Solving a first O.D.E












0












$begingroup$


I'm solving a rocket problem with the following O.D.E:



$$frac{dv}{dt}=-g-frac{k}{m}, v^2$$



with $g$ as gravitational constant, and $m$ (no changing mass) and $k$ are also constants. Given the initial condition $v(0)=V_0$ m/s, how to go about fnding its analytical solution?






ordinary-differential-equations







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  • 2




    $begingroup$
    Welcome to MSE. What are your attempts solving this problem?
    $endgroup$
    – James
    Jan 2 at 9:57










  • $begingroup$
    It's a separable equation
    $endgroup$
    – Yuriy S
    Jan 2 at 9:59










  • $begingroup$
    I was looking at this ODE to see if I can treat it as a separable equation: link
    $endgroup$
    – macy
    Jan 2 at 10:14










  • $begingroup$
    The physics of this equation is only valid for $v>0$. In general it should be $dot v=-g-frac km v|v|$, as the air friction always acts to reduce the velocity, that is as a force in the opposite direction.
    $endgroup$
    – LutzL
    Feb 20 at 9:49
















0












$begingroup$


I'm solving a rocket problem with the following O.D.E:



$$frac{dv}{dt}=-g-frac{k}{m}, v^2$$



with $g$ as gravitational constant, and $m$ (no changing mass) and $k$ are also constants. Given the initial condition $v(0)=V_0$ m/s, how to go about fnding its analytical solution?






ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Welcome to MSE. What are your attempts solving this problem?
    $endgroup$
    – James
    Jan 2 at 9:57










  • $begingroup$
    It's a separable equation
    $endgroup$
    – Yuriy S
    Jan 2 at 9:59










  • $begingroup$
    I was looking at this ODE to see if I can treat it as a separable equation: link
    $endgroup$
    – macy
    Jan 2 at 10:14










  • $begingroup$
    The physics of this equation is only valid for $v>0$. In general it should be $dot v=-g-frac km v|v|$, as the air friction always acts to reduce the velocity, that is as a force in the opposite direction.
    $endgroup$
    – LutzL
    Feb 20 at 9:49














0












0








0





$begingroup$


I'm solving a rocket problem with the following O.D.E:



$$frac{dv}{dt}=-g-frac{k}{m}, v^2$$



with $g$ as gravitational constant, and $m$ (no changing mass) and $k$ are also constants. Given the initial condition $v(0)=V_0$ m/s, how to go about fnding its analytical solution?






ordinary-differential-equations







share|cite|improve this question











$endgroup$




I'm solving a rocket problem with the following O.D.E:



$$frac{dv}{dt}=-g-frac{k}{m}, v^2$$



with $g$ as gravitational constant, and $m$ (no changing mass) and $k$ are also constants. Given the initial condition $v(0)=V_0$ m/s, how to go about fnding its analytical solution?






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 8 at 8:39







macy

















asked Jan 2 at 9:53









macymacy

125




125








  • 2




    $begingroup$
    Welcome to MSE. What are your attempts solving this problem?
    $endgroup$
    – James
    Jan 2 at 9:57










  • $begingroup$
    It's a separable equation
    $endgroup$
    – Yuriy S
    Jan 2 at 9:59










  • $begingroup$
    I was looking at this ODE to see if I can treat it as a separable equation: link
    $endgroup$
    – macy
    Jan 2 at 10:14










  • $begingroup$
    The physics of this equation is only valid for $v>0$. In general it should be $dot v=-g-frac km v|v|$, as the air friction always acts to reduce the velocity, that is as a force in the opposite direction.
    $endgroup$
    – LutzL
    Feb 20 at 9:49














  • 2




    $begingroup$
    Welcome to MSE. What are your attempts solving this problem?
    $endgroup$
    – James
    Jan 2 at 9:57










  • $begingroup$
    It's a separable equation
    $endgroup$
    – Yuriy S
    Jan 2 at 9:59










  • $begingroup$
    I was looking at this ODE to see if I can treat it as a separable equation: link
    $endgroup$
    – macy
    Jan 2 at 10:14










  • $begingroup$
    The physics of this equation is only valid for $v>0$. In general it should be $dot v=-g-frac km v|v|$, as the air friction always acts to reduce the velocity, that is as a force in the opposite direction.
    $endgroup$
    – LutzL
    Feb 20 at 9:49








2




2




$begingroup$
Welcome to MSE. What are your attempts solving this problem?
$endgroup$
– James
Jan 2 at 9:57




$begingroup$
Welcome to MSE. What are your attempts solving this problem?
$endgroup$
– James
Jan 2 at 9:57












