How to solve the Lienard equation $v'' = av -bvv'$?












1












$begingroup$


For a Physics project that I'm doing in school, I have the following differential equation:



$$v'' = av -bvv'$$



where $v$ is the velocity, $v'$ is the rate of change of velocity with respect to time, and $v''$ is the second derivative of the velocity with respect to time.
$a$ and $b$ are both positive constants.
I also know the initial velocity and the initial rate of change of velocity when $t=0$.



It's really important that I solve it, and I'm not able to find the solution anywhere.



This is really urgent, and I hope one of you guys can solve this equation soon.
Thanks






ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What have you tried so far? I believe the intention of whoever assigned this problem to you was for you to do it yourself, not find a solution online. Please include any work you have tried and what you are having difficulty with in your question. Thanks!
    $endgroup$
    – Aniruddh Venkatesan
    Jan 19 at 4:50












  • $begingroup$
    The equation has not been given to me. I created it for a physics paper that I am writing, but I'm stuck and am not able to solve it. Any help from you would be much appreciated
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:12










  • $begingroup$
    I see. Have you taken a class in Differential Equations or is the first time you're seeing them?
    $endgroup$
    – Aniruddh Venkatesan
    Jan 19 at 5:14










  • $begingroup$
    No, I have not taken any classes on it. But I have studied basic first order differential equations
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:19










  • $begingroup$
    Because of the $v v_{}^prime$ term, this is not a linear differential equation.
    $endgroup$
    – John Wayland Bales
    Jan 19 at 5:21


















1












$begingroup$


For a Physics project that I'm doing in school, I have the following differential equation:



$$v'' = av -bvv'$$



where $v$ is the velocity, $v'$ is the rate of change of velocity with respect to time, and $v''$ is the second derivative of the velocity with respect to time.
$a$ and $b$ are both positive constants.
I also know the initial velocity and the initial rate of change of velocity when $t=0$.



It's really important that I solve it, and I'm not able to find the solution anywhere.



This is really urgent, and I hope one of you guys can solve this equation soon.
Thanks






ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What have you tried so far? I believe the intention of whoever assigned this problem to you was for you to do it yourself, not find a solution online. Please include any work you have tried and what you are having difficulty with in your question. Thanks!
    $endgroup$
    – Aniruddh Venkatesan
    Jan 19 at 4:50












  • $begingroup$
    The equation has not been given to me. I created it for a physics paper that I am writing, but I'm stuck and am not able to solve it. Any help from you would be much appreciated
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:12










  • $begingroup$
    I see. Have you taken a class in Differential Equations or is the first time you're seeing them?
    $endgroup$
    – Aniruddh Venkatesan
    Jan 19 at 5:14










  • $begingroup$
    No, I have not taken any classes on it. But I have studied basic first order differential equations
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:19










  • $begingroup$
    Because of the $v v_{}^prime$ term, this is not a linear differential equation.
    $endgroup$
    – John Wayland Bales
    Jan 19 at 5:21
















1












1








1





$begingroup$


For a Physics project that I'm doing in school, I have the following differential equation:



$$v'' = av -bvv'$$



where $v$ is the velocity, $v'$ is the rate of change of velocity with respect to time, and $v''$ is the second derivative of the velocity with respect to time.
$a$ and $b$ are both positive constants.
I also know the initial velocity and the initial rate of change of velocity when $t=0$.



It's really important that I solve it, and I'm not able to find the solution anywhere.



This is really urgent, and I hope one of you guys can solve this equation soon.
Thanks






ordinary-differential-equations







share|cite|improve this question











$endgroup$




For a Physics project that I'm doing in school, I have the following differential equation:



$$v'' = av -bvv'$$



where $v$ is the velocity, $v'$ is the rate of change of velocity with respect to time, and $v''$ is the second derivative of the velocity with respect to time.
$a$ and $b$ are both positive constants.
I also know the initial velocity and the initial rate of change of velocity when $t=0$.



It's really important that I solve it, and I'm not able to find the solution anywhere.



