Evaluate $dotv=-fracgammamv+g$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How to solve this ODE: $dfracdotysqrt1+doty^2=c$?solution of a ODE with a function of $dotx$Analytic approximation of $ddot x+gamma sign(dot x)+x=0$Second Order Inhomogenous Differential EquationEvaluate function for observed $x, dotx$.Solving $rmdot v = frac q m v times B + frac q m E$Solving system of two inhomogenous ODEsSolve for $theta$ in $sum M_y = - I dotTheta(dotpsi cos(theta)+p) $Solution of differential equation with two variableSolution of a differential equations
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Evaluate $dotv=-fracgammamv+g$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How to solve this ODE: $dfracdotysqrt1+doty^2=c$?solution of a ODE with a function of $dotx$Analytic approximation of $ddot x+gamma sign(dot x)+x=0$Second Order Inhomogenous Differential EquationEvaluate function for observed $x, dotx$.Solving $rmdot v = frac q m v times B + frac q m E$Solving system of two inhomogenous ODEsSolve for $theta$ in $sum M_y = - I dotTheta(dotpsi cos(theta)+p) $Solution of differential equation with two variableSolution of a differential equations
$begingroup$
Solve: $$dotv=-fracgammamv+g$$
Where $gamma,m,g$ are constants
I have to to separate the equation as $dotv=fracdvdt$
and got to:
$$fracdv-gv=-fracgammamdt$$
ordinary-differential-equations physics
asked Apr 2 at 15:45
newherenewhere
885411
$endgroup$
add a comment |
$begingroup$
Solve: $$dotv=-fracgammamv+g$$
Where $gamma,m,g$ are constants
I have to to separate the equation as $dotv=fracdvdt$
and got to:
$$fracdv-gv=-fracgammamdt$$
ordinary-differential-equations physics
asked Apr 2 at 15:45
newherenewhere
885411
$endgroup$
3
$begingroup$
What about dividing by r.h.s? $$ fractext d vg - fracgammam v = text d t $$
$endgroup$
– Harry49
Apr 2 at 15:50
$begingroup$
@Harry49 so we get $fracln(g-fracgammamv)-fracgammam=dt$? integrating both sides and taking exponent gives $v=fracmgamma(g-e^t)$?
$endgroup$
– newhere
Apr 2 at 15:54
1
$begingroup$
@newhere close, but don't forget to do the reverse chain rule when you integrate. You're missing some constants in from of the $ln$. Also don't forget you get a constant of integration somewhere
$endgroup$
– John Doe
Apr 2 at 15:56
$begingroup$
I think it's easier to use the method of integrating factor. Or $v=v_h+v_p$.
$endgroup$
– Hans Lundmark
Apr 2 at 16:20
add a comment |
$begingroup$
Solve: $$dotv=-fracgammamv+g$$
Where $gamma,m,g$ are constants
I have to to separate the equation as $dotv=fracdvdt$
and got to:
$$fracdv-gv=-fracgammamdt$$
ordinary-differential-equations physics
asked Apr 2 at 15:45
newherenewhere
885411
$endgroup$
Solve: $$dotv=-fracgammamv+g$$
Where $gamma,m,g$ are constants
I have to to separate the equation as $dotv=fracdvdt$
and got to:
$$fracdv-gv=-fracgammamdt$$
ordinary-differential-equations physics
ordinary-differential-equations physics
asked Apr 2 at 15:45
newherenewhere
885411
asked Apr 2 at 15:45
newherenewhere
885411
asked Apr 2 at 15:45
newherenewhere
885411
asked Apr 2 at 15:45
newherenewhere
885411
asked Apr 2 at 15:45
newherenewhere
885411
885411
3
$begingroup$
What about dividing by r.h.s? $$ fractext d vg - fracgammam v = text d t $$
$endgroup$
– Harry49
Apr 2 at 15:50
$begingroup$
@Harry49 so we get $fracln(g-fracgammamv)-fracgammam=dt$? integrating both sides and taking exponent gives $v=fracmgamma(g-e^t)$?
