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System of linear differential equation
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How to solve the system of linear differential equation of the form $$x' = Ax + b$$ I can solve the homogeneous form by finding the eigenvalues and respective eigenvectors, but how to find the particular solution part. Also is there any limitation from getting eigenvalues positive, negative or complex. Any other different method involving matrix algebra is also welcome. Thank you.
linear-algebra matrices differential-equations
edited Jul 21 at 17:39
Rodrigo de Azevedo
12.6k41751
asked Jul 21 at 17:36
Mridul Dey
765
add a comment |Â
up vote
3
down vote
favorite
How to solve the system of linear differential equation of the form $$x' = Ax + b$$ I can solve the homogeneous form by finding the eigenvalues and respective eigenvectors, but how to find the particular solution part. Also is there any limitation from getting eigenvalues positive, negative or complex. Any other different method involving matrix algebra is also welcome. Thank you.
linear-algebra matrices differential-equations
edited Jul 21 at 17:39
Rodrigo de Azevedo
12.6k41751
asked Jul 21 at 17:36
Mridul Dey
765
Thank you very much. All the answers are really helpful.
– Mridul Dey
Jul 21 at 18:08
You're welcome! If you feel like one of the answers is what you were looking for, you should mark it as accepted!
– Davide Morgante
Jul 21 at 19:36
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
How to solve the system of linear differential equation of the form $$x' = Ax + b$$ I can solve the homogeneous form by finding the eigenvalues and respective eigenvectors, but how to find the particular solution part. Also is there any limitation from getting eigenvalues positive, negative or complex. Any other different method involving matrix algebra is also welcome. Thank you.
linear-algebra matrices differential-equations
edited Jul 21 at 17:39
Rodrigo de Azevedo
12.6k41751
asked Jul 21 at 17:36
Mridul Dey
765
How to solve the system of linear differential equation of the form $$x' = Ax + b$$ I can solve the homogeneous form by finding the eigenvalues and respective eigenvectors, but how to find the particular solution part. Also is there any limitation from getting eigenvalues positive, negative or complex. Any other different method involving matrix algebra is also welcome. Thank you.
linear-algebra matrices differential-equations
edited Jul 21 at 17:39
Rodrigo de Azevedo
12.6k41751
asked Jul 21 at 17:36
Mridul Dey
765
edited Jul 21 at 17:39
Rodrigo de Azevedo
12.6k41751
edited Jul 21 at 17:39
Rodrigo de Azevedo
12.6k41751
edited Jul 21 at 17:39
Rodrigo de Azevedo
12.6k41751
12.6k41751
asked Jul 21 at 17:36
Mridul Dey
765
asked Jul 21 at 17:36
Mridul Dey
765
asked Jul 21 at 17:36
Mridul Dey
765
765
Thank you very much. All the answers are really helpful.
– Mridul Dey
Jul 21 at 18:08
You're welcome! If you feel like one of the answers is what you were looking for, you should mark it as accepted!
– Davide Morgante
Jul 21 at 19:36
add a comment |Â
Thank you very much. All the answers are really helpful.
– Mridul Dey
Jul 21 at 18:08
You're welcome! If you feel like one of the answers is what you were looking for, you should mark it as accepted!
– Davide Morgante
Jul 21 at 19:36
Thank you very much. All the answers are really helpful.
– Mridul Dey
Jul 21 at 18:08
Thank you very much. All the answers are really helpful.
– Mridul Dey
Jul 21 at 18:08
You're welcome! If you feel like one of the answers is what you were looking for, you should mark it as accepted!
– Davide Morgante
Jul 21 at 19:36
You're welcome! If you feel like one of the answers is what you were looking for, you should mark it as accepted!
– Davide Morgante
Jul 21 at 19:36
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
Being $b$ a fixed constant (suppose $det Aneq0$), we can choose a new variable $x_*$ that solves $$Ax_* + b = 0 Rightarrow x_* = -A^-1b$$ The now we can solve the equation in the fixed point $z = x-x_* = x+A^-1b$. This variable solves the following equation $$dotz = Az$$
In fact, form the definition of $z$ $$dotz=dotx\ Az = A(x+A^-1b) = Ax+b$$
So we're saying that, if $x$ is a solution to $dotx=Ax+b$ then the variable $z = x+A^-1b$ is a solution to the simpler equation $dotz=Az$.
The solution to this simpler equation is just $$z(t)= [e^At]z(0)$$ where $[e^At]$ is the exponential of the matrix $A$. At this point you can go back to find $x(t)$ simply by using the definition of $z$: $$x(t) = z(t)-A^-1b = [e^At](x(0)+A^-1b)-A^-1b = [e^At]x(0)+A^-1b([e^At]-1)$$ where we have used the fact that $$z(0) = x(0) + A^-1b$$
Easier way
You can even use this simpler method: given the system of equations $$dotx = Ax+f(t)$$ the solution to this is as follows $$x(t) = [e^At]x(0)+int_0^t e^A(t-t')f(t')dt'$$
in your case the function is just a constant function $f(t)=b$ for which you can see that gives the same solution as before! These methods are good because they let you solve every system of equations of this type because all the informations about the system is embedded into the exponential of the matrix
answered Jul 21 at 17:49
Davide Morgante
1,812220
add a comment |Â
up vote
3
down vote
If $b$ is the columnspace of $A$, define $y$ so that $Ay = -b$. Then note that $x(t) = e^Atx(0) + y$ solves the differential equation since $x' = A e^Atx(0)$ and $Ax = A e^Atx(0) - b$.
answered Jul 21 at 17:48
Marcus M
8,1731847
add a comment |Â
up vote
2
down vote
Take Laplace transform therefore $$sX(s)-x(0)=AX(s)+b\(sI-A)X(s)=b+x(0)$$or$$X(s)=(sI-A)^-1(b+x(0))$$but we can't proceed any further. We need to know the characteristics of the system.
answered Jul 21 at 17:44
Mostafa Ayaz
8,5773630
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Being $b$ a fixed constant (suppose $det Aneq0$), we can choose a new variable $x_*$ that solves $$Ax_* + b = 0 Rightarrow x_* = -A^-1b$$ The now we can solve the equation in the fixed point $z = x-x_* = x+A^-1b$. This variable solves the following equation $$dotz = Az$$
In fact, form the definition of $z$ $$dotz=dotx\ Az = A(x+A^-1b) = Ax+b$$
So we're saying that, if $x$ is a solution to $dotx=Ax+b$ then the variable $z = x+A^-1b$ is a solution to the simpler equation $dotz=Az$.
The solution to this simpler equation is just $$z(t)= [e^At]z(0)$$ where $[e^At]$ is the exponential of the matrix $A$. At this point you can go back to find $x(t)$ simply by using the definition of $z$: $$x(t) = z(t)-A^-1b = [e^At](x(0)+A^-1b)-A^-1b = [e^At]x(0)+A^-1b([e^At]-1)$$ where we have used the fact that $$z(0) = x(0) + A^-1b$$
Easier way
You can even use this simpler method: given the system of equations $$dotx = Ax+f(t)$$ the solution to this is as follows $$x(t) = [e^At]x(0)+int_0^t e^A(t-t')f(t')dt'$$
in your case the function is just a constant function $f(t)=b$ for which you can see that gives the same solution as before! These methods are good because they let you solve every system of equations of this type because all the informations about the system is embedded into the exponential of the matrix
answered Jul 21 at 17:49
Davide Morgante
1,812220
add a comment |Â
up vote
4
down vote
Being $b$ a fixed constant (suppose $det Aneq0$), we can choose a new variable $x_*$ that solves $$Ax_* + b = 0 Rightarrow x_* = -A^-1b$$ The now we can solve the equation in the fixed point $z = x-x_* = x+A^-1b$. This variable solves the following equation $$dotz = Az$$
In fact, form the definition of $z$ $$dotz=dotx\ Az = A(x+A^-1b) = Ax+b$$
So we're saying that, if $x$ is a solution to $dotx=Ax+b$ then the variable $z = x+A^-1b$ is a solution to the simpler equation $dotz=Az$.
The solution to this simpler equation is just $$z(t)= [e^At]z(0)$$ where $[e^At]$ is the exponential of the matrix $A$. At this point you can go back to find $x(t)$ simply by using the definition of $z$: $$x(t) = z(t)-A^-1b = [e^At](x(0)+A^-1b)-A^-1b = [e^At]x(0)+A^-1b([e^At]-1)$$ where we have used the fact that $$z(0) = x(0) + A^-1b$$
Easier way
You can even use this simpler method: given the system of equations $$dotx = Ax+f(t)$$ the solution to this is as follows $$x(t) = [e^At]x(0)+int_0^t e^A(t-t')f(t')dt'$$
in your case the function is just a constant function $f(t)=b$ for which you can see that gives the same solution as before! These methods are good because they let you solve every system of equations of this type because all the informations about the system is embedded into the exponential of the matrix
answered Jul 21 at 17:49
Davide Morgante
1,812220
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Being $b$ a fixed constant (suppose $det Aneq0$), we can choose a new variable $x_*$ that solves $$Ax_* + b = 0 Rightarrow x_* = -A^-1b$$ The now we can solve the equation in the fixed point $z = x-x_* = x+A^-1b$. This variable solves the following equation $$dotz = Az$$
In fact, form the definition of $z$ $$dotz=dotx\ Az = A(x+A^-1b) = Ax+b$$
So we're saying that, if $x$ is a solution to $dotx=Ax+b$ then the variable $z = x+A^-1b$ is a solution to the simpler equation $dotz=Az$.
The solution to this simpler equation is just $$z(t)= [e^At]z(0)$$ where $[e^At]$ is the exponential of the matrix $A$. At this point you can go back to find $x(t)$ simply by using the definition of $z$: $$x(t) = z(t)-A^-1b = [e^At](x(0)+A^-1b)-A^-1b = [e^At]x(0)+A^-1b([e^At]-1)$$ where we have used the fact that $$z(0) = x(0) + A^-1b$$
Easier way
You can even use this simpler method: given the system of equations $$dotx = Ax+f(t)$$ the solution to this is as follows $$x(t) = [e^At]x(0)+int_0^t e^A(t-t')f(t')dt'$$
in your case the function is just a constant function $f(t)=b$ for which you can see that gives the same solution as before! These methods are good because they let you solve every system of equations of this type because all the informations about the system is embedded into the exponential of the matrix
answered Jul 21 at 17:49
Davide Morgante
1,812220
Being $b$ a fixed constant (suppose $det Aneq0$), we can choose a new variable $x_*$ that solves $$Ax_* + b = 0 Rightarrow x_* = -A^-1b$$ The now we can solve the equation in the fixed point $z = x-x_* = x+A^-1b$. This variable solves the following equation $$dotz = Az$$
In fact, form the definition of $z$ $$dotz=dotx\ Az = A(x+A^-1b) = Ax+b$$
So we're saying that, if $x$ is a solution to $dotx=Ax+b$ then the variable $z = x+A^-1b$ is a solution to the simpler equation $dotz=Az$.
The solution to this simpler equation is just $$z(t)= [e^At]z(0)$$ where $[e^At]$ is the exponential of the matrix $A$. At this point you can go back to find $x(t)$ simply by using the definition of $z$: $$x(t) = z(t)-A^-1b = [e^At](x(0)+A^-1b)-A^-1b = [e^At]x(0)+A^-1b([e^At]-1)$$ where we have used the fact that $$z(0) = x(0) + A^-1b$$
Easier way
You can even use this simpler method: given the system of equations $$dotx = Ax+f(t)$$ the solution to this is as follows $$x(t) = [e^At]x(0)+int_0^t e^A(t-t')f(t')dt'$$
in your case the function is just a constant function $f(t)=b$ for which you can see that gives the same solution as before! These methods are good because they let you solve every system of equations of this type because all the informations about the system is embedded into the exponential of the matrix
answered Jul 21 at 17:49
Davide Morgante
1,812220
edited Jul 21 at 17:55
answered Jul 21 at 17:49
Davide Morgante
1,812220
answered Jul 21 at 17:49
Davide Morgante
1,812220
answered Jul 21 at 17:49
Davide Morgante
1,812220
1,812220
add a comment |Â
add a comment |Â
up vote
3
down vote
If $b$ is the columnspace of $A$, define $y$ so that $Ay = -b$. Then note that $x(t) = e^Atx(0) + y$ solves the differential equation since $x' = A e^Atx(0)$ and $Ax = A e^Atx(0) - b$.
