Question

Verify that $\mathit{X}\mathbf{=}\left(\begin{array}{c}{c}_1\\ c_2\end{array}\right){\mathit{e}}^{\mathbf{t}}$is a solution of the linear system

$\mathit{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\mathit{X}$

for arbitrary constants ${\mathbf{c}}_{\mathbf{1}}$and ${\mathbf{c}}_{\mathbf{2}}$. By hand, draw a phase portrait of the system.

Short answer

It is verified that $x=\left(\begin{array}{c}{c}_1\\ c_2\end{array}\right)e^t$is a solution of the linear system $X\text{'}=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)X$and a phase portrait of the system is sketched as:

Step-by-step solution

Definition of Matrix

• The matrix exponential, like the regular exponential function, is a matrix function on square matrices. It's used to solve linear differential equation systems.

• The matrix exponential gives the relationship between a matrix Lie algebra and the appropriate Lie group in Lie group theory.

Verify the given matrix

The given information is written as:

$X\text{'}=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)X$and$X=\left(\begin{array}{c}{c}_1\\ c_2\end{array}\right)e^t$

Differentiate $x=\left(\begin{array}{c}{c}_1\\ c_2\end{array}\right)e^t$on both sides as follows:

$X=\left(\begin{array}{c}{c}_1\\ c_2\end{array}\right)e^t$

Now, substitute this derivative into $X\text{'}=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)X$as:

$$\begin{gathered} \left(\begin{array}{c}{c}_1\\ c_2\end{array}\right)e^t=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}{c}_1\\ c_2\end{array}\right)e^t \\ =\left(\begin{array}{c}{c}_1\\ c_2\end{array}\right)e^t \end{gathered}$$

Therefore, $X=\left(\begin{array}{c}{c}_1\\ c_2\end{array}\right)e^t$is a solution of the linear system $X\text{'}=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)X$and a phase portrait of the system is shown below: