Question

In Problems 47and 48solve the given initial-value problem.

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

Short answer

The initial value for $X\text{'}=\left(\begin{array}{cc}6& -1\\ 5& 4\end{array}\right)X$is $X=-2\left(\begin{array}{c}\cos 2t\\ \cos 2t+2\sin 2t\end{array}\right)e^{5t}-5\left(\begin{array}{c}\sin 2t\\ -2\cos 2t+\sin 2t\end{array}\right)e^{5t}$

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

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

Where

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

Obtaining the eigenvalues of the coefficient matrix,

$$\begin{gathered} det\left(A-\lambda I\right)=0 \\ \left|\begin{array}{cc}6-\lambda & -1\\ 5& 4-\lambda \end{array}\right|=0 \\ (6-\lambda )(4-\lambda )+5=0 \\ 24-6\lambda -4\lambda +{\lambda }^2+5=0 \\ {\lambda }^2-10\lambda +29=0 \end{gathered}$$

Where,

$$\begin{gathered} \lambda =\frac{10\pm \sqrt{{\left(10\right)}^2-4\left(1\right)\left(29\right)}}{2} \\ =\frac{10\pm 4i}{2} \\ =5\pm 2i \end{gathered}$$

So our eigenvalues are ${\lambda }_1=5+2i$, and${\lambda }_2=\overline{{\lambda }_1}=5-2i.$

Determine the Eigen vector and corresponding solution vector

For${\lambda }_1=5+2i$:

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

Apply row operation $\left(\frac{1}{1+2i}\right)R_2-R_1\to R_2$ :

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

Here we have

$(1-2i)k_1-k_2=0\to k_1=\frac{1}{1-2i}{k}_2$

Choosing k₂ = 1-2i yields k₁= 1.

This gives an eigenvector

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

And corresponding solution vectors

$B_1=\left(\begin{array}{c}1\\ 1\end{array}\right)and{B}_2=\left(\begin{array}{c}0\\ -2\end{array}\right)$

Determine the general solution of the system

Also,

$\lambda =\alpha +\beta i\to \lambda =5+2i$

Where $\alpha =5and\beta =2$

So the corresponding solution vectors are

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

And,

$$\begin{gathered} X2=\left[B_2\cos \beta t+B_1\sin \beta t\right]e^{\alpha t} \\ =\left[\left(\begin{array}{c}0\\ -2\end{array}\right)\cos 2t+\left(\begin{array}{c}1\\ 1\end{array}\right)\sin 2t\right]e^{5t} \\ =\left(\begin{array}{c}\sin 2t\\ -2\cos 2t+\sin 2t\end{array}\right)e^{5t} \end{gathered}$$

The general solution of the given system is

$$\begin{gathered} X=c_1{X}_1+c_2{X}_2 \\ =c_1\left(\begin{array}{c}\cos 2t\\ \cos 2t+2\sin 2t\end{array}\right)e^{5t}+c_2\left(\begin{array}{c}\sin 2t\\ -2\cos 2t+\sin 2t\end{array}\right)e^{5t} \end{gathered}$$

Determine the initial value condition

Applying the initial value condition

$$\begin{gathered} X\left(0\right)=\left(\begin{array}{c}-2\\ 8\end{array}\right) \\ \left(\begin{array}{c}-2\\ 8\end{array}\right)=c_1\left(\begin{array}{c}1\\ 1\end{array}\right)+c_2\left(\begin{array}{c}0\\ -2\end{array}\right) \end{gathered}$$

Here we have

c₁= -2

And

$c_1-2c_2=8\to -2-2c_2=8$

c₂ = -5

Hence, the solution that satisfies the initial condition is

$X=-2\left(\begin{array}{c}\cos 2t\\ \cos 2t+2\sin 2t\end{array}\right)e^{5t}-5\left(\begin{array}{c}\sin 2t\\ -2\cos 2t+\sin 2t\end{array}\right)e^{5t}$

Therefore, the initial value condition for $X\text{'}=\left(\begin{array}{cc}6& -1\\ 5& 4\end{array}\right)X$is.

$X=-2\left(\begin{array}{c}\cos 2t\\ \cos 2t+2\sin 2t\end{array}\right)e^{5t}-5\left(\begin{array}{c}\sin 2t\\ -2\cos 2t+\sin 2t\end{array}\right)e^{5t}$