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Solution of differential equation $d/dt langle N(t) rangle=k-Gamma langle N(t) rangle$ where brackets indicate average
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I have the following ODE yielded as a steady state solution to a more complicated ODE. The important part is just to consider the following:
$$d/dt langle N(t) rangle=k-Gamma langle N(t) rangle$$
where $langle N(t) rangle$ can be seen as $y$ or an average of $y$ (brackets indicating avg). I am not sure if that is the reason for the result.
The solution is given by
$$langle N(t) rangle=k/Gamma (1-e^-Gamma t)+langle N(t) rangle e^-Gamma t$$
I do not see that however. So far, I've solved first for
$$d/dt <N(t)>=Gamma <N(t)>=0$$
Yielding
$$langle N(t) rangle>=Ccdot e^-Gamma t$$
and $C$
$$C=k/Gamma (1-e^-Gamma t)+c_0$$
I do not see how it solves the equation by inserting c.
Any help is highly appreciated :)
edit: I had a thought it may have to do with the average being additive, and starting and $langle N(0) rangle$ being the same as $langle N(t) rangle$ since it is steady state. I am not completely sure however.
differential-equations
asked Jul 20 at 18:11
user469216
445
add a comment |Â
up vote
2
down vote
favorite
I have the following ODE yielded as a steady state solution to a more complicated ODE. The important part is just to consider the following:
$$d/dt langle N(t) rangle=k-Gamma langle N(t) rangle$$
where $langle N(t) rangle$ can be seen as $y$ or an average of $y$ (brackets indicating avg). I am not sure if that is the reason for the result.
The solution is given by
$$langle N(t) rangle=k/Gamma (1-e^-Gamma t)+langle N(t) rangle e^-Gamma t$$
I do not see that however. So far, I've solved first for
$$d/dt <N(t)>=Gamma <N(t)>=0$$
Yielding
$$langle N(t) rangle>=Ccdot e^-Gamma t$$
and $C$
$$C=k/Gamma (1-e^-Gamma t)+c_0$$
I do not see how it solves the equation by inserting c.
Any help is highly appreciated :)
edit: I had a thought it may have to do with the average being additive, and starting and $langle N(0) rangle$ being the same as $langle N(t) rangle$ since it is steady state. I am not completely sure however.
differential-equations
asked Jul 20 at 18:11
user469216
445
Uselangleandranglefor angle brackets: $langle$ $rangle$
– Dylan
Jul 20 at 18:43
What is the $Gamma$?
– Botond
Jul 20 at 18:52
$Gamma$ is just a constant :)
– user469216
Jul 20 at 18:53
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have the following ODE yielded as a steady state solution to a more complicated ODE. The important part is just to consider the following:
$$d/dt langle N(t) rangle=k-Gamma langle N(t) rangle$$
where $langle N(t) rangle$ can be seen as $y$ or an average of $y$ (brackets indicating avg). I am not sure if that is the reason for the result.
The solution is given by
$$langle N(t) rangle=k/Gamma (1-e^-Gamma t)+langle N(t) rangle e^-Gamma t$$
I do not see that however. So far, I've solved first for
$$d/dt <N(t)>=Gamma <N(t)>=0$$
Yielding
$$langle N(t) rangle>=Ccdot e^-Gamma t$$
and $C$
$$C=k/Gamma (1-e^-Gamma t)+c_0$$
I do not see how it solves the equation by inserting c.
Any help is highly appreciated :)
edit: I had a thought it may have to do with the average being additive, and starting and $langle N(0) rangle$ being the same as $langle N(t) rangle$ since it is steady state. I am not completely sure however.
differential-equations
asked Jul 20 at 18:11
user469216
445
I have the following ODE yielded as a steady state solution to a more complicated ODE. The important part is just to consider the following:
$$d/dt langle N(t) rangle=k-Gamma langle N(t) rangle$$
where $langle N(t) rangle$ can be seen as $y$ or an average of $y$ (brackets indicating avg). I am not sure if that is the reason for the result.
The solution is given by
$$langle N(t) rangle=k/Gamma (1-e^-Gamma t)+langle N(t) rangle e^-Gamma t$$
I do not see that however. So far, I've solved first for
$$d/dt <N(t)>=Gamma <N(t)>=0$$
Yielding
$$langle N(t) rangle>=Ccdot e^-Gamma t$$
and $C$
$$C=k/Gamma (1-e^-Gamma t)+c_0$$
I do not see how it solves the equation by inserting c.
Any help is highly appreciated :)
edit: I had a thought it may have to do with the average being additive, and starting and $langle N(0) rangle$ being the same as $langle N(t) rangle$ since it is steady state. I am not completely sure however.
differential-equations
asked Jul 20 at 18:11
user469216
445
edited Jul 20 at 18:43
asked Jul 20 at 18:11
user469216
445
asked Jul 20 at 18:11
user469216
445
asked Jul 20 at 18:11
user469216
445
445
Uselangleandranglefor angle brackets: $langle$ $rangle$
– Dylan
Jul 20 at 18:43
What is the $Gamma$?
– Botond
Jul 20 at 18:52
$Gamma$ is just a constant :)
– user469216
Jul 20 at 18:53
add a comment |Â
Uselangleandranglefor angle brackets: $langle$ $rangle$
– Dylan
Jul 20 at 18:43
What is the $Gamma$?