$begingroup$
It's a separable equation
$endgroup$
– Yuriy S
Jan 2 at 9:59




$begingroup$
It's a separable equation
$endgroup$
– Yuriy S
Jan 2 at 9:59












$begingroup$
I was looking at this ODE to see if I can treat it as a separable equation: link
$endgroup$
– macy
Jan 2 at 10:14




$begingroup$
I was looking at this ODE to see if I can treat it as a separable equation: link
$endgroup$
– macy
Jan 2 at 10:14












$begingroup$
The physics of this equation is only valid for $v>0$. In general it should be $dot v=-g-frac km v|v|$, as the air friction always acts to reduce the velocity, that is as a force in the opposite direction.
$endgroup$
– LutzL
Feb 20 at 9:49




$begingroup$
The physics of this equation is only valid for $v>0$. In general it should be $dot v=-g-frac km v|v|$, as the air friction always acts to reduce the velocity, that is as a force in the opposite direction.
$endgroup$
– LutzL
Feb 20 at 9:49










2 Answers
2






active

oldest

votes


















2












$begingroup$

Write the equation as $$frac{dt}{dv}=-frac 1{g+frac{k}{m}, v^2}$$



$$-t+C=int frac {dv}{g+frac{k}{m}, v^2}$$ Now, a suitable change of variable will make you facing a very well know integral.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @macy. What is the simplest integral (you know it) where the numerator is $color{red}{1}$ and the denominator contains $color{red}{(text{something}+x^2)}$ ?
    $endgroup$
    – Claude Leibovici
    Jan 2 at 10:21












  • $begingroup$
    it would be $arctan(x)$ is that something is unity.
    $endgroup$
    – macy
    Jan 2 at 10:24












  • $begingroup$
    @macy. Good bet !
    $endgroup$
    – Claude Leibovici
    Jan 2 at 10:30



















1












$begingroup$

As a Riccati equation one approach is to set $v=frac{m}{k}frac{u'}{u}$ so that you get to
$$
frac{m}{k}frac{u''}{u}-frac{m}{k}frac{u'^2}{u^2}=-g-frac{m}{k}frac{u'^2}{u^2}
implies
u''+frac{kg}{m}u=0
$$
and this is now a harmonic oscillator equation with frequency $omega=sqrt{frac{kg}{m}}$ and thus solutions $u=Acosωt+Bsinωt$ from which you can reconstruct the function $v$.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Write the equation as $$frac{dt}{dv}=-frac 1{g+frac{k}{m}, v^2}$$



    $$-t+C=int frac {dv}{g+frac{k}{m}, v^2}$$ Now, a suitable change of variable will make you facing a very well know integral.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @macy. What is the simplest integral (you know it) where the numerator is $color{red}{1}$ and the denominator contains $color{red}{(text{something}+x^2)}$ ?
      $endgroup$
      – Claude Leibovici
      Jan 2 at 10:21












    • $begingroup$
      it would be $arctan(x)$ is that something is unity.
      $endgroup$
      – macy
      Jan 2 at 10:24












    • $begingroup$
      @macy. Good bet !
      $endgroup$
      – Claude Leibovici
      Jan 2 at 10:30
















    2












    $begingroup$

    Write the equation as $$frac{dt}{dv}=-frac 1{g+frac{k}{m}, v^2}$$



    $$-t+C=int frac {dv}{g+frac{k}{m}, v^2}$$ Now, a suitable change of variable will make you facing a very well know integral.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @macy. What is the simplest integral (you know it) where the numerator is $color{red}{1}$ and the denominator contains $color{red}{(text{something}+x^2)}$ ?
      $endgroup$
      – Claude Leibovici
      Jan 2 at 10:21












    • $begingroup$
      it would be $arctan(x)$ is that something is unity.
      $endgroup$
      – macy
      Jan 2 at 10:24












    • $begingroup$
      @macy. Good bet !
      $endgroup$
      – Claude Leibovici
      Jan 2 at 10:30














    2












    2








    2





    $begingroup$

    Write the equation as $$frac{dt}{dv}=-frac 1{g+frac{k}{m}, v^2}$$



    $$-t+C=int frac {dv}{g+frac{k}{m}, v^2}$$ Now, a suitable change of variable will make you facing a very well know integral.






    share|cite|improve this answer









    $endgroup$



    Write the equation as $$frac{dt}{dv}=-frac 1{g+frac{k}{m}, v^2}$$



    $$-t+C=int frac {dv}{g+frac{k}{m}, v^2}$$ Now, a suitable change of variable will make you facing a very well know integral.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 2 at 10:07









    Claude LeiboviciClaude Leibovici

    123k1157134




    123k1157134












    • $begingroup$
      @macy. What is the simplest integral (you know it) where the numerator is $color{red}{1}$ and the denominator contains $color{red}{(text{something}+x^2)}$ ?
      $endgroup$
      – Claude Leibovici
      Jan 2 at 10:21












    • $begingroup$
      it would be $arctan(x)$ is that something is unity.
      $endgroup$
      – macy
      Jan 2 at 10:24












    • $begingroup$
      @macy. Good bet !
      $endgroup$
      – Claude Leibovici
      Jan 2 at 10:30


















    • $begingroup$
      @macy. What is the simplest integral (you know it) where the numerator is $color{red}{1}$ and the denominator contains $color{red}{(text{something}+x^2)}$ ?
      $endgroup$
      – Claude Leibovici
      Jan 2 at 10:21












    • $begingroup$
      it would be $arctan(x)$ is that something is unity.
      $endgroup$
      – macy
      Jan 2 at 10:24












    • $begingroup$
      @macy. Good bet !
      $endgroup$
      – Claude Leibovici
      Jan 2 at 10:30
















    $begingroup$
    @macy. What is the simplest integral (you know it) where the numerator is $color{red}{1}$ and the denominator contains $color{red}{(text{something}+x^2)}$ ?
    $endgroup$
    – Claude Leibovici
    Jan 2 at 10:21






    $begingroup$
    @macy. What is the simplest integral (you know it) where the numerator is $color{red}{1}$ and the denominator contains $color{red}{(text{something}+x^2)}$ ?
    $endgroup$
    – Claude Leibovici
    Jan 2 at 10:21














    $begingroup$
    it would be $arctan(x)$ is that something is unity.
    $endgroup$
    – macy
    Jan 2 at 10:24






    $begingroup$
    it would be $arctan(x)$ is that something is unity.
    $endgroup$
    – macy
    Jan 2 at 10:24














    $begingroup$
    @macy. Good bet !
    $endgroup$
    – Claude Leibovici
    Jan 2 at 10:30




    $begingroup$
    @macy. Good bet !
    $endgroup$
    – Claude Leibovici
    Jan 2 at 10:30











    1












    $begingroup$

    As a Riccati equation one approach is to set $v=frac{m}{k}frac{u'}{u}$ so that you get to
    $$
    frac{m}{k}frac{u''}{u}-frac{m}{k}frac{u'^2}{u^2}=-g-frac{m}{k}frac{u'^2}{u^2}
    implies
    u''+frac{kg}{m}u=0
    $$
    and this is now a harmonic oscillator equation with frequency $omega=sqrt{frac{kg}{m}}$ and thus solutions $u=Acosωt+Bsinωt$ from which you can reconstruct the function $v$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      As a Riccati equation one approach is to set $v=frac{m}{k}frac{u'}{u}$ so that you get to
      $$
      frac{m}{k}frac{u''}{u}-frac{m}{k}frac{u'^2}{u^2}=-g-frac{m}{k}frac{u'^2}{u^2}
      implies
      u''+frac{kg}{m}u=0
      $$
      and this is now a harmonic oscillator equation with frequency $omega=sqrt{frac{kg}{m}}$ and thus solutions $u=Acosωt+Bsinωt$ from which you can reconstruct the function $v$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        As a Riccati equation one approach is to set $v=frac{m}{k}frac{u'}{u}$ so that you get to
        $$
        frac{m}{k}frac{u''}{u}-frac{m}{k}frac{u'^2}{u^2}=-g-frac{m}{k}frac{u'^2}{u^2}
        implies
        u''+frac{kg}{m}u=0
        $$
        and this is now a harmonic oscillator equation with frequency $omega=sqrt{frac{kg}{m}}$ and thus solutions $u=Acosωt+Bsinωt$ from which you can reconstruct the function $v$.






        share|cite|improve this answer









        $endgroup$



        As a Riccati equation one approach is to set $v=frac{m}{k}frac{u'}{u}$ so that you get to
        $$
        frac{m}{k}frac{u''}{u}-frac{m}{k}frac{u'^2}{u^2}=-g-frac{m}{k}frac{u'^2}{u^2}
        implies
        u''+frac{kg}{m}u=0
        $$
        and this is now a harmonic oscillator equation with frequency $omega=sqrt{frac{kg}{m}}$ and thus solutions $u=Acosωt+Bsinωt$ from which you can reconstruct the function $v$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 10:40









        LutzLLutzL

        59.2k42056




        59.2k42056






























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