$endgroup$
– newhere
Apr 2 at 15:54
1
$begingroup$
@newhere close, but don't forget to do the reverse chain rule when you integrate. You're missing some constants in from of the $ln$. Also don't forget you get a constant of integration somewhere
$endgroup$
– John Doe
Apr 2 at 15:56
$begingroup$
I think it's easier to use the method of integrating factor. Or $v=v_h+v_p$.
$endgroup$
– Hans Lundmark
Apr 2 at 16:20
add a comment |
3
$begingroup$
What about dividing by r.h.s? $$ fractext d vg - fracgammam v = text d t $$
$endgroup$
– Harry49
Apr 2 at 15:50
$begingroup$
@Harry49 so we get $fracln(g-fracgammamv)-fracgammam=dt$? integrating both sides and taking exponent gives $v=fracmgamma(g-e^t)$?
$endgroup$
– newhere
Apr 2 at 15:54
1
$begingroup$
@newhere close, but don't forget to do the reverse chain rule when you integrate. You're missing some constants in from of the $ln$. Also don't forget you get a constant of integration somewhere
$endgroup$
– John Doe
Apr 2 at 15:56
$begingroup$
I think it's easier to use the method of integrating factor. Or $v=v_h+v_p$.
$endgroup$
– Hans Lundmark
Apr 2 at 16:20
3
3
$begingroup$
What about dividing by r.h.s? $$ fractext d vg - fracgammam v = text d t $$
$endgroup$
– Harry49
Apr 2 at 15:50
$begingroup$
What about dividing by r.h.s? $$ fractext d vg - fracgammam v = text d t $$
$endgroup$
– Harry49
Apr 2 at 15:50
$begingroup$
@Harry49 so we get $fracln(g-fracgammamv)-fracgammam=dt$? integrating both sides and taking exponent gives $v=fracmgamma(g-e^t)$?
$endgroup$
– newhere
Apr 2 at 15:54
$begingroup$
@Harry49 so we get $fracln(g-fracgammamv)-fracgammam=dt$? integrating both sides and taking exponent gives $v=fracmgamma(g-e^t)$?
$endgroup$
– newhere
Apr 2 at 15:54
1
1
$begingroup$
@newhere close, but don't forget to do the reverse chain rule when you integrate. You're missing some constants in from of the $ln$. Also don't forget you get a constant of integration somewhere
$endgroup$
– John Doe
Apr 2 at 15:56
$begingroup$
@newhere close, but don't forget to do the reverse chain rule when you integrate. You're missing some constants in from of the $ln$. Also don't forget you get a constant of integration somewhere
$endgroup$
– John Doe
Apr 2 at 15:56
$begingroup$
I think it's easier to use the method of integrating factor. Or $v=v_h+v_p$.
$endgroup$
– Hans Lundmark
Apr 2 at 16:20
$begingroup$
I think it's easier to use the method of integrating factor. Or $v=v_h+v_p$.
$endgroup$
– Hans Lundmark
Apr 2 at 16:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your rearrangement is incorrect; what you can condlude is that $fracdv/dt-gv=-fracgammam$, but note that cross-multiplying by $dt$ changes $g$ to $gdt$. Harry49 has already explained how to solve the problem by separation, so I'll mention another method, that of integration factors.
Our original equation is $dotv+Pv=Q$ with $P:=fracgammam,,Q:=g$, neither of which depend on $v$. The procedure I'm about to describe does require this, but it allows for a $t$-dependence.
Write $R:=expint Pdt$ so $R^prime=RP,,(Rv)^prime=RQ,,v=R^-1int RQdt$. With indefinite integration $P$ ($R$) is determined up to an additive (multiplicative) constant, while $v$ is determined up to $CR$ with $C$ an integration constant, regardless of which antiderivative of $P$ is chosen in defining $R$.
In the case at hand $v=e^-gamma t/mint ge^gamma t/m dt=Ce^-gamma t/m+fracmggamma$. If $v_0$ denotes the $t=0$ value of $v$, $v=(v_0-fracmggamma)e^-gamma t/m+fracmggamma$.
Edit: as for separation, we get $t-t_0=-fracmgammaln |g-fracgamma vm|implies v=gmpfracmgammaexpfracgamma (t-t_0)m$, which can be manipulated in the same way.
answered Apr 2 at 16:03
J.G.J.G.
34.1k23252
$endgroup$
$begingroup$
Taking $R=e^fracgammamt$ I got $v=fracmggamma$ is it correct?
$endgroup$
– newhere
Apr 3 at 13:43
$begingroup$
@newhere Your result is just a constant. You should get $e^-gamma t/mint ge^gamma t/mdt$, but bear in mind the integral includes an integration constant.
$endgroup$
– J.G.
Apr 3 at 13:50
$begingroup$
$dotv+fracgammamv=g$ taking integration factor $R=e^int fracgammamdt=e^fracgammamt$ so $e^ fracgammamtdotv+e^fracgammamtfracgammamv=e^ fracgammamtg$ setting $k=fracgammam$ we have $e^ktdotv+e^ktkv=e^ ktg$ integrating both sides we get $e^ktv=frace^ ktkg+c$
$endgroup$
– newhere
Apr 3 at 14:10
1
$begingroup$
@newhere With $R=expfracgamma tm$ we have $int RQ dt=fracmggammaexpfracgamma tm+C$ so $v=fracmggamma+Cexpfrac-gamma tm$.
$endgroup$
– J.G.
Apr 3 at 14:12
1
$begingroup$
Just found where I lost the constant, thanks!!!
$endgroup$
– newhere
Apr 3 at 14:13
add a comment |
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$begingroup$
Your rearrangement is incorrect; what you can condlude is that $fracdv/dt-gv=-fracgammam$, but note that cross-multiplying by $dt$ changes $g$ to $gdt$. Harry49 has already explained how to solve the problem by separation, so I'll mention another method, that of integration factors.
Our original equation is $dotv+Pv=Q$ with $P:=fracgammam,,Q:=g$, neither of which depend on $v$. The procedure I'm about to describe does require this, but it allows for a $t$-dependence.
Write $R:=expint Pdt$ so $R^prime=RP,,(Rv)^prime=RQ,,v=R^-1int RQdt$. With indefinite integration $P$ ($R$) is determined up to an additive (multiplicative) constant, while $v$ is determined up to $CR$ with $C$ an integration constant, regardless of which antiderivative of $P$ is chosen in defining $R$.
In the case at hand $v=e^-gamma t/mint ge^gamma t/m dt=Ce^-gamma t/m+fracmggamma$. If $v_0$ denotes the $t=0$ value of $v$, $v=(v_0-fracmggamma)e^-gamma t/m+fracmggamma$.
Edit: as for separation, we get $t-t_0=-fracmgammaln |g-fracgamma vm|implies v=gmpfracmgammaexpfracgamma (t-t_0)m$, which can be manipulated in the same way.
answered Apr 2 at 16:03
J.G.J.G.
34.1k23252
$endgroup$
$begingroup$
Taking $R=e^fracgammamt$ I got $v=fracmggamma$ is it correct?
$endgroup$
– newhere
Apr 3 at 13:43
$begingroup$
@newhere Your result is just a constant. You should get $e^-gamma t/mint ge^gamma t/mdt$, but bear in mind the integral includes an integration constant.
$endgroup$
– J.G.
Apr 3 at 13:50
$begingroup$
$dotv+fracgammamv=g$ taking integration factor $R=e^int fracgammamdt=e^fracgammamt$ so $e^ fracgammamtdotv+e^fracgammamtfracgammamv=e^ fracgammamtg$ setting $k=fracgammam$ we have $e^ktdotv+e^ktkv=e^ ktg$ integrating both sides we get $e^ktv=frace^ ktkg+c$
$endgroup$
– newhere
Apr 3 at 14:10
1
$begingroup$
@newhere With $R=expfracgamma tm$ we have $int RQ dt=fracmggammaexpfracgamma tm+C$ so $v=fracmggamma+Cexpfrac-gamma tm$.
