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Found in: Page 425

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Use the concept of a continuous dynamical system$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}{\mathbf{=}}{\mathbf{-}}{\mathbit{k}}{\mathbit{x}}$ .Solve the differential equation $\frac{\mathbf{d}\stackrel{\mathbf{\to }}{\mathbf{x}}}{\mathbf{d}\mathbf{t}}{\mathbf{=}}{\mathbit{A}}\stackrel{\mathbf{\to }}{\mathbf{x}}$. Solvethe system when A is diagonalizable over R,and sketch the phase portrait for 2 × 2 matrices A.Solve the initial value problems posed in Exercises 1through 5. Graph the solution.3. $\frac{\mathbf{d}\mathbf{P}}{\mathbf{d}\mathbf{t}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{03}}{\mathbit{P}}$ with ${\mathbit{P}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}{\mathbf{=}}{\mathbf{7}}$.

The solution is $y=7{e}^{0.03t}$ .

See the step by step solution

## Step 1: Definition of the differential equation

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

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

## Step 2: Calculation of the Solution

Given the differential equation $\frac{dP}{dt}=0.03P$ with the initial condition $P\left(0\right)=7$ .

Substitute in the solution $y=C{e}^{kx}$ as follows:

$\begin{array}{l}y=C{e}^{kx}\\ y=7{e}^{0.03t}\end{array}$

Hence, the solution for the differential equation $\frac{dP}{dt}=0.03P$ is $y=7{e}^{0.03t}$.

## Step 3: Graphical representation of the solution

The graph of the equation $y={e}^{0.03t}$ is sketched below as follows: