Systems of DE (Eigenvectors) [closed]

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In a $2times 2$ system of differential equations with initial conditions how are the unknowns in the general solution ($C_1$ and $C_2$) calculated given that there are an infinite number of eigenvectors possible? Won't that fact, which i'm assuming is true, yield infinitely many solutions, all affecting how the solution trajectories look?

Thanks.

differential-equations systems-of-equations

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edited Jul 30 at 12:47

Harry Peter

5,44311438

asked Jul 29 at 23:56

MisterOH

193

closed as off-topic by amWhy, Taroccoesbrocco, Mostafa Ayaz, Rhys Steele, José Carlos Santos Jul 30 at 22:39

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Taroccoesbrocco, Mostafa Ayaz, Rhys Steele, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  D.E. problems require initial conditions in order to find particular solutions.
  – é«˜ç”°èˆª
  Jul 29 at 23:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I understand that but the general form of the solution for these types of equations have the eigenvector for each eigenvalue in them. I'm asking since there are many possible eigenvectors does any constant multiple of them give a correct answer?
  – MisterOH
  Jul 30 at 0:01
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

In a $2times 2$ system of differential equations with initial conditions how are the unknowns in the general solution ($C_1$ and $C_2$) calculated given that there are an infinite number of eigenvectors possible? Won't that fact, which i'm assuming is true, yield infinitely many solutions, all affecting how the solution trajectories look?

Thanks.

differential-equations systems-of-equations

share|cite|improve this question

edited Jul 30 at 12:47

Harry Peter

5,44311438

asked Jul 29 at 23:56

MisterOH

193

closed as off-topic by amWhy, Taroccoesbrocco, Mostafa Ayaz, Rhys Steele, José Carlos Santos Jul 30 at 22:39

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Taroccoesbrocco, Mostafa Ayaz, Rhys Steele, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  D.E. problems require initial conditions in order to find particular solutions.
  – é«˜ç”°èˆª
  Jul 29 at 23:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I understand that but the general form of the solution for these types of equations have the eigenvector for each eigenvalue in them. I'm asking since there are many possible eigenvectors does any constant multiple of them give a correct answer?
  – MisterOH
  Jul 30 at 0:01
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

In a $2times 2$ system of differential equations with initial conditions how are the unknowns in the general solution ($C_1$ and $C_2$) calculated given that there are an infinite number of eigenvectors possible? Won't that fact, which i'm assuming is true, yield infinitely many solutions, all affecting how the solution trajectories look?

Thanks.

differential-equations systems-of-equations

share|cite|improve this question

edited Jul 30 at 12:47

Harry Peter

5,44311438

asked Jul 29 at 23:56

MisterOH

193

In a $2times 2$ system of differential equations with initial conditions how are the unknowns in the general solution ($C_1$ and $C_2$) calculated given that there are an infinite number of eigenvectors possible? Won't that fact, which i'm assuming is true, yield infinitely many solutions, all affecting how the solution trajectories look?

Thanks.

differential-equations systems-of-equations

share|cite|improve this question

edited Jul 30 at 12:47

Harry Peter

5,44311438

asked Jul 29 at 23:56

MisterOH

193

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 12:47

Harry Peter

5,44311438

edited Jul 30 at 12:47

Harry Peter

5,44311438

edited Jul 30 at 12:47

Harry Peter

5,44311438

5,44311438

asked Jul 29 at 23:56

MisterOH

193

asked Jul 29 at 23:56

MisterOH

193

asked Jul 29 at 23:56

MisterOH

193

193

closed as off-topic by amWhy, Taroccoesbrocco, Mostafa Ayaz, Rhys Steele, José Carlos Santos Jul 30 at 22:39

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Taroccoesbrocco, Mostafa Ayaz, Rhys Steele, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, Taroccoesbrocco, Mostafa Ayaz, Rhys Steele, José Carlos Santos Jul 30 at 22:39

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Taroccoesbrocco, Mostafa Ayaz, Rhys Steele, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  D.E. problems require initial conditions in order to find particular solutions.
  – é«˜ç”°èˆª
  Jul 29 at 23:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I understand that but the general form of the solution for these types of equations have the eigenvector for each eigenvalue in them. I'm asking since there are many possible eigenvectors does any constant multiple of them give a correct answer?
  – MisterOH
  Jul 30 at 0:01
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  D.E. problems require initial conditions in order to find particular solutions.
  – é«˜ç”°èˆª
  Jul 29 at 23:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I understand that but the general form of the solution for these types of equations have the eigenvector for each eigenvalue in them. I'm asking since there are many possible eigenvectors does any constant multiple of them give a correct answer?
  – MisterOH
  Jul 30 at 0:01
  

  
  
  
  

D.E. problems require initial conditions in order to find particular solutions.
– é«˜ç”°èˆª
Jul 29 at 23:57

D.E. problems require initial conditions in order to find particular solutions.
– é«˜ç”°èˆª
Jul 29 at 23:57

I understand that but the general form of the solution for these types of equations have the eigenvector for each eigenvalue in them. I'm asking since there are many possible eigenvectors does any constant multiple of them give a correct answer?
– MisterOH
Jul 30 at 0:01

I understand that but the general form of the solution for these types of equations have the eigenvector for each eigenvalue in them. I'm asking since there are many possible eigenvectors does any constant multiple of them give a correct answer?
– MisterOH
Jul 30 at 0:01

add a comment | 

2 Answers
2

active

oldest

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0
down vote



accepted

The general form solution should be (with real eigenvalues) $$vecx(t)=c_1e^lambda_1tveceta_1+c_2e^lambda_2tveceta_2$$ where $lambda$ are the eigenvalues and $veceta$ are the eigenvectors. The arbitrary constants $c_1$ and $c_2$ are there because, as you say, there are an infinite amount of eigenvectors that satisfy the system, which is why we need the initial conditions.

share|cite|improve this answer

edited Jul 30 at 0:18

answered Jul 30 at 0:10

高田航

1,116318

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought I was getting wrong answers because my eigenvectors didn't match those in my textbook examples, but after simplifying solutions (multiplying through) I see that my solutions are correct. Thanks.
  – MisterOH
  Jul 30 at 0:36
  

  
  
  
  

add a comment | 

up vote
0
down vote

In order to find the general solution you only need two linearly independent eigenvectors along with the eigenvalues. The general solution is $$y= C_1 e^lambda _1tV_1 + C_2 e^lambda _2tV_2$$ where the coefficients are found from the initial values.

share|cite|improve this answer

answered Jul 30 at 0:12

Mohammad Riazi-Kermani

27.3k41851

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

The general form solution should be (with real eigenvalues) $$vecx(t)=c_1e^lambda_1tveceta_1+c_2e^lambda_2tveceta_2$$ where $lambda$ are the eigenvalues and $veceta$ are the eigenvectors. The arbitrary constants $c_1$ and $c_2$ are there because, as you say, there are an infinite amount of eigenvectors that satisfy the system, which is why we need the initial conditions.

share|cite|improve this answer

edited Jul 30 at 0:18

answered Jul 30 at 0:10

高田航

1,116318

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought I was getting wrong answers because my eigenvectors didn't match those in my textbook examples, but after simplifying solutions (multiplying through) I see that my solutions are correct. Thanks.
  – MisterOH
  Jul 30 at 0:36
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

The general form solution should be (with real eigenvalues) $$vecx(t)=c_1e^lambda_1tveceta_1+c_2e^lambda_2tveceta_2$$ where $lambda$ are the eigenvalues and $veceta$ are the eigenvectors. The arbitrary constants $c_1$ and $c_2$ are there because, as you say, there are an infinite amount of eigenvectors that satisfy the system, which is why we need the initial conditions.

share|cite|improve this answer

edited Jul 30 at 0:18

answered Jul 30 at 0:10

高田航

1,116318

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought I was getting wrong answers because my eigenvectors didn't match those in my textbook examples, but after simplifying solutions (multiplying through) I see that my solutions are correct. Thanks.
  – MisterOH
  Jul 30 at 0:36
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

