Short Answer

Use the concept of a continuous dynamical system $\frac{d x}{d t} = - k x$ .Solve the differential equation $\frac{d \vec{x}}{d t} = A \vec{x}$ . 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. $\frac{d P}{d t} = 0.03 P$ with $P (0) = 7$ .

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition of the differential equation

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

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

Step 2: Calculation of the Solution

Given the differential equation $\frac{d P}{d t} = 0.03 P$ with the initial condition $P (0) = 7$ .

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

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

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

Step 3: Graphical representation of the solution

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

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Linear Algebra With Applications

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Short Answer

Step by Step Solution

Step 1: Definition of the differential equation

Step 2: Calculation of the Solution

Step 3: Graphical representation of the solution

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Linear Algebra With Applications

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Step 1: Definition of the differential equation

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Step 3: Graphical representation of the solution

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