This is really urgent, and I hope one of you guys can solve this equation soon.
Thanks






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 19 at 9:04









Rodrigo de Azevedo

13k41958




13k41958










asked Jan 19 at 4:43









Rama VelidandaRama Velidanda

111




111








  • 2




    $begingroup$
    What have you tried so far? I believe the intention of whoever assigned this problem to you was for you to do it yourself, not find a solution online. Please include any work you have tried and what you are having difficulty with in your question. Thanks!
    $endgroup$
    – Aniruddh Venkatesan
    Jan 19 at 4:50












  • $begingroup$
    The equation has not been given to me. I created it for a physics paper that I am writing, but I'm stuck and am not able to solve it. Any help from you would be much appreciated
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:12










  • $begingroup$
    I see. Have you taken a class in Differential Equations or is the first time you're seeing them?
    $endgroup$
    – Aniruddh Venkatesan
    Jan 19 at 5:14










  • $begingroup$
    No, I have not taken any classes on it. But I have studied basic first order differential equations
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:19










  • $begingroup$
    Because of the $v v_{}^prime$ term, this is not a linear differential equation.
    $endgroup$
    – John Wayland Bales
    Jan 19 at 5:21
















  • 2




    $begingroup$
    What have you tried so far? I believe the intention of whoever assigned this problem to you was for you to do it yourself, not find a solution online. Please include any work you have tried and what you are having difficulty with in your question. Thanks!
    $endgroup$
    – Aniruddh Venkatesan
    Jan 19 at 4:50












  • $begingroup$
    The equation has not been given to me. I created it for a physics paper that I am writing, but I'm stuck and am not able to solve it. Any help from you would be much appreciated
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:12










  • $begingroup$
    I see. Have you taken a class in Differential Equations or is the first time you're seeing them?
    $endgroup$
    – Aniruddh Venkatesan
    Jan 19 at 5:14










  • $begingroup$
    No, I have not taken any classes on it. But I have studied basic first order differential equations
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:19










  • $begingroup$
    Because of the $v v_{}^prime$ term, this is not a linear differential equation.
    $endgroup$
    – John Wayland Bales
    Jan 19 at 5:21










2




2




$begingroup$
What have you tried so far? I believe the intention of whoever assigned this problem to you was for you to do it yourself, not find a solution online. Please include any work you have tried and what you are having difficulty with in your question. Thanks!
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 4:50






$begingroup$
What have you tried so far? I believe the intention of whoever assigned this problem to you was for you to do it yourself, not find a solution online. Please include any work you have tried and what you are having difficulty with in your question. Thanks!
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 4:50














$begingroup$
The equation has not been given to me. I created it for a physics paper that I am writing, but I'm stuck and am not able to solve it. Any help from you would be much appreciated
$endgroup$
– Rama Velidanda
Jan 19 at 5:12




$begingroup$
The equation has not been given to me. I created it for a physics paper that I am writing, but I'm stuck and am not able to solve it. Any help from you would be much appreciated
$endgroup$
– Rama Velidanda
Jan 19 at 5:12












$begingroup$
I see. Have you taken a class in Differential Equations or is the first time you're seeing them?
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 5:14




$begingroup$
I see. Have you taken a class in Differential Equations or is the first time you're seeing them?
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 5:14












$begingroup$
No, I have not taken any classes on it. But I have studied basic first order differential equations
$endgroup$
– Rama Velidanda
Jan 19 at 5:19




$begingroup$
No, I have not taken any classes on it. But I have studied basic first order differential equations
$endgroup$
– Rama Velidanda
Jan 19 at 5:19












$begingroup$
Because of the $v v_{}^prime$ term, this is not a linear differential equation.
$endgroup$
– John Wayland Bales
Jan 19 at 5:21






$begingroup$
Because of the $v v_{}^prime$ term, this is not a linear differential equation.
$endgroup$
– John Wayland Bales
Jan 19 at 5:21












2 Answers
2






active

oldest

votes


















1












$begingroup$

Let $w(v) = v'(t)$. Then the equation should become $ww' = av-bwv$ (show this). Then you should be able to take it from here.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, But I'm not sure how to take it further as there are now 3 variables in the equation. Sorry if I'm making an obvious mistake, I am new to this
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:10










  • $begingroup$
    No no, there are just two variables. Maybe it's safer to write the equation in the form $w frac{dw}{dv} = - b v w + a v$. Here, $v$ is the new independent variable and $w$ is the new dependent variable.
    $endgroup$
    – Christoph
    Jan 19 at 5:28