The general form solution should be (with real eigenvalues) $$vecx(t)=c_1e^lambda_1tveceta_1+c_2e^lambda_2tveceta_2$$ where $lambda$ are the eigenvalues and $veceta$ are the eigenvectors. The arbitrary constants $c_1$ and $c_2$ are there because, as you say, there are an infinite amount of eigenvectors that satisfy the system, which is why we need the initial conditions.

share|cite|improve this answer

edited Jul 30 at 0:18

answered Jul 30 at 0:10

高田航

1,116318

The general form solution should be (with real eigenvalues) $$vecx(t)=c_1e^lambda_1tveceta_1+c_2e^lambda_2tveceta_2$$ where $lambda$ are the eigenvalues and $veceta$ are the eigenvectors. The arbitrary constants $c_1$ and $c_2$ are there because, as you say, there are an infinite amount of eigenvectors that satisfy the system, which is why we need the initial conditions.

share|cite|improve this answer

edited Jul 30 at 0:18

answered Jul 30 at 0:10

高田航

1,116318

share|cite|improve this answer

share|cite|improve this answer

edited Jul 30 at 0:18

edited Jul 30 at 0:18

edited Jul 30 at 0:18

answered Jul 30 at 0:10

高田航

1,116318

answered Jul 30 at 0:10

高田航

1,116318

answered Jul 30 at 0:10

高田航

1,116318

1,116318

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought I was getting wrong answers because my eigenvectors didn't match those in my textbook examples, but after simplifying solutions (multiplying through) I see that my solutions are correct. Thanks.
  – MisterOH
  Jul 30 at 0:36
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I thought I was getting wrong answers because my eigenvectors didn't match those in my textbook examples, but after simplifying solutions (multiplying through) I see that my solutions are correct. Thanks.
  – MisterOH
  Jul 30 at 0:36
  

  
  
  
  

I thought I was getting wrong answers because my eigenvectors didn't match those in my textbook examples, but after simplifying solutions (multiplying through) I see that my solutions are correct. Thanks.
– MisterOH
Jul 30 at 0:36

I thought I was getting wrong answers because my eigenvectors didn't match those in my textbook examples, but after simplifying solutions (multiplying through) I see that my solutions are correct. Thanks.
– MisterOH
Jul 30 at 0:36

add a comment | 

up vote
0
down vote

In order to find the general solution you only need two linearly independent eigenvectors along with the eigenvalues. The general solution is $$y= C_1 e^lambda _1tV_1 + C_2 e^lambda _2tV_2$$ where the coefficients are found from the initial values.

share|cite|improve this answer

answered Jul 30 at 0:12

Mohammad Riazi-Kermani

27.3k41851

add a comment | 

up vote
0
down vote

In order to find the general solution you only need two linearly independent eigenvectors along with the eigenvalues. The general solution is $$y= C_1 e^lambda _1tV_1 + C_2 e^lambda _2tV_2$$ where the coefficients are found from the initial values.

share|cite|improve this answer

answered Jul 30 at 0:12

Mohammad Riazi-Kermani

27.3k41851

add a comment | 

up vote
0
down vote

up vote
0
down vote

In order to find the general solution you only need two linearly independent eigenvectors along with the eigenvalues. The general solution is $$y= C_1 e^lambda _1tV_1 + C_2 e^lambda _2tV_2$$ where the coefficients are found from the initial values.

share|cite|improve this answer

answered Jul 30 at 0:12

Mohammad Riazi-Kermani

27.3k41851

In order to find the general solution you only need two linearly independent eigenvectors along with the eigenvalues. The general solution is $$y= C_1 e^lambda _1tV_1 + C_2 e^lambda _2tV_2$$ where the coefficients are found from the initial values.

share|cite|improve this answer

answered Jul 30 at 0:12

Mohammad Riazi-Kermani

27.3k41851

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 0:12

Mohammad Riazi-Kermani

27.3k41851

answered Jul 30 at 0:12

Mohammad Riazi-Kermani

27.3k41851

answered Jul 30 at 0:12

Mohammad Riazi-Kermani

27.3k41851

27.3k41851

add a comment | 

add a comment |