$endgroup$
– J.G.
Apr 3 at 14:12
1
$begingroup$
Just found where I lost the constant, thanks!!!
$endgroup$
– newhere
Apr 3 at 14:13
add a comment |
$begingroup$
Your rearrangement is incorrect; what you can condlude is that $fracdv/dt-gv=-fracgammam$, but note that cross-multiplying by $dt$ changes $g$ to $gdt$. Harry49 has already explained how to solve the problem by separation, so I'll mention another method, that of integration factors.
Our original equation is $dotv+Pv=Q$ with $P:=fracgammam,,Q:=g$, neither of which depend on $v$. The procedure I'm about to describe does require this, but it allows for a $t$-dependence.
Write $R:=expint Pdt$ so $R^prime=RP,,(Rv)^prime=RQ,,v=R^-1int RQdt$. With indefinite integration $P$ ($R$) is determined up to an additive (multiplicative) constant, while $v$ is determined up to $CR$ with $C$ an integration constant, regardless of which antiderivative of $P$ is chosen in defining $R$.
In the case at hand $v=e^-gamma t/mint ge^gamma t/m dt=Ce^-gamma t/m+fracmggamma$. If $v_0$ denotes the $t=0$ value of $v$, $v=(v_0-fracmggamma)e^-gamma t/m+fracmggamma$.
Edit: as for separation, we get $t-t_0=-fracmgammaln |g-fracgamma vm|implies v=gmpfracmgammaexpfracgamma (t-t_0)m$, which can be manipulated in the same way.
answered Apr 2 at 16:03
J.G.J.G.
34.1k23252
$endgroup$
$begingroup$
Taking $R=e^fracgammamt$ I got $v=fracmggamma$ is it correct?
$endgroup$
– newhere
Apr 3 at 13:43
$begingroup$
@newhere Your result is just a constant. You should get $e^-gamma t/mint ge^gamma t/mdt$, but bear in mind the integral includes an integration constant.
$endgroup$
– J.G.
Apr 3 at 13:50
$begingroup$
$dotv+fracgammamv=g$ taking integration factor $R=e^int fracgammamdt=e^fracgammamt$ so $e^ fracgammamtdotv+e^fracgammamtfracgammamv=e^ fracgammamtg$ setting $k=fracgammam$ we have $e^ktdotv+e^ktkv=e^ ktg$ integrating both sides we get $e^ktv=frace^ ktkg+c$
$endgroup$
– newhere
Apr 3 at 14:10
1
$begingroup$
@newhere With $R=expfracgamma tm$ we have $int RQ dt=fracmggammaexpfracgamma tm+C$ so $v=fracmggamma+Cexpfrac-gamma tm$.
$endgroup$
– J.G.
Apr 3 at 14:12
1
$begingroup$
Just found where I lost the constant, thanks!!!
$endgroup$
– newhere
Apr 3 at 14:13
add a comment |
$begingroup$
Your rearrangement is incorrect; what you can condlude is that $fracdv/dt-gv=-fracgammam$, but note that cross-multiplying by $dt$ changes $g$ to $gdt$. Harry49 has already explained how to solve the problem by separation, so I'll mention another method, that of integration factors.
Our original equation is $dotv+Pv=Q$ with $P:=fracgammam,,Q:=g$, neither of which depend on $v$. The procedure I'm about to describe does require this, but it allows for a $t$-dependence.
Write $R:=expint Pdt$ so $R^prime=RP,,(Rv)^prime=RQ,,v=R^-1int RQdt$. With indefinite integration $P$ ($R$) is determined up to an additive (multiplicative) constant, while $v$ is determined up to $CR$ with $C$ an integration constant, regardless of which antiderivative of $P$ is chosen in defining $R$.