  • $begingroup$
    The next step would be to define another new independent variable $z := -frac{1}{2}b v^2$, $frac{d}{dv} = - b v frac{d}{dz}$. This will give you a separable first-order ODE for the function $w(z)$.
    $endgroup$
    – Christoph
    Jan 19 at 5:31












  • $begingroup$
    thank you, this helps. However, I still do not know how to solve an Abel equation. If you could solve it for me and show me your method, that would be great. If not, I'll try working with this.
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:39






  • 1




    $begingroup$
    Actually, the equation written down by AlexanderJ93 is already separable … I did one step too much.
    $endgroup$
    – Christoph
    Jan 19 at 5:52





















0












$begingroup$

First of all, let me remark that it seems impossible to obtain a closed-form solution in terms of elementary functions to this ordinary differential equation (ODE) in the general case.



Writing $v' = w(v)$ with a new function $w$, $v'' = w frac{dw}{dv}$, we obtain the first-order separable ODE
begin{equation}
w frac{dw}{dv} = a v - b v w = v (a - b w).
end{equation}
Assuming that $w not equiv frac{a}{b}$ and using separation of variables we obtain
begin{equation}
frac{w}{a-bw} , mathrm{d}w = v , mathrm{d}v quad Leftrightarrow quad lnleft|frac{b}{a}w-1right| + frac{b}{a} w - 1 = - frac{b^2}{2 a} v^2 + C_1, quad C_1 in mathbb{R}.
end{equation}
Therefore, the function $w(v)$ may be obtained in terms of the Lambert W function, which cannot be expressed in terms of elementary functions.



Finally, the solution of the separable ODE $frac{dv}{dt} = w(v)$ yields an implicit representation
begin{equation}
int frac{1}{w(v)} , mathrm{d}v = t
end{equation}
for $v$, with the function $w(v)$ from above, where the indefinite integral contains the second constant of integration. The values of the two constants of integration will be determined from the given initial values $v(0), v'(0)$.



Edit: The solution of the initial-value problem $v'' = a v - b v v'$, $v(0) = v_0$, $v'(0) = frac{a}{b}$, is given by the linear function $v(t) = frac{a}{b} t + v_0$. This is the special case of a constant $w equiv frac{a}{b}$, for which we don't need to do the separation of variables.






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Let $w(v) = v'(t)$. Then the equation should become $ww' = av-bwv$ (show this). Then you should be able to take it from here.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you, But I'm not sure how to take it further as there are now 3 variables in the equation. Sorry if I'm making an obvious mistake, I am new to this
      $endgroup$
      – Rama Velidanda
      Jan 19 at 5:10










    • $begingroup$
      No no, there are just two variables. Maybe it's safer to write the equation in the form $w frac{dw}{dv} = - b v w + a v$. Here, $v$ is the new independent variable and $w$ is the new dependent variable.
      $endgroup$
      – Christoph
      Jan 19 at 5:28










    • $begingroup$
      The next step would be to define another new independent variable $z := -frac{1}{2}b v^2$, $frac{d}{dv} = - b v frac{d}{dz}$. This will give you a separable first-order ODE for the function $w(z)$.
      $endgroup$
      – Christoph
      Jan 19 at 5:31












    • $begingroup$
      thank you, this helps. However, I still do not know how to solve an Abel equation. If you could solve it for me and show me your method, that would be great. If not, I'll try working with this.
      $endgroup$
      – Rama Velidanda
      Jan 19 at 5:39






    • 1




      $begingroup$
      Actually, the equation written down by AlexanderJ93 is already separable … I did one step too much.
      $endgroup$
      – Christoph
      Jan 19 at 5:52


















    1












    $begingroup$

    Let $w(v) = v'(t)$. Then the equation should become $ww' = av-bwv$ (show this). Then you should be able to take it from here.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you, But I'm not sure how to take it further as there are now 3 variables in the equation. Sorry if I'm making an obvious mistake, I am new to this
      $endgroup$
      – Rama Velidanda
      Jan 19 at 5:10










    • $begingroup$
      No no, there are just two variables. Maybe it's safer to write the equation in the form $w frac{dw}{dv} = - b v w + a v$. Here, $v$ is the new independent variable and $w$ is the new dependent variable.
      $endgroup$
      – Christoph
      Jan 19 at 5:28










