Additional Problems Verify that x 1 ( t ) = [ e t 2 − t − 1 ] and x 2 ( t ) = [ 0 2 e t ] are linearly independent solutions to x ′ = A x , where A = [ 2 t − 1 0 e t − t 2 1 ] . Write the general solution to the system x ′ = A x .

Question

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Answer 1

Textbook Question

Chapter 9.11, Problem 1AP

Additional Problems

Verify that x 1 ( t ) = [ e t 2 − t − 1 ] and x 2 ( t ) = [ 0 2 e t ] are linearly independent solutions to x ′ = A x , where

A = [ 2 t − 1 0 e t − t 2 1 ] .

Write the general solution to the system x ′ = A x .

Expert Solution & Answer

To determine

To verify:

That x1(t) and x2(t) are linearly independent solutions to x′=Ax.

Answer to Problem 1AP

Solution:

The general solution of the system is x(t)=[c1et2−t−c1+2c2et].

Explanation of Solution

Given:

The solutions are given as,

x1(t)=[et2−t−1], x2(t)=[02et] and A=[2t−10et−t21].

Approach:

If Wronskian of the solutions is non zero, then the solutions are linearly independent on any interval.

The general solution to the system is given as,

x(t)=c1x1(t)+c2x2(t).

Calculation:

Differentiate the given functions with respect to t.

x′1(t)=[et2−t(2t−1)0] and x′2(t)=[02et]

Find Ax1.

Ax1=[2t−10et−t21][et2−t−1]=[et2−t(2t−1)0]

Find Ax2.

Ax2=[2t−10et−t21][02et]=[02et]

Since, x′1=Ax1 and x′2=Ax2, therefore, the given functions are the solutions to the system.

Find the Wronskian of the solutions.

W[x1,x2](t)=|et2−t0−12et|=2et2

Here, the Wronskian is nonzero, the solutions are linearly independent on any interval.

Find the general solution to the system.

x(t)=c1x1(t)+c2x2(t)=c1[et2−t−1]+c2[02et]=[c1et2−t−c1+2c2et]

Therefore, the general solution of the system is x(t)=[c1et2−t−c1+2c2et].

Conclusion:

Hence, the general solution of the system is x(t)=[c1et2−t−c1+2c2et].

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Answer 2

Textbook Question

Chapter 9.11, Problem 1AP

Additional Problems

Verify that x 1 ( t ) = [ e t 2 − t − 1 ] and x 2 ( t ) = [ 0 2 e t ] are linearly independent solutions to x ′ = A x , where

A = [ 2 t − 1 0 e t − t 2 1 ] .

Write the general solution to the system x ′ = A x .

Expert Solution & Answer

To determine

To verify:

That x1(t) and x2(t) are linearly independent solutions to x′=Ax.

Answer to Problem 1AP

Solution:

The general solution of the system is x(t)=[c1et2−t−c1+2c2et].

Explanation of Solution

Given:

The solutions are given as,

x1(t)=[et2−t−1], x2(t)=[02et] and A=[2t−10et−t21].

Approach:

If Wronskian of the solutions is non zero, then the solutions are linearly independent on any interval.

The general solution to the system is given as,

x(t)=c1x1(t)+c2x2(t).

Calculation:

Differentiate the given functions with respect to t.

x′1(t)=[et2−t(2t−1)0] and x′2(t)=[02et]

Find Ax1.

Ax1=[2t−10et−t21][et2−t−1]=[et2−t(2t−1)0]

Find Ax2.

Ax2=[2t−10et−t21][02et]=[02et]

Since, x′1=Ax1 and x′2=Ax2, therefore, the given functions are the solutions to the system.

Find the Wronskian of the solutions.

W[x1,x2](t)=|et2−t0−12et|=2et2

Here, the Wronskian is nonzero, the solutions are linearly independent on any interval.

Find the general solution to the system.

x(t)=c1x1(t)+c2x2(t)=c1[et2−t−1]+c2[02et]=[c1et2−t−c1+2c2et]

Therefore, the general solution of the system is x(t)=[c1et2−t−c1+2c2et].

Conclusion:

Hence, the general solution of the system is x(t)=[c1et2−t−c1+2c2et].

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Answer 3

Textbook Question

Chapter 9.11, Problem 1AP

Additional Problems

Verify that x 1 ( t ) = [ e t 2 − t − 1 ] and x 2 ( t ) = [ 0 2 e t ] are linearly independent solutions to x ′ = A x , where

A = [ 2 t − 1 0 e t − t 2 1 ] .

Write the general solution to the system x ′ = A x .

Expert Solution & Answer

To determine

To verify:

That x1(t) and x2(t) are linearly independent solutions to x′=Ax.

Answer to Problem 1AP

Solution:

The general solution of the system is x(t)=[c1et2−t−c1+2c2et].

Explanation of Solution

Given:

The solutions are given as,

x1(t)=[et2−t−1], x2(t)=[02et] and A=[2t−10et−t21].

Approach:

If Wronskian of the solutions is non zero, then the solutions are linearly independent on any interval.

The general solution to the system is given as,

x(t)=c1x1(t)+c2x2(t).

Calculation:

Differentiate the given functions with respect to t.

x′1(t)=[et2−t(2t−1)0] and x′2(t)=[02et]

Find Ax1.

Ax1=[2t−10et−t21][et2−t−1]=[et2−t(2t−1)0]

Find Ax2.

Ax2=[2t−10et−t21][02et]=[02et]

Since, x′1=Ax1 and x′2=Ax2, therefore, the given functions are the solutions to the system.

Find the Wronskian of the solutions.

W[x1,x2](t)=|et2−t0−12et|=2et2

Here, the Wronskian is nonzero, the solutions are linearly independent on any interval.

Find the general solution to the system.

x(t)=c1x1(t)+c2x2(t)=c1[et2−t−1]+c2[02et]=[c1et2−t−c1+2c2et]

Therefore, the general solution of the system is x(t)=[c1et2−t−c1+2c2et].

Conclusion:

Hence, the general solution of the system is x(t)=[c1et2−t−c1+2c2et].

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Answer 4

Textbook Question

Chapter 9.11, Problem 1AP

Additional Problems

Verify that x 1 ( t ) = [ e t 2 − t − 1 ] and x 2 ( t ) = [ 0 2 e t ] are linearly independent solutions to x ′ = A x , where

A = [ 2 t − 1 0 e t − t 2 1 ] .

Write the general solution to the system x ′ = A x .

Expert Solution & Answer

To determine

To verify:

That x1(t) and x2(t) are linearly independent solutions to x′=Ax.

Answer to Problem 1AP

Solution:

The general solution of the system is x(t)=[c1et2−t−c1+2c2et].

Explanation of Solution

Given:

The solutions are given as,

x1(t)=[et2−t−1], x2(t)=[02et] and A=[2t−10et−t21].

Approach:

If Wronskian of the solutions is non zero, then the solutions are linearly independent on any interval.

The general solution to the system is given as,

x(t)=c1x1(t)+c2x2(t).

Calculation:

Differentiate the given functions with respect to t.

x′1(t)=[et2−t(2t−1)0] and x′2(t)=[02et]

Find Ax1.

Ax1=[2t−10et−t21][et2−t−1]=[et2−t(2t−1)0]

Find Ax2.

Ax2=[2t−10et−t21][02et]=[02et]

Since, x′1=Ax1 and x′2=Ax2, therefore, the given functions are the solutions to the system.

Find the Wronskian of the solutions.

W[x1,x2](t)=|et2−t0−12et|=2et2

Here, the Wronskian is nonzero, the solutions are linearly independent on any interval.

Find the general solution to the system.

x(t)=c1x1(t)+c2x2(t)=c1[et2−t−1]+c2[02et]=[c1et2−t−c1+2c2et]

Therefore, the general solution of the system is x(t)=[c1et2−t−c1+2c2et].

Conclusion:

Hence, the general solution of the system is x(t)=[c1et2−t−c1+2c2et].

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

To verify:

That x1(t) and x2(t) are linearly independent solutions to x′=Ax.

Answer to Problem 1AP

Solution:

The general solution of the system is x(t)=[c1et2−t−c1+2c2et].

Explanation of Solution

Given:

The solutions are given as,

x1(t)=[et2−t−1], x2(t)=[02et] and A=[2t−10et−t21].

Approach:

If Wronskian of the solutions is non zero, then the solutions are linearly independent on any interval.

The general solution to the system is given as,

x(t)=c1x1(t)+c2x2(t).

Calculation:

Differentiate the given functions with respect to t.

x′1(t)=[et2−t(2t−1)0] and x′2(t)=[02et]

Find Ax1.

Ax1=[2t−10et−t21][et2−t−1]=[et2−t(2t−1)0]

Find Ax2.

Ax2=[2t−10et−t21][02et]=[02et]

Since, x′1=Ax1 and x′2=Ax2, therefore, the given functions are the solutions to the system.

Find the Wronskian of the solutions.

W[x1,x2](t)=|et2−t0−12et|=2et2

Here, the Wronskian is nonzero, the solutions are linearly independent on any interval.

Find the general solution to the system.

x(t)=c1x1(t)+c2x2(t)=c1[et2−t−1]+c2[02et]=[c1et2−t−c1+2c2et]

Therefore, the general solution of the system is x(t)=[c1et2−t−c1+2c2et].

Conclusion:

Hence, the general solution of the system is x(t)=[c1et2−t−c1+2c2et].

Answer 8

Expert Solution & Answer

To determine

To verify:

That x1(t) and x2(t) are linearly independent solutions to x′=Ax.

Answer to Problem 1AP

Solution:

The general solution of the system is x(t)=[c1et2−t−c1+2c2et].

Explanation of Solution

Given:

The solutions are given as,

x1(t)=[et2−t−1], x2(t)=[02et] and A=[2t−10et−t21].

Approach:

If Wronskian of the solutions is non zero, then the solutions are linearly independent on any interval.

The general solution to the system is given as,

x(t)=c1x1(t)+c2x2(t).

Calculation:

Differentiate the given functions with respect to t.

x′1(t)=[et2−t(2t−1)0] and x′2(t)=[02et]

Find Ax1.

Ax1=[2t−10et−t21][et2−t−1]=[et2−t(2t−1)0]

Find Ax2.

Ax2=[2t−10et−t21][02et]=[02et]

Since, x′1=Ax1 and x′2=Ax2, therefore, the given functions are the solutions to the system.

Find the Wronskian of the solutions.

W[x1,x2](t)=|et2−t0−12et|=2et2

Here, the Wronskian is nonzero, the solutions are linearly independent on any interval.

Find the general solution to the system.

x(t)=c1x1(t)+c2x2(t)=c1[et2−t−1]+c2[02et]=[c1et2−t−c1+2c2et]

Therefore, the general solution of the system is x(t)=[c1et2−t−c1+2c2et].

Conclusion:

Hence, the general solution of the system is x(t)=[c1et2−t−c1+2c2et].

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

To verify:

That x1(t) and x2(t) are linearly independent solutions to x′=Ax.

Answer 12

Answer to Problem 1AP

Solution:

The general solution of the system is x(t)=[c1et2−t−c1+2c2et].

Answer 13

Explanation of Solution

Given:

The solutions are given as,

x1(t)=[et2−t−1], x2(t)=[02et] and A=[2t−10et−t21].

Approach:

If Wronskian of the solutions is non zero, then the solutions are linearly independent on any interval.

The general solution to the system is given as,

x(t)=c1x1(t)+c2x2(t).

Calculation:

Differentiate the given functions with respect to t.

x′1(t)=[et2−t(2t−1)0] and x′2(t)=[02et]

Find Ax1.

Ax1=[2t−10et−t21][et2−t−1]=[et2−t(2t−1)0]

Find Ax2.

Ax2=[2t−10et−t21][02et]=[02et]

Since, x′1=Ax1 and x′2=Ax2, therefore, the given functions are the solutions to the system.

Find the Wronskian of the solutions.

W[x1,x2](t)=|et2−t0−12et|=2et2

Here, the Wronskian is nonzero, the solutions are linearly independent on any interval.

Find the general solution to the system.

x(t)=c1x1(t)+c2x2(t)=c1[et2−t−1]+c2[02et]=[c1et2−t−c1+2c2et]

Therefore, the general solution of the system is x(t)=[c1et2−t−c1+2c2et].

Conclusion:

Hence, the general solution of the system is x(t)=[c1et2−t−c1+2c2et].

Answer 14

Ch. 9.1 - For Question a-j, decide if the given statement is...Ch. 9.1 - For Question a-j, decide if the given statement is...Ch. 9.1 - For Question a-j, decide if the given statement is...Ch. 9.1 - For Question a-j, decide if the given statement is...Ch. 9.1 - Prob. 5TFR Ch. 9.1 - For Question a-j, decide if the given statement is...Ch. 9.1 - Prob. 7TFR Ch. 9.1 - For Question a-j, decide if the given statement is...Ch. 9.1 - Prob. 9TFR Ch. 9.1 - Prob. 10TFR

Additional Problems Verify that x 1 ( t ) = [ e t 2 − t − 1 ] and x 2 ( t ) = [ 0 2 e t ] are linearly independent solutions to x ′ = A x , where A = [ 2 t − 1 0 e t − t 2 1 ] . Write the general solution to the system x ′ = A x .

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Answer to Problem 1AP

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