Question

In Problems 47and 48 find the given Initial – value problem.

$X\text{'}=\left(\begin{array}{ccc}1& -12& -14\\ 1& 2& -3\\ 1& 1& -2\end{array}\right)X$

Short answer

The initial value for $X\text{'}=\left(\begin{array}{ccc}1& -12& -14\\ 1& 2& -3\\ 1& 1& -2\end{array}\right)X$is $X=\left(\begin{array}{c}-25\\ 7\\ -6\end{array}\right)e^t-\left(\begin{array}{c}\cos 5t-5\sin 5t\\ \cos 5t\\ \cos 5t\end{array}\right)+6\left(\begin{array}{c}5\cos 5t+\sin 5t\\ \sin 5t\\ \sin 5t\end{array}\right)$

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.

$p\left(\lambda \right)=det(\lambda I-A)$

Its root is the eigenvalue of A.

Determine the eigen values

Given

$X\text{'}=\left(\begin{array}{ccc}1& -12& -14\\ 1& 2& -3\\ 1& 1& -2\end{array}\right)X$

Where

$A=\left(\begin{array}{ccc}1& -12& -14\\ 1& 2& -3\\ 1& 1& -2\end{array}\right)$

Obtaining the eigenvalues of the coefficient matrix,

$det(A-\lambda I)=0$

$$\begin{gathered} \left|\begin{array}{ccc}1-\lambda & -12& -14\\ 1& 2-\lambda & -3\\ 1& 1& -2-\lambda \end{array}\right|=0 \\ (1-\lambda )\left|\begin{array}{cc}2-\lambda & -3\\ 1& -2-\lambda \end{array}\right|+12\left|\begin{array}{cc}1& -3\\ 1& -2-\lambda \end{array}\right|-14\left|\begin{array}{cc}1& 2-\lambda \\ 1& 1\end{array}\right|=0 \end{gathered}$$

$$\begin{gathered} (1-\lambda )\left[\right(2-\lambda \left)\right(-2-\lambda )+3]+12\left[\right(-2-\lambda )+3]-14[1-(2-\lambda \left)\right]=0 \\ (1-\lambda )\left[-4-2\lambda +2\lambda +{\lambda }^2+3\right]+12(1-\lambda )+14(1-\lambda )=0 \\ (1-\lambda )\left({\lambda }^2-1\right)+12(1-\lambda )+14(1-\lambda )=0 \\ (1-\lambda )\left[\left({\lambda }^2-1\right)+12+14\right]=0 \\ (1-\lambda )\left[{\lambda }^2+25\right]=0 \end{gathered}$$

so our eigenvalues are ${\lambda }_1=1,{\lambda }_2=5i$and${\lambda }_3=\overline{{\lambda }_2}=-5i$

Determine the eigenvector and the corresponding vector

For${\lambda }_1=1:$

$$\begin{gathered} (A-I|0)=\left(\begin{array}{cccc}1-1& -12& -14& 0\\ 1& 2-1& -3& 0\\ 1& 1& -2-1& 0\end{array}\right) \\ =\left(\begin{array}{cccc}0& -12& -14& 0\\ 1& 1& -3& 0\\ 1& 1& -3& 0\end{array}\right) \end{gathered}$$

Apply row operation$R_3-R_2\to R_3$:

$=\left(\begin{array}{cccc}0& -12& -14& 0\\ 1& 1& -3& 0\\ 0& 0& 0& 0\end{array}\right)$

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

$=\left(\begin{array}{cccc}0& 6& 7& 0\\ 1& 1& -3& 0\\ 0& 0& 0& 0\end{array}\right)$

from the first row we have

$6k_2+7k_3=0$

$k_2=-\frac{7}{6}{k}_3$

choosing k₃=-6 yields k₂=7.

To solve for c, we substitute the values we choose for k₂ and k₃ into the equation we obtain from the second row,

$$\begin{gathered} k_1+k_2-3k_3=0 \\ {k}_1+7-3(-6)=0 \\ {k}_1=-25 \end{gathered}$$

This gives an eigenvector and a corresponding solution vector

$K_1=\left(\begin{array}{c}-25\\ 7\\ -6\end{array}\right),X_1=\left(\begin{array}{c}-25\\ 7\\ -6\end{array}\right)e^t$

For${\lambda }_2=5i$:

$(A-(5i\left)I\right|0)=\left(\begin{array}{ccc}1-5i& -12& -14\\ 1& 2-5i& -3\\ 1& 1& -2-5i\end{array}\begin{array}{c}0\\ 0\\ 0\end{array}\right)$

Apply row operation$R_3-R_2\to R_3$:

$=\left(\begin{array}{ccc}1-5i& -12& -14\\ 1& 2-5i& -3\\ 1& -1+5i& 1-5i\end{array}\begin{array}{c}0\\ 0\\ 0\end{array}\right)$

Apply row operation$\frac{1}{-1+5i}{R}_3\to R_3$:

$=\left(\begin{array}{ccc}1-5i& -12& -14\\ 1& 2-5i& -3\\ 0& 1& -1\end{array}\begin{array}{c}0\\ 0\\ 0\end{array}\right)$

from the third row we have

$$\begin{gathered} k_2-k_3=0 \\ {k}_2=k_3 \end{gathered}$$

from the first row we have

$$\begin{gathered} (1-5i)k_1-12k_2-14k_3=0 \\ (1-5i)k_1-12k_3-14k_3=0 \\ (1-5i)k_1-26k_3 \\ =0(1-5i)k_1 \\ =26k_3 \\ {k}_1=(1+5i)k_3 \end{gathered}$$

choosing k₃=1 yields k₂=1 and k₁=1+5i.