    • $begingroup$
      The next step would be to define another new independent variable $z := -frac{1}{2}b v^2$, $frac{d}{dv} = - b v frac{d}{dz}$. This will give you a separable first-order ODE for the function $w(z)$.
      $endgroup$
      – Christoph
      Jan 19 at 5:31












    • $begingroup$
      thank you, this helps. However, I still do not know how to solve an Abel equation. If you could solve it for me and show me your method, that would be great. If not, I'll try working with this.
      $endgroup$
      – Rama Velidanda
      Jan 19 at 5:39






    • 1




      $begingroup$
      Actually, the equation written down by AlexanderJ93 is already separable … I did one step too much.
      $endgroup$
      – Christoph
      Jan 19 at 5:52
















    1












    1








    1





    $begingroup$

    Let $w(v) = v'(t)$. Then the equation should become $ww' = av-bwv$ (show this). Then you should be able to take it from here.






    share|cite|improve this answer









    $endgroup$



    Let $w(v) = v'(t)$. Then the equation should become $ww' = av-bwv$ (show this). Then you should be able to take it from here.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 19 at 4:54









    AlexanderJ93AlexanderJ93

    6,173823




    6,173823












    • $begingroup$
      Thank you, But I'm not sure how to take it further as there are now 3 variables in the equation. Sorry if I'm making an obvious mistake, I am new to this
      $endgroup$
      – Rama Velidanda
      Jan 19 at 5:10










    • $begingroup$
      No no, there are just two variables. Maybe it's safer to write the equation in the form $w frac{dw}{dv} = - b v w + a v$. Here, $v$ is the new independent variable and $w$ is the new dependent variable.
      $endgroup$
      – Christoph
      Jan 19 at 5:28










    • $begingroup$
      The next step would be to define another new independent variable $z := -frac{1}{2}b v^2$, $frac{d}{dv} = - b v frac{d}{dz}$. This will give you a separable first-order ODE for the function $w(z)$.
      $endgroup$
      – Christoph
      Jan 19 at 5:31












    • $begingroup$
      thank you, this helps. However, I still do not know how to solve an Abel equation. If you could solve it for me and show me your method, that would be great. If not, I'll try working with this.
      $endgroup$
      – Rama Velidanda
      Jan 19 at 5:39






    • 1




      $begingroup$
      Actually, the equation written down by AlexanderJ93 is already separable … I did one step too much.
      $endgroup$
      – Christoph
      Jan 19 at 5:52




















    • $begingroup$
      Thank you, But I'm not sure how to take it further as there are now 3 variables in the equation. Sorry if I'm making an obvious mistake, I am new to this
      $endgroup$
      – Rama Velidanda
      Jan 19 at 5:10










    • $begingroup$
      No no, there are just two variables. Maybe it's safer to write the equation in the form $w frac{dw}{dv} = - b v w + a v$. Here, $v$ is the new independent variable and $w$ is the new dependent variable.
      $endgroup$
      – Christoph
      Jan 19 at 5:28










    • $begingroup$
      The next step would be to define another new independent variable $z := -frac{1}{2}b v^2$, $frac{d}{dv} = - b v frac{d}{dz}$. This will give you a separable first-order ODE for the function $w(z)$.
      $endgroup$
      – Christoph
      Jan 19 at 5:31












    • $begingroup$
      thank you, this helps. However, I still do not know how to solve an Abel equation. If you could solve it for me and show me your method, that would be great. If not, I'll try working with this.
      $endgroup$
      – Rama Velidanda
      Jan 19 at 5:39






    • 1




      $begingroup$
      Actually, the equation written down by AlexanderJ93 is already separable … I did one step too much.
      $endgroup$
      – Christoph
      Jan 19 at 5:52


















    $begingroup$
    Thank you, But I'm not sure how to take it further as there are now 3 variables in the equation. Sorry if I'm making an obvious mistake, I am new to this
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:10




    $begingroup$
    Thank you, But I'm not sure how to take it further as there are now 3 variables in the equation. Sorry if I'm making an obvious mistake, I am new to this
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:10












    $begingroup$
    No no, there are just two variables. Maybe it's safer to write the equation in the form $w frac{dw}{dv} = - b v w + a v$. Here, $v$ is the new independent variable and $w$ is the new dependent variable.
    $endgroup$
    – Christoph
    Jan 19 at 5:28




