Question

In Problemsfind the general solution of the given system.

$\mathit{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{ccc}2& 5& 1\\ -5& -6& 4\\ 0& 0& 2\end{array}\right)\mathit{X}$

Short answer

The general solution of the given system for $X\text{'}=\left(\begin{array}{ccc}2& 5& 1\\ -5& -6& 4\\ 0& 0& 2\end{array}\right)X$is role="math" localid="1668055410636" $X=c_1\left(\begin{array}{c}28\\ -5\\ 25\end{array}\right)e^{2t}+c_2\left(\begin{array}{c}4\cos 3t+3\sin 3t\\ -5\cos 3t\\ 0\end{array}\right)e^{-2t}+c_3\left(\begin{array}{c}4\sin 3t-3\cos 3t\\ -5\sin 3t\\ 0\end{array}\right)e^{-2t}$

Step-by-step solution

Determine the complex eigen values

The polynomial of degree n of the variable $\lambda$is the characteristic polynomial of the n x n matrix A.

role="math" localid="1668055561186" $p\left(\lambda \right)=det(\lambda I-A)$

Its root is the eigenvalue of A .

Determine the eigen values

Given

$X\text{'}=\left(\begin{array}{ccc}2& 5& 1\\ -5& -6& 4\\ 0& 0& 2\end{array}\right)X$

Where

$A=\left(\begin{array}{ccc}2& 5& 1\\ -5& -6& 4\\ 0& 0& 2\end{array}\right)$

Obtaining the eigenvalues,

$$\begin{gathered} det(A-\lambda I)=0\to \left(\begin{array}{ccc}2-\lambda & 5& 1\\ -5& -6-\lambda & 4\\ 0& 0& 2-\lambda \end{array}\right)=0 \\ (2-\lambda )\left|\begin{array}{cc}-6-\lambda & 4\\ 0& 2-\lambda \end{array}\right|-5\left|\begin{array}{cc}-5& 4\\ 0& 2-\lambda \end{array}\right|+1\left|\begin{array}{cc}-5& -6-\lambda \\ 0& 0\end{array}\right| \end{gathered}$$

$(2-\lambda )(-6-\lambda )(2-\lambda )+25(2-\lambda )=0$

$(2-\lambda )\left[\right(-6-\lambda \left)\right(2-\lambda )+25]=0$

$(2-\lambda )\left[-12+6\lambda -2\lambda +{\lambda }^2+25\right]=0$

$(2-\lambda )\left[{\lambda }^2+4\lambda +13\right]=0$

For ${\lambda }^2+4\lambda +13=0$, we have

$\lambda =\frac{-4\pm \sqrt{4^2-4\left(1\right)\left(13\right)}}{2\left(1\right)}$

$$\begin{gathered} =\frac{-4\pm \sqrt{-36}}{2} \\ =\frac{-4\pm 6i}{2} \\ =-2\pm 3i \end{gathered}$$

So our eigenvalues are ${\lambda }_1=2,{\lambda }_2=-2+3i$and ${\lambda }_3=\overline{{\lambda }_2}=-2-3i$

Determine the eigenvector and corresponding vector

For ${\lambda }_{-}1=2:$

$$\begin{gathered} (A-2I|0)=0\to \left(\begin{array}{ccc}2-2& 5& 1\\ -5& -6-2& 4\\ 0& 0& 2-2\end{array}\begin{array}{c}0\\ 0\\ 0\end{array}\right)=0 \\ =\left(\begin{array}{ccc}0& 5& 1\\ -5& -8& 4\\ 0& 0& 0\end{array}\begin{array}{c}0\\ 0\\ 0\end{array}\right) \end{gathered}$$

from the first row we have

$5k_2+k_3=0$

$k_2=-\frac{1}{5}{k}_3$

choosing k₃=25 yields k₂=-5b, to solve for k₁ we substitute these into the equation we get from the second row,

$-5k_1-8k_2+4k_3=0$

$k_1=\frac{4}{5}{k}_3-\frac{8}{5}{k}_2$

$$\begin{gathered} =\frac{4}{5}\left(25\right)-\frac{8}{5}(-5) \\ =28 \end{gathered}$$

This gives an eigenvector and a corresponding solution vector

$K_1=\left(\begin{array}{c}28\\ -5\\ 25\end{array}\right),X_1=\left(\begin{array}{c}28\\ -5\\ 25\end{array}\right)e^{2t}$

For ${\lambda }_{-}2=-2+3i$:

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

From the third row we can see that k₃=0, as for the first row we have

$(4-3i)k_1+5k_2+k_3=0$

$k_2=-\frac{4-3i}{5}{k}_1$

choosing k₁ =4-3i yields k₂=-5.

This gives an eigenvector

$$\begin{gathered} K_2=\left(\begin{array}{c}4-3i\\ -5\\ 0\end{array}\right) \\ =\left(\begin{array}{c}4\\ -5\\ 0\end{array}\right)+i\left(\begin{array}{c}-3\\ 0\\ 0\end{array}\right) \end{gathered}$$

and corresponding column vectors

$B_1=\left(\begin{array}{c}4\\ -5\\ 0\end{array}\right)and{B}_2=\left(\begin{array}{c}-3\\ 0\\ 0\end{array}\right)$

Determine the general solution of the given system

Also,

$\lambda =\alpha +\beta i$

$\lambda =-2+3i$

where $\alpha =-2and\beta =3$.

Therefore,

$$\begin{gathered} X_{\mathbf{2}}=\left[B_1\cos \beta t-B_{\mathbf{2}}\sin \beta t\right]e^{\alpha t} \\ =\left[\left(\begin{array}{c}4\\ -5\\ 0\end{array}\right)\cos 3t-\left(\begin{array}{c}-3\\ 0\\ 0\end{array}\right)\sin 3t\right]e^{-2t} \\ =\left(\begin{array}{c}4\cos 3t+3\sin 3t\\ -5\cos 3t\\ 0\end{array}\right)e^{-2t} \end{gathered}$$

And

$$\begin{gathered} X_3=\left[B_2\cos \beta t+B_1\sin \beta t\right]e^{\alpha t} \\ =\left[\left(\begin{array}{c}-3\\ 0\\ 0\end{array}\right)\cos 3t+\left(\begin{array}{c}4\\ -5\\ 0\end{array}\right)\sin 3t\right]e^{-2t} \\ =\left(\begin{array}{c}4\sin 3t-3\cos 3t\\ -5\sin 3t\\ 0\end{array}\right)e^{-2t} \end{gathered}$$

Therefore, The general solution of the given system for $X\text{'}=\left(\begin{array}{ccc}2& 5& 1\\ -5& -6& 4\\ 0& 0& 2\end{array}\right)X$is

$X=c_1\left(\begin{array}{c}28\\ -5\\ 25\end{array}\right)e^{2t}+c_2\left(\begin{array}{c}4\cos 3t+3\sin 3t\\ -5\cos 3t\\ 0\end{array}\right)e^{-2t}+c_3\left(\begin{array}{c}4\sin 3t-3\cos 3t\\ -5\sin 3t\\ 0\end{array}\right)e^{-2t}$