In the case at hand $v=e^-gamma t/mint ge^gamma t/m dt=Ce^-gamma t/m+fracmggamma$. If $v_0$ denotes the $t=0$ value of $v$, $v=(v_0-fracmggamma)e^-gamma t/m+fracmggamma$.
Edit: as for separation, we get $t-t_0=-fracmgammaln |g-fracgamma vm|implies v=gmpfracmgammaexpfracgamma (t-t_0)m$, which can be manipulated in the same way.
answered Apr 2 at 16:03
J.G.J.G.
34.1k23252
$endgroup$
Your rearrangement is incorrect; what you can condlude is that $fracdv/dt-gv=-fracgammam$, but note that cross-multiplying by $dt$ changes $g$ to $gdt$. Harry49 has already explained how to solve the problem by separation, so I'll mention another method, that of integration factors.
Our original equation is $dotv+Pv=Q$ with $P:=fracgammam,,Q:=g$, neither of which depend on $v$. The procedure I'm about to describe does require this, but it allows for a $t$-dependence.
Write $R:=expint Pdt$ so $R^prime=RP,,(Rv)^prime=RQ,,v=R^-1int RQdt$. With indefinite integration $P$ ($R$) is determined up to an additive (multiplicative) constant, while $v$ is determined up to $CR$ with $C$ an integration constant, regardless of which antiderivative of $P$ is chosen in defining $R$.
In the case at hand $v=e^-gamma t/mint ge^gamma t/m dt=Ce^-gamma t/m+fracmggamma$. If $v_0$ denotes the $t=0$ value of $v$, $v=(v_0-fracmggamma)e^-gamma t/m+fracmggamma$.
Edit: as for separation, we get $t-t_0=-fracmgammaln |g-fracgamma vm|implies v=gmpfracmgammaexpfracgamma (t-t_0)m$, which can be manipulated in the same way.
answered Apr 2 at 16:03
J.G.J.G.
34.1k23252
edited Apr 3 at 13:51
answered Apr 2 at 16:03
J.G.J.G.
34.1k23252
answered Apr 2 at 16:03
J.G.J.G.
34.1k23252
answered Apr 2 at 16:03
J.G.J.G.
34.1k23252
34.1k23252
$begingroup$
Taking $R=e^fracgammamt$ I got $v=fracmggamma$ is it correct?
$endgroup$
– newhere
Apr 3 at 13:43
$begingroup$
@newhere Your result is just a constant. You should get $e^-gamma t/mint ge^gamma t/mdt$, but bear in mind the integral includes an integration constant.
$endgroup$
– J.G.
Apr 3 at 13:50
$begingroup$
$dotv+fracgammamv=g$ taking integration factor $R=e^int fracgammamdt=e^fracgammamt$ so $e^ fracgammamtdotv+e^fracgammamtfracgammamv=e^ fracgammamtg$ setting $k=fracgammam$ we have $e^ktdotv+e^ktkv=e^ ktg$ integrating both sides we get $e^ktv=frace^ ktkg+c$
$endgroup$
– newhere
Apr 3 at 14:10
1
$begingroup$
@newhere With $R=expfracgamma tm$ we have $int RQ dt=fracmggammaexpfracgamma tm+C$ so $v=fracmggamma+Cexpfrac-gamma tm$.
$endgroup$
– J.G.
Apr 3 at 14:12
1
$begingroup$
Just found where I lost the constant, thanks!!!
$endgroup$
– newhere
Apr 3 at 14:13
add a comment |
$begingroup$
Taking $R=e^fracgammamt$ I got $v=fracmggamma$ is it correct?
$endgroup$
– newhere
Apr 3 at 13:43
$begingroup$
@newhere Your result is just a constant. You should get $e^-gamma t/mint ge^gamma t/mdt$, but bear in mind the integral includes an integration constant.
$endgroup$
– J.G.