– Botond
Jul 20 at 18:52
$Gamma$ is just a constant :)
– user469216
Jul 20 at 18:53
Use
langle and rangle for angle brackets: $langle$ $rangle$– Dylan
Jul 20 at 18:43
Use
langle and rangle for angle brackets: $langle$ $rangle$– Dylan
Jul 20 at 18:43
What is the $Gamma$?
– Botond
Jul 20 at 18:52
What is the $Gamma$?
– Botond
Jul 20 at 18:52
$Gamma$ is just a constant :)
– user469216
Jul 20 at 18:53
$Gamma$ is just a constant :)
– user469216
Jul 20 at 18:53
add a comment |Â
1 Answer
1
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up vote
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accepted
Let $y = langle N(t)rangle$, so that we may write
$$fracdydt +Gamma y = k$$ This equation can be solved by method of integrating factors. An integrating factor here is $e^int_0^t Gamma, ds = e^Gamma t$. Multiply the equation by $e^Gamma t$ and note that by the product rule, we obtain $$fracddt(e^Gamma t y) = ke^Gamma t$$ Integrate both sides to get $$e^Gamma ty = frackGammae^Gamma t + C$$where $C$ is a constant. Setting $t = 0$, $y(0) = k/Gamma + C$, so then $C = y(0) - k/Gamma$. Therefore $$y = frackGamma + left(y(0) - frackGammaright)e^-Gamma t= frackGammaleft(1 - e^-Gamma tright) + y(0)e^-Gamma t$$ as desired.
answered Jul 20 at 18:51
kobe
33.9k22146
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $y = langle N(t)rangle$, so that we may write
$$fracdydt +Gamma y = k$$ This equation can be solved by method of integrating factors. An integrating factor here is $e^int_0^t Gamma, ds = e^Gamma t$. Multiply the equation by $e^Gamma t$ and note that by the product rule, we obtain $$fracddt(e^Gamma t y) = ke^Gamma t$$ Integrate both sides to get $$e^Gamma ty = frackGammae^Gamma t + C$$where $C$ is a constant. Setting $t = 0$, $y(0) = k/Gamma + C$, so then $C = y(0) - k/Gamma$. Therefore $$y = frackGamma + left(y(0) - frackGammaright)e^-Gamma t= frackGammaleft(1 - e^-Gamma tright) + y(0)e^-Gamma t$$ as desired.
answered Jul 20 at 18:51
kobe
33.9k22146
add a comment |Â
up vote
2
down vote
accepted
Let $y = langle N(t)rangle$, so that we may write
$$fracdydt +Gamma y = k$$ This equation can be solved by method of integrating factors. An integrating factor here is $e^int_0^t Gamma, ds = e^Gamma t$. Multiply the equation by $e^Gamma t$ and note that by the product rule, we obtain $$fracddt(e^Gamma t y) = ke^Gamma t$$ Integrate both sides to get $$e^Gamma ty = frackGammae^Gamma t + C$$where $C$ is a constant. Setting $t = 0$, $y(0) = k/Gamma + C$, so then $C = y(0) - k/Gamma$. Therefore $$y = frackGamma + left(y(0) - frackGammaright)e^-Gamma t= frackGammaleft(1 - e^-Gamma tright) + y(0)e^-Gamma t$$ as desired.
answered Jul 20 at 18:51
kobe
33.9k22146
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $y = langle N(t)rangle$, so that we may write
$$fracdydt +Gamma y = k$$ This equation can be solved by method of integrating factors. An integrating factor here is $e^int_0^t Gamma, ds = e^Gamma t$. Multiply the equation by $e^Gamma t$ and note that by the product rule, we obtain $$fracddt(e^Gamma t y) = ke^Gamma t$$ Integrate both sides to get $$e^Gamma ty = frackGammae^Gamma t + C$$where $C$ is a constant. Setting $t = 0$, $y(0) = k/Gamma + C$, so then $C = y(0) - k/Gamma$. Therefore $$y = frackGamma + left(y(0) - frackGammaright)e^-Gamma t= frackGammaleft(1 - e^-Gamma tright) + y(0)e^-Gamma t$$ as desired.
answered Jul 20 at 18:51
kobe
33.9k22146
Let $y = langle N(t)rangle$, so that we may write
$$fracdydt +Gamma y = k$$ This equation can be solved by method of integrating factors. An integrating factor here is $e^int_0^t Gamma, ds = e^Gamma t$. Multiply the equation by $e^Gamma t$ and note that by the product rule, we obtain $$fracddt(e^Gamma t y) = ke^Gamma t$$ Integrate both sides to get $$e^Gamma ty = frackGammae^Gamma t + C$$where $C$ is a constant. Setting $t = 0$, $y(0) = k/Gamma + C$, so then $C = y(0) - k/Gamma$. Therefore $$y = frackGamma + left(y(0) - frackGammaright)e^-Gamma t= frackGammaleft(1 - e^-Gamma tright) + y(0)e^-Gamma t$$ as desired.
answered Jul 20 at 18:51
kobe
33.9k22146
answered Jul 20 at 18:51
kobe
33.9k22146
answered Jul 20 at 18:51
kobe
33.9k22146
answered Jul 20 at 18:51
kobe
33.9k22146
33.9k22146
add a comment |Â
add a comment |Â
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Use
langleandranglefor angle brackets: $langle$ $rangle$– Dylan
Jul 20 at 18:43
What is the $Gamma$?
– Botond
Jul 20 at 18:52
$Gamma$ is just a constant :)
– user469216
Jul 20 at 18:53