This gives an eigenvector

$$\begin{gathered} K_2=\left(\begin{array}{c}1+5i\\ 1\\ 1\end{array}\right) \\ =\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)+i\left(\begin{array}{c}5\\ 0\\ 0\end{array}\right) \end{gathered}$$

and column vectors

$B_1=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)and{B}_2=\left(\begin{array}{c}5\\ 0\\ 0\end{array}\right)$

Determine the general solution of the system

Also,

$$\begin{gathered} \lambda =\alpha +\beta i \\ \lambda =0+5i \end{gathered}$$

where$\alpha =0and\beta =5$

Therefore,

$$\begin{gathered} X_{\mathbf{2}}=\left[B_{\mathbf{1}}\cos \beta t-B_{\mathbf{2}}\sin \beta t\right]e^{\alpha t} \\ =\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)\cos 5t-\left(\begin{array}{c}5\\ 0\\ 0\end{array}\right)\sin 5t \\ =\left(\begin{array}{c}\cos 5t-5\sin 5t\\ \cos 5t\\ \cos 5t\end{array}\right) \end{gathered}$$

and,

$$\begin{gathered} X_3=\left[B_2\cos \beta t+B_1\sin \beta t\right]e^{\alpha t} \\ =\left(\begin{array}{c}5\\ 0\\ 0\end{array}\right)\cos 5t+\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)\sin 5t \\ =\left(\begin{array}{c}5\cos 5t+5\sin 5t\\ \sin 5t\\ \sin 5t\end{array}\right) \end{gathered}$$

The general solution of the system is$$\begin{gathered} X=c_1{X}_{\mathbf{1}}+c_2{X}_{\mathbf{2}}+c_3{X}_3 \\ X=c_1\left(\begin{array}{c}-25\\ 7\\ -6\end{array}\right)e^t+c_2\left(\begin{array}{c}\cos 5t-5\sin 5t\\ cos5t\\ \cos 5t\end{array}\right)+c_3\left(\begin{array}{c}5\cos 5t+5\sin 5t\\ \sin 5t\\ \sin 5t\end{array}\right) \end{gathered}$$

Determine the initial value condition

Applying the initial value condition

$$\begin{gathered} X\left(0\right)=\left(\begin{array}{c}4\\ 6\\ -7\end{array}\right) \\ \left(\begin{array}{c}4\\ 6\\ -7\end{array}\right)=c_1\left(\begin{array}{c}-25\\ 7\\ -6\end{array}\right)+c_2\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)+c_3\left(\begin{array}{c}5\\ 0\\ 0\end{array}\right) \end{gathered}$$

here we have

$$\begin{gathered} -25c_1+c_2+5c_3=4......\left(1\right) \\ 7c_1+c_2=6..........\left(2\right) \\ -6c_1+c_2=-7.....\left(3\right) \end{gathered}$$

substract equation (3) from (2)

$$\begin{gathered} 13c_1=13 \\ {c}_1=1 \end{gathered}$$

To solve for c₂ we substitute this value we obtained for c₁ into equation (2) or (3),

$$\begin{gathered} 7\left(1\right)+c_2=6 \\ {c}_2=-1 \end{gathered}$$

And finally to solve for c₃ we substitute the values we obtained for c₁ and c₂ into (1),

$$\begin{gathered} -25\left(1\right)-1+5c_3=4 \\ {c}_3=6 \end{gathered}$$

therefore, the solution satisfying the given initial condition is

$X=\left(\begin{array}{c}-25\\ 7\\ -6\end{array}\right)e^t-\left(\begin{array}{c}\cos 5t-5\sin 5t\\ \cos 5t\\ \cos 5t\end{array}\right)+6\left(\begin{array}{c}5\cos 5t+\sin 5t\\ \sin 5t\\ \sin 5t\end{array}\right)$

Therefore, the initial value for $X\text{'}=\left(\begin{array}{ccc}1& -12& -14\\ 1& 2& -3\\ 1& 1& -2\end{array}\right)X$is

$X=\left(\begin{array}{c}-25\\ 7\\ -6\end{array}\right)e^t-\left(\begin{array}{c}\cos 5t-5\sin 5t\\ \cos 5t\\ \cos 5t\end{array}\right)+6\left(\begin{array}{c}5\cos 5t+\sin 5t\\ \sin 5t\\ \sin 5t\end{array}\right)$.