathered}$$

From the third row we have,

$-4k_1+\left(-4-2i\right)k_3=0\to k_1=\frac{-2-i}{2}{k}_3$

Choosing k3=2 yields k1=-2-i. We can substitute the these values into the equation we have from the first row,

$$\begin{gathered} k_1+k_2+k_3=0\to \left(\right)\left(-2-i\right)+k_2+2\left(2\right)=0 \\ -4+3i+k_2+4=0 \\ {k}_2=-3i \end{gathered}$$

This gives an eigenvector,

$k_2=\begin{array}{c}-2-i\\ -3i\\ 2\end{array}=\left(\begin{array}{c}-2\\ 0\\ 2\end{array}\right)+i\left(\begin{array}{c}-1\\ -3\\ 0\end{array}\right)$

And corresponding column vectors,

$B_1=\begin{array}{c}-2\\ 0\\ 2\end{array},B_2=\begin{array}{c}-1\\ -3\\ 0\end{array}$

Also,$\lambda =\alpha +\beta i\to \lambda =1+2i$

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

Therefore,

$$\begin{gathered} X_2=\left[B_1\cos \beta t-B_2\sin \beta t\right]e^{\alpha t} \\ =\left(\begin{array}{cc}\left(\begin{array}{c}-2\\ 0\\ 2\end{array}\right)\cos 2t-\left(\begin{array}{c}-1\\ -3\\ 0\end{array}\right)& \sin 2t\end{array}\right)e^t=\left(\begin{array}{c}-2\cos 2t+\sin 2t\\ 3\sin 2t\\ 2\cos 2t\end{array}\right)e^t \end{gathered}$$

$$\begin{gathered} X_3=\left[B_2\cos \beta t+B_1\sin \beta t\right]e^{\alpha t} \\ =\left(\begin{array}{cc}\left(\begin{array}{c}-1\\ -3\\ 0\end{array}\right)\cos 2t+\left(\begin{array}{c}-2\\ 0\\ 2\end{array}\right)& \sin 2t\end{array}\right)e^t=\left(\begin{array}{c}-\cos 2t-2\sin 2t\\ -3\cos 2t\\ 2\sin 2t\end{array}\right)e^t \end{gathered}$$

Thus the general solution is,

$$\begin{gathered} X=c_1{X}_1+c_2{X}_2+c_3{X}_3 \\ =c_1\left(\begin{array}{c}0\\ -2\\ 1\end{array}\right)e^{-3t}+c_2\left(\begin{array}{c}-2\cos 3t+\sin 3t\\ 3\sin 2t\\ 2\cos 2t\end{array}\right)e^t+c_3\left(\begin{array}{c}-\cos 2t-2\sin 2t\\ -3\cos 2t\\ 2\sin 2t\end{array}\right)e^t \end{gathered}$$