Results 1 to 7 of 7

Math Help - Ordinary Differential Equation Help

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    15

    Ordinary Differential Equation Help

    Hi, can anyone help me with the following question:


    Find the solution of the ODE:

    m(dv/dt)=mg-kv^2

    with the initial condition v=v0 at t=0

    Is it integration by parts that should be used?
    Thanks in advance for any help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
     kv^2 + m\dfrac{dv}{dt} = mg

    Separate the variables: \dfrac{k}{m}v^2\, dv = g\, dt

    If, as I assume m is mass then it is non-zero and can be cancelled. I have also assumed that v << c so we can ignore relativity
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,319
    Thanks
    1241
    Quote Originally Posted by