answered Jul 21 at 17:48
Marcus M
8,1731847
add a comment |Â
up vote
3
down vote
If $b$ is the columnspace of $A$, define $y$ so that $Ay = -b$. Then note that $x(t) = e^Atx(0) + y$ solves the differential equation since $x' = A e^Atx(0)$ and $Ax = A e^Atx(0) - b$.
answered Jul 21 at 17:48
Marcus M
8,1731847
add a comment |Â
up vote
3
down vote
up vote
3
down vote
If $b$ is the columnspace of $A$, define $y$ so that $Ay = -b$. Then note that $x(t) = e^Atx(0) + y$ solves the differential equation since $x' = A e^Atx(0)$ and $Ax = A e^Atx(0) - b$.
answered Jul 21 at 17:48
Marcus M
8,1731847
If $b$ is the columnspace of $A$, define $y$ so that $Ay = -b$. Then note that $x(t) = e^Atx(0) + y$ solves the differential equation since $x' = A e^Atx(0)$ and $Ax = A e^Atx(0) - b$.
answered Jul 21 at 17:48
Marcus M
8,1731847
answered Jul 21 at 17:48
Marcus M
8,1731847
answered Jul 21 at 17:48
Marcus M
8,1731847
answered Jul 21 at 17:48
Marcus M
8,1731847
8,1731847
add a comment |Â
add a comment |Â
up vote
2
down vote
Take Laplace transform therefore $$sX(s)-x(0)=AX(s)+b\(sI-A)X(s)=b+x(0)$$or$$X(s)=(sI-A)^-1(b+x(0))$$but we can't proceed any further. We need to know the characteristics of the system.
answered Jul 21 at 17:44
Mostafa Ayaz
8,5773630
add a comment |Â
up vote
2
down vote
Take Laplace transform therefore $$sX(s)-x(0)=AX(s)+b\(sI-A)X(s)=b+x(0)$$or$$X(s)=(sI-A)^-1(b+x(0))$$but we can't proceed any further. We need to know the characteristics of the system.
answered Jul 21 at 17:44
Mostafa Ayaz
8,5773630
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Take Laplace transform therefore $$sX(s)-x(0)=AX(s)+b\(sI-A)X(s)=b+x(0)$$or$$X(s)=(sI-A)^-1(b+x(0))$$but we can't proceed any further. We need to know the characteristics of the system.
answered Jul 21 at 17:44
Mostafa Ayaz
8,5773630
Take Laplace transform therefore $$sX(s)-x(0)=AX(s)+b\(sI-A)X(s)=b+x(0)$$or$$X(s)=(sI-A)^-1(b+x(0))$$but we can't proceed any further. We need to know the characteristics of the system.
answered Jul 21 at 17:44
Mostafa Ayaz
8,5773630
answered Jul 21 at 17:44
Mostafa Ayaz
8,5773630
answered Jul 21 at 17:44
Mostafa Ayaz
8,5773630
answered Jul 21 at 17:44
Mostafa Ayaz
8,5773630
8,5773630
add a comment |Â
add a comment |Â
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Thank you very much. All the answers are really helpful.
– Mridul Dey
Jul 21 at 18:08
You're welcome! If you feel like one of the answers is what you were looking for, you should mark it as accepted!
– Davide Morgante
Jul 21 at 19:36