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General and particular solution for a matrix

1

$begingroup$

I'm given the matrix A and a vector b:

beginequation
A= beginpmatrix
9 &-12 \
6 & -9
endpmatrix,; b= binom75
endequation
I have calculated the two eigenvalues and eigenvectors as:

beginequation
lambda_1=3,; lambda_2=-3 ; ; and ; ;
v_1= binom21,; v_2= binom11
endequation

The I found the general solution to A by:
beginequation
x'(t)=Ax(t)
endequation

This gives me:
beginequation
x=c_1binom21e^3t+c_2binom11e^-3t
endequation

Is that correct so far?

Then I have to give the particular solution that satisfies:

beginequation
x(0)=b
endequation

And now I'm stuck as I don't know how to proceed (b is the vector given above).

Thanks

ordinary-differential-equations matrix-equations

share|cite|improve this question

asked Mar 31 at 13:48

Donatello V.Donatello V.

174

$endgroup$

add a comment | 

1

$begingroup$

I'm given the matrix A and a vector b:

beginequation
A= beginpmatrix
9 &-12 \
6 & -9
endpmatrix,; b= binom75
endequation
I have calculated the two eigenvalues and eigenvectors as:

beginequation
lambda_1=3,; lambda_2=-3 ; ; and ; ;
v_1= binom21,; v_2= binom11
endequation

The I found the general solution to A by:
beginequation
x'(t)=Ax(t)
endequation

This gives me:
beginequation
x=c_1binom21e^3t+c_2binom11e^-3t
endequation

Is that correct so far?

Then I have to give the particular solution that satisfies:

beginequation
x(0)=b
endequation

And now I'm stuck as I don't know how to proceed (b is the vector given above).

Thanks

ordinary-differential-equations matrix-equations

share|cite|improve this question

asked Mar 31 at 13:48

Donatello V.Donatello V.

174

$endgroup$

add a comment | 

$begingroup$

I'm given the matrix A and a vector b:

beginequation
A= beginpmatrix
9 &-12 \
6 & -9
endpmatrix,; b= binom75
endequation
I have calculated the two eigenvalues and eigenvectors as:

beginequation
lambda_1=3,; lambda_2=-3 ; ; and ; ;
v_1= binom21,; v_2= binom11
endequation

The I found the general solution to A by:
beginequation
x'(t)=Ax(t)
endequation

This gives me:
beginequation
x=c_1binom21e^3t+c_2binom11e^-3t
endequation

Is that correct so far?

Then I have to give the particular solution that satisfies:

beginequation
x(0)=b
endequation

And now I'm stuck as I don't know how to proceed (b is the vector given above).

Thanks

ordinary-differential-equations matrix-equations

share|cite|improve this question

asked Mar 31 at 13:48

Donatello V.Donatello V.

174

$endgroup$

share|cite|improve this question

asked Mar 31 at 13:48

Donatello V.Donatello V.

174

share|cite|improve this question

asked Mar 31 at 13:48

Donatello V.Donatello V.

174

asked Mar 31 at 13:48

Donatello V.Donatello V.

174

asked Mar 31 at 13:48

Donatello V.Donatello V.

174

1 Answer
1

active

oldest

votes

0

$begingroup$

$$x(0)=c_1binom21+c_2binom11=binom2c_1+c_2c_1+c_2=binom75$$
$$therefore c_1=2, c_2=3$$

share|cite|improve this answer

answered Mar 31 at 13:51

Peter ForemanPeter Foreman

7,2711318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for this - didn't think it was that easy. So my general solution was correct?
  $endgroup$
  – Donatello V.
  Mar 31 at 13:59
  

  
  
  
  

add a comment | 

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0

$begingroup$

$$x(0)=c_1binom21+c_2binom11=binom2c_1+c_2c_1+c_2=binom75$$
$$therefore c_1=2, c_2=3$$

share|cite|improve this answer

answered Mar 31 at 13:51

Peter ForemanPeter Foreman

7,2711318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for this - didn't think it was that easy. So my general solution was correct?
  $endgroup$
  – Donatello V.
  Mar 31 at 13:59
  

  
  
  
  

add a comment | 

0

$begingroup$

$$x(0)=c_1binom21+c_2binom11=binom2c_1+c_2c_1+c_2=binom75$$
$$therefore c_1=2, c_2=3$$

share|cite|improve this answer

answered Mar 31 at 13:51

Peter ForemanPeter Foreman

7,2711318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for this - didn't think it was that easy. So my general solution was correct?
  $endgroup$
  – Donatello V.
  Mar 31 at 13:59
  

  
  
  
  

add a comment | 

$begingroup$

$$x(0)=c_1binom21+c_2binom11=binom2c_1+c_2c_1+c_2=binom75$$
$$therefore c_1=2, c_2=3$$

share|cite|improve this answer

answered Mar 31 at 13:51

Peter ForemanPeter Foreman

7,2711318

$endgroup$

share|cite|improve this answer

answered Mar 31 at 13:51

Peter ForemanPeter Foreman

7,2711318

answered Mar 31 at 13:51

Peter ForemanPeter Foreman

7,2711318

answered Mar 31 at 13:51

Peter ForemanPeter Foreman

7,2711318