    $begingroup$
    No no, there are just two variables. Maybe it's safer to write the equation in the form $w frac{dw}{dv} = - b v w + a v$. Here, $v$ is the new independent variable and $w$ is the new dependent variable.
    $endgroup$
    – Christoph
    Jan 19 at 5:28












    $begingroup$
    The next step would be to define another new independent variable $z := -frac{1}{2}b v^2$, $frac{d}{dv} = - b v frac{d}{dz}$. This will give you a separable first-order ODE for the function $w(z)$.
    $endgroup$
    – Christoph
    Jan 19 at 5:31






    $begingroup$
    The next step would be to define another new independent variable $z := -frac{1}{2}b v^2$, $frac{d}{dv} = - b v frac{d}{dz}$. This will give you a separable first-order ODE for the function $w(z)$.
    $endgroup$
    – Christoph
    Jan 19 at 5:31














    $begingroup$
    thank you, this helps. However, I still do not know how to solve an Abel equation. If you could solve it for me and show me your method, that would be great. If not, I'll try working with this.
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:39




    $begingroup$
    thank you, this helps. However, I still do not know how to solve an Abel equation. If you could solve it for me and show me your method, that would be great. If not, I'll try working with this.
    $endgroup$
    – Rama Velidanda
    Jan 19 at 5:39




    1




    1




    $begingroup$
    Actually, the equation written down by AlexanderJ93 is already separable … I did one step too much.
    $endgroup$
    – Christoph
    Jan 19 at 5:52






    $begingroup$
    Actually, the equation written down by AlexanderJ93 is already separable … I did one step too much.
    $endgroup$
    – Christoph
    Jan 19 at 5:52













    0












    $begingroup$

    First of all, let me remark that it seems impossible to obtain a closed-form solution in terms of elementary functions to this ordinary differential equation (ODE) in the general case.



    Writing $v' = w(v)$ with a new function $w$, $v'' = w frac{dw}{dv}$, we obtain the first-order separable ODE
    begin{equation}
    w frac{dw}{dv} = a v - b v w = v (a - b w).
    end{equation}
    Assuming that $w not equiv frac{a}{b}$ and using separation of variables we obtain
    begin{equation}
    frac{w}{a-bw} , mathrm{d}w = v , mathrm{d}v quad Leftrightarrow quad lnleft|frac{b}{a}w-1right| + frac{b}{a} w - 1 = - frac{b^2}{2 a} v^2 + C_1, quad C_1 in mathbb{R}.
    end{equation}
    Therefore, the function $w(v)$ may be obtained in terms of the Lambert W function, which cannot be expressed in terms of elementary functions.



    Finally, the solution of the separable ODE $frac{dv}{dt} = w(v)$ yields an implicit representation
    begin{equation}
    int frac{1}{w(v)} , mathrm{d}v = t
    end{equation}
    for $v$, with the function $w(v)$ from above, where the indefinite integral contains the second constant of integration. The values of the two constants of integration will be determined from the given initial values $v(0), v'(0)$.



    Edit: The solution of the initial-value problem $v'' = a v - b v v'$, $v(0) = v_0$, $v'(0) = frac{a}{b}$, is given by the linear function $v(t) = frac{a}{b} t + v_0$. This is the special case of a constant $w equiv frac{a}{b}$, for which we don't need to do the separation of variables.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      First of all, let me remark that it seems impossible to obtain a closed-form solution in terms of elementary functions to this ordinary differential equation (ODE) in the general case.



      Writing $v' = w(v)$ with a new function $w$, $v'' = w frac{dw}{dv}$, we obtain the first-order separable ODE
      begin{equation}
      w frac{dw}{dv} = a v - b v w = v (a - b w).
      end{equation}
      Assuming that $w not equiv frac{a}{b}$ and using separation of variables we obtain
      begin{equation}
      frac{w}{a-bw} , mathrm{d}w = v , mathrm{d}v quad Leftrightarrow quad lnleft|frac{b}{a}w-1right| + frac{b}{a} w - 1 = - frac{b^2}{2 a} v^2 + C_1, quad C_1 in mathbb{R}.
      end{equation}
      Therefore, the function $w(v)$ may be obtained in terms of the Lambert W function, which cannot be expressed in terms of elementary functions.



