Question

Use the concept of a continuous dynamical system.Solve the differential equation $\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\mathit{k}\mathit{x}$. Solvethe system$\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}$ whenAis diagonalizable overR,and sketch the phase portrait for 2×2 matricesA.

Solve the initial value problems posed in Exercises 1through 5. Graph the solution.

5.$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8}\mathit{y}$with$\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{8}$

Short answer

The solution is$y\left(t\right)=-0.8e^{0.8t}$.

Step-by-step solution

Definition of the differential equation          

Consider the differential equation $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{k}\mathit{x}$with initial value ${\mathit{x}}_{\mathbf{0}}$(k is an arbitrary constant). The solution is $\mathit{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathit{x}}_{\mathbf{0}}{\mathit{e}}^{\mathbf{k}\mathbf{t}}$.

The solution of the linear differential equation$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{k}\mathit{x}$and $\mathbf{\text{y}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{\text{C}}$is $\mathbf{\text{y}}\mathbf{=}{\mathbf{\text{Ce}}}^{\mathbf{\text{kx}}}$.

Calculation of the solution 

Given the differential equation $\frac{dy}{dt}=0.8y$with the initial condition $y\left(0\right)=-0.8$.

Substitute in the solution$y=C{e}^{kt}$as follows.

$\begin{array}{l}y=C{e}^{kt}\\ y=-0.8e^{0.8t}\end{array}$

Hence, the solution for the differential equation $\frac{dy}{dt}=0.8y$is$y=-0.8e^{0.8t}$ .

Graphical representation of  the solution

The graph of the equation$y=-0.8e^{0.8t}$ is sketched below as follows: