Question

In Problems 5-14solve the given linear system.

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

Short answer

The solution for the linear system $$\begin{gathered} \frac{dx}{dt}=2x+y \\ \frac{dy}{dt}=-x \end{gathered}$$is $X=c_1\left(\begin{array}{c}1\\ -1\end{array}\right)e^t+c_2\left(\begin{array}{c}1\\ -1\end{array}\right)t{e}^t+\left(\begin{array}{c}0\\ 1\end{array}\right)e^t$.

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

It can be written in matrix form as,

localid="1664104672373" $X\text{'}=\left(\begin{array}{cc}2& 1\\ -1& 0\end{array}\right)X$

where, the matrix is $A=\left(\begin{array}{cc}2& 1\\ -1& 0\end{array}\right)$.

Now, the eigenvalues of the matrix is,

$$\begin{gathered} detA-\lambda I=0 \\ \left|\begin{array}{cc}2-\lambda & 1\\ -1& 0-\lambda \end{array}\right|=0 \end{gathered}$$

$$\begin{gathered} -\lambda (2-\lambda )+1=0 \\ {\lambda }^2-2\lambda +1=0 \\ (\lambda -1)(\lambda -1)=0 \end{gathered}$$

So, it has eigenvalue of $(\lambda -1)$which has algebraic multiplicity 2.

Find the eigenvector and solution vector of the given linear system.

Further solve the expression $(\mathbf{A}-\lambda \mathbf{I})\mathbf{K}=0$, it gives

$$\begin{gathered} \left|\begin{array}{cc}2-1& 1\\ -1& 0-1\end{array}\right|\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ \left(\begin{array}{cc}1& 1\\ -1& -1\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \end{gathered}$$

Apply row operation,

$$\begin{gathered} R_2+R_1\to R_2:\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ {k}_1+k_2=0\to k_1=-k_2 \end{gathered}$$

Choosing k₂= -1 it has k₁=1 .

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

Find the general solution of the given linear system.

Solve, further the expression $(\mathbf{A}-\lambda \mathbf{I})\mathbf{P}=\mathbf{K}$,

localid="1664105897852" $\left(\begin{array}{cc}1& 1\\ -1& -1\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\end{array}\right)=0$

Apply row operation,

localid="1664106062407" $$\begin{gathered} R_2+R_1\to R_2: \\ \left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\end{array}\right)=\left(\begin{array}{c}1\\ 0\end{array}\right) \\ {p}_1+p_2=0 \end{gathered}$$

Choosing p₁ = 0 it has p₂ =1.

This gives an eigenvector and a corresponding solution vector, $P=\left(\begin{array}{c}0\\ 1\end{array}\right),X_2=\left(\begin{array}{c}1\\ -1\end{array}\right)t{e}^t+\left(\begin{array}{c}0\\ 1\end{array}\right)e^t$.

Therefore, the general solution of the system is .

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