Question

In Problems 5-14solve the given linear system.

$$\begin{gathered} \frac{dx}{dt}=-4x+2y \\ \frac{dy}{dt}=2x-4y \end{gathered}$$

Short answer

The solution for the linear system $\frac{dx}{dt}=-4x+2y$and$\frac{dy}{dt}=2x-4y$is $X=c_1\left(\begin{array}{c}1\\ 1\end{array}\right)e^{-2t}+c_2\left(\begin{array}{c}1\\ -1\end{array}\right)e^{-6t}$.

Step-by-step solution

Define Matrix Exponential

Consider a square matrix A of size n*n. This matrix can contain either complex numbers or real numbers. The matrix can be calculated as:

$A^0=I,A^1=A,A^2=A.A,A^3=A^2.A,.....,A^k=A.\underset{ktimes}{A}.....$

where I is the unit matrix of order n.

Therefore, the infinite matrix power series is $I+\frac{t}{1!}A+\frac{t^2}{2!}{A}^2+\frac{t^3}{3!}{A}^3+\cdots +\frac{t^k}{k!}{A}^k+\cdots$

Now, the matrix exponential is defined as the sum of the infinite matrix power series.

It is denoted by the expression e^At. It is given by the formula,$e^{tA}=\sum _{k=0}^{\infty }\frac{t^k}{k!}{A}^k$.

Find the Eigenvalue of the given Linear System

The given information is written as:

$$\begin{gathered} \frac{dx}{dt}=-4x+2y \\ \frac{dy}{dt}=2x-4y \end{gathered}$$

It can be written in the matrix form as:

$X\text{'}=\left(\begin{array}{cc}-4& 2\\ 2& -4\end{array}\right)X$where$A=\left(\begin{array}{cc}-4& 2\\ 2& -4\end{array}\right)$

Find $\det \left(\mathbf{A}-\lambda \mathbf{I}\right)$as:

$\det \left(\mathbf{A}-\lambda \mathbf{I}\right)=0$

role="math" localid="1664175171675" $$\begin{gathered} \left|\begin{array}{cc}-4-\lambda & 2\\ 2& -4-\lambda \end{array}\right|=0 \\ (-4-\lambda )(-4-\lambda )-4=0 \end{gathered}$$

Simplify the equation as:

$$\begin{gathered} (-4-\lambda )(-4-\lambda )-4=0 \\ 16+4\lambda +4\lambda +{\lambda }^2-4=0 \\ {\lambda }^2+8\lambda +12=0 \\ (\lambda +2)(\lambda +6)=0 \end{gathered}$$

So, we have eigenvalues ${\lambda }_1=-2and{\lambda }_2=-6$.

Find the Eigenvector and Solution Vector of the given Linear System

For ${\lambda }_1=-2$, find the eigenvector and solution vector as:

$$\begin{gathered} (A+2I/0=\left(\begin{array}{ccc}-4-(-2)& 2& 0\\ 2& -4-(-2)& 0\end{array}\right) \\ =\left(\begin{array}{ccc}-2& 2& 0\\ 2& -2& 0\end{array}\right) \end{gathered}$$

Apply row operationwidth="96">R2+R1→R2as:

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

Therefore, $-2k_1+2k_2=0\Rightarrow k_1=k_2$.

Choosing k₂ = 1 yields k₁ = 1.

This gives an eigenvector and a corresponding solution vector $K_1=\left(\begin{array}{c}1\\ 1\end{array}\right),X_1=\left(\begin{array}{c}1\\ 1\end{array}\right)e^{-2t}$, respectively.

For, ${\lambda }_2=-6$find the eigenvector and solution vector as:

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

Apply row operation $R_2-R_1\to R_2$as:

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

Therefore,$2k_1+2k_2=0\Rightarrow k_1=-k_2$.

Choosing k₂=-1 yields k₁ =1.

This gives an eigenvector and a corresponding solution vector, $K_2=\left(\begin{array}{c}1\\ -1\end{array}\right),X_2=\left(\begin{array}{c}1\\ -1\end{array}\right)e^{-6t}$respectively.

Find the General Solution of the given Linear System

The general solution of the given system using and can be written as:

X=c₁X₁+c₂X₂

$=c_1\left(\begin{array}{c}1\\ 1\end{array}\right)e^{-2t}+c_2\left(\begin{array}{c}1\\ -1\end{array}\right)e^{-6t}$

Therefore, the general solution of the system is $X=c_1\left(\begin{array}{c}1\\ 1\end{array}\right)e^{-2t}+c_2\left(\begin{array}{c}1\\ -1\end{array}\right)e^{-6t}$$X=c_1\left(\begin{array}{c}1\\ 1\end{array}\right)e^{-2t}+c_2\left(\begin{array}{c}1\\ -1\end{array}\right)e^{-6t}$.