Apr 3 at 13:50
$begingroup$
$dotv+fracgammamv=g$ taking integration factor $R=e^int fracgammamdt=e^fracgammamt$ so $e^ fracgammamtdotv+e^fracgammamtfracgammamv=e^ fracgammamtg$ setting $k=fracgammam$ we have $e^ktdotv+e^ktkv=e^ ktg$ integrating both sides we get $e^ktv=frace^ ktkg+c$
$endgroup$
– newhere
Apr 3 at 14:10
1
$begingroup$
@newhere With $R=expfracgamma tm$ we have $int RQ dt=fracmggammaexpfracgamma tm+C$ so $v=fracmggamma+Cexpfrac-gamma tm$.
$endgroup$
– J.G.
Apr 3 at 14:12
1
$begingroup$
Just found where I lost the constant, thanks!!!
$endgroup$
– newhere
Apr 3 at 14:13
$begingroup$
Taking $R=e^fracgammamt$ I got $v=fracmggamma$ is it correct?
$endgroup$
– newhere
Apr 3 at 13:43
$begingroup$
Taking $R=e^fracgammamt$ I got $v=fracmggamma$ is it correct?
$endgroup$
– newhere
Apr 3 at 13:43
$begingroup$
@newhere Your result is just a constant. You should get $e^-gamma t/mint ge^gamma t/mdt$, but bear in mind the integral includes an integration constant.
$endgroup$
– J.G.
Apr 3 at 13:50
$begingroup$
@newhere Your result is just a constant. You should get $e^-gamma t/mint ge^gamma t/mdt$, but bear in mind the integral includes an integration constant.
$endgroup$
– J.G.
Apr 3 at 13:50
$begingroup$
$dotv+fracgammamv=g$ taking integration factor $R=e^int fracgammamdt=e^fracgammamt$ so $e^ fracgammamtdotv+e^fracgammamtfracgammamv=e^ fracgammamtg$ setting $k=fracgammam$ we have $e^ktdotv+e^ktkv=e^ ktg$ integrating both sides we get $e^ktv=frace^ ktkg+c$
$endgroup$
– newhere
Apr 3 at 14:10
$begingroup$
$dotv+fracgammamv=g$ taking integration factor $R=e^int fracgammamdt=e^fracgammamt$ so $e^ fracgammamtdotv+e^fracgammamtfracgammamv=e^ fracgammamtg$ setting $k=fracgammam$ we have $e^ktdotv+e^ktkv=e^ ktg$ integrating both sides we get $e^ktv=frace^ ktkg+c$
$endgroup$
– newhere
Apr 3 at 14:10
1
1
$begingroup$
@newhere With $R=expfracgamma tm$ we have $int RQ dt=fracmggammaexpfracgamma tm+C$ so $v=fracmggamma+Cexpfrac-gamma tm$.
$endgroup$
– J.G.
Apr 3 at 14:12
$begingroup$
@newhere With $R=expfracgamma tm$ we have $int RQ dt=fracmggammaexpfracgamma tm+C$ so $v=fracmggamma+Cexpfrac-gamma tm$.
$endgroup$
– J.G.
Apr 3 at 14:12
1
1
$begingroup$
Just found where I lost the constant, thanks!!!
$endgroup$
– newhere
Apr 3 at 14:13
$begingroup$
Just found where I lost the constant, thanks!!!
$endgroup$
– newhere
Apr 3 at 14:13
add a comment |
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3
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What about dividing by r.h.s? $$ fractext d vg - fracgammam v = text d t $$
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– Harry49
Apr 2 at 15:50
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@Harry49 so we get $fracln(g-fracgammamv)-fracgammam=dt$? integrating both sides and taking exponent gives $v=fracmgamma(g-e^t)$?
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– newhere
Apr 2 at 15:54
1
$begingroup$
@newhere close, but don't forget to do the reverse chain rule when you integrate. You're missing some constants in from of the $ln$. Also don't forget you get a constant of integration somewhere
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– John Doe
Apr 2 at 15:56
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I think it's easier to use the method of integrating factor. Or $v=v_h+v_p$.
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– Hans Lundmark
Apr 2 at 16:20