Short Answer

Let be an matrix and a scalar. Consider the following two systems:

$\frac{d \vec{x}}{dt} = A \vec{x} (l) \frac{d \vec{c}}{dt} = (A + {kl}_{n}) \vec{c} (ll)$

Show that if $\vec{x} (t)$ is a solution of the system (l) then role="math" localid="1659701582223" $\vec{c} (t) = e^{k t} \vec{x} (t)$ is a solution of the system (ll).

Yes, $\vec{c} (t) = e^{k t} \vec{x} (t)$ is a solution of the system $\frac{d \vec{c}}{dt} = (A + {kl}_{n}) \vec{c}$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Determine the equation of the solution.

Consider $\vec{x} (t)$ and $\vec{c} (t)$ is a solution of the system $\frac{d \vec{x}}{d t} = A \vec{x}$ and $\frac{d \vec{c}}{dt} = (A + {kl}_{n}) \vec{c}$ respectively.

Assume $λ_{1}, λ_{2}, λ_{3}, \dots, λ_{n}$ be the Eigen values of the matrix A then there exist Eigen vectors ${\vec{v}}_{1}, {\vec{v}}_{2} . {\vec{v}}_{3}, \dots, {\vec{v}}_{n}$ such that ${\vec{x}}_{1} (t) = c_{1} e^{λ_{1} t} {\vec{v}}_{1} + c_{2} e^{λ_{2} t} {\vec{v}}_{2} + . . . + c_{n} e^{λ_{n} t} {\vec{v}}_{n}$ and $\vec{c} (t) = c_{1} e^{(λ_{1} + k)} {\vec{v}}_{1} + c_{2} e^{(λ_{2} + k)} {\vec{v}}_{2} + . . . + c_{n} e^{(λ_{n} + k)} {\vec{v}}_{n}$ where $c_{1}, c_{2}, \dots, c_{n}$ is constant.

If x(t) is the solution of the linear system localid="1659702528208" $\frac{d \vec{x}}{d t} = A \vec{x}$ then ${\vec{x}}_{1} (t) = c_{1} e^{λ_{1} t} {\vec{v}}_{1} + c_{2} e^{λ_{2} t} {\vec{v}}_{2} + . . . + c_{n} e^{λ_{n} t} {\vec{v}}_{n}$ where $λ_{1}, λ_{2}, λ_{3}, \dots, λ_{n}$ be the Eigen values and of $n \times n$ matrix A.

Step 2: Show that c→(t)=ektx→(t) is a solution of the system.

Simplify the equation ${\vec{x}}_{1} (t) = c_{1} e^{λ_{1} t} {\vec{v}}_{1} + c_{2} e^{λ_{2} t} {\vec{v}}_{2} + . . . + c_{n} e^{λ_{n} t} {\vec{v}}_{n}$ as follows.

$\vec{c} (t) = c_{1} e^{(λ_{1} + k)} {\vec{v}}_{1} + c_{2} e^{(λ_{2} + k)} {\vec{v}}_{2} + . . . + c_{n} e^{(λ_{n} + k)} {\vec{v}}_{n} \vec{c} (t) = c_{1} e^{λ_{1} t} e^{k t} {\vec{v}}_{1} + c_{2} e^{λ_{2} t} e^{k t} {\vec{v}}_{2} + . . . + c_{n} e^{λ_{n} t} e^{k t} {\vec{v}}_{n} \vec{c} (t) = e^{k t} \{c_{1} e^{λ_{1} t} {\vec{v}}_{1} + c_{2} e^{λ_{2} t} {\vec{v}}_{2} + . . . + c_{n} e^{λ_{n} t} {\vec{v}}_{n}\} \vec{c} (t) = e^{k t} \vec{x} (t)$

By the definition of solution of the linear system, $\vec{c} (t) = e^{k t} \vec{x} (t)$ is a solution.

Hence, if $\vec{x} (t)$ and $\vec{c} (t)$ is a solution of the system $\frac{d \vec{x}}{d t} = A \vec{x}$ and $\frac{d \vec{c}}{dt} = (A + {kl}_{n}) \vec{c}$ respectively then is a solution.

Find Study Materials for

Create Study Materials

Select your language

Linear Algebra With Applications

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Determine the equation of the solution.

Step 2: Show that c→(t)=ektx→(t) is a solution of the system.

Most popular questions for Math Textbooks

Want to see more solutions like these?

Recommended explanations on Math Textbooks

Select your language

Linear Algebra With Applications

Answers without the blur.

Short Answer

Let be an matrix and a scalar. Consider the following two systems: dx→dt=Ax→ (l)dc→dt=(A+kln)c→ (ll) Show that if x→(t) is a solution of the system (l) then role="math" localid="1659701582223" c→(t)=ektx→(t) is a solution of the system (ll).

Step by Step Solution

Step 1: Determine the equation of the solution.

Step 2: Show that c→(t)=ektx→(t) is a solution of the system.

Most popular questions for Math Textbooks

Want to see more solutions like these?

Recommended explanations on Math Textbooks

Decision Maths

Calculus

Probability and Statistics

Statistics

Mechanics Maths

Geometry

Pure Maths