      Finally, the solution of the separable ODE $frac{dv}{dt} = w(v)$ yields an implicit representation
      begin{equation}
      int frac{1}{w(v)} , mathrm{d}v = t
      end{equation}
      for $v$, with the function $w(v)$ from above, where the indefinite integral contains the second constant of integration. The values of the two constants of integration will be determined from the given initial values $v(0), v'(0)$.



      Edit: The solution of the initial-value problem $v'' = a v - b v v'$, $v(0) = v_0$, $v'(0) = frac{a}{b}$, is given by the linear function $v(t) = frac{a}{b} t + v_0$. This is the special case of a constant $w equiv frac{a}{b}$, for which we don't need to do the separation of variables.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        First of all, let me remark that it seems impossible to obtain a closed-form solution in terms of elementary functions to this ordinary differential equation (ODE) in the general case.



        Writing $v' = w(v)$ with a new function $w$, $v'' = w frac{dw}{dv}$, we obtain the first-order separable ODE
        begin{equation}
        w frac{dw}{dv} = a v - b v w = v (a - b w).
        end{equation}
        Assuming that $w not equiv frac{a}{b}$ and using separation of variables we obtain
        begin{equation}
        frac{w}{a-bw} , mathrm{d}w = v , mathrm{d}v quad Leftrightarrow quad lnleft|frac{b}{a}w-1right| + frac{b}{a} w - 1 = - frac{b^2}{2 a} v^2 + C_1, quad C_1 in mathbb{R}.
        end{equation}
        Therefore, the function $w(v)$ may be obtained in terms of the Lambert W function, which cannot be expressed in terms of elementary functions.



        Finally, the solution of the separable ODE $frac{dv}{dt} = w(v)$ yields an implicit representation
        begin{equation}
        int frac{1}{w(v)} , mathrm{d}v = t
        end{equation}
        for $v$, with the function $w(v)$ from above, where the indefinite integral contains the second constant of integration. The values of the two constants of integration will be determined from the given initial values $v(0), v'(0)$.



        Edit: The solution of the initial-value problem $v'' = a v - b v v'$, $v(0) = v_0$, $v'(0) = frac{a}{b}$, is given by the linear function $v(t) = frac{a}{b} t + v_0$. This is the special case of a constant $w equiv frac{a}{b}$, for which we don't need to do the separation of variables.






        share|cite|improve this answer











        $endgroup$



        First of all, let me remark that it seems impossible to obtain a closed-form solution in terms of elementary functions to this ordinary differential equation (ODE) in the general case.



        Writing $v' = w(v)$ with a new function $w$, $v'' = w frac{dw}{dv}$, we obtain the first-order separable ODE
        begin{equation}
        w frac{dw}{dv} = a v - b v w = v (a - b w).
        end{equation}
        Assuming that $w not equiv frac{a}{b}$ and using separation of variables we obtain
        begin{equation}
        frac{w}{a-bw} , mathrm{d}w = v , mathrm{d}v quad Leftrightarrow quad lnleft|frac{b}{a}w-1right| + frac{b}{a} w - 1 = - frac{b^2}{2 a} v^2 + C_1, quad C_1 in mathbb{R}.
        end{equation}
        Therefore, the function $w(v)$ may be obtained in terms of the Lambert W function, which cannot be expressed in terms of elementary functions.



        Finally, the solution of the separable ODE $frac{dv}{dt} = w(v)$ yields an implicit representation
        begin{equation}
        int frac{1}{w(v)} , mathrm{d}v = t
        end{equation}
        for $v$, with the function $w(v)$ from above, where the indefinite integral contains the second constant of integration. The values of the two constants of integration will be determined from the given initial values $v(0), v'(0)$.



        Edit: The solution of the initial-value problem $v'' = a v - b v v'$, $v(0) = v_0$, $v'(0) = frac{a}{b}$, is given by the linear function $v(t) = frac{a}{b} t + v_0$. This is the special case of a constant $w equiv frac{a}{b}$, for which we don't need to do the separation of variables.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 19 at 9:47

























        answered Jan 19 at 8:35









        ChristophChristoph

        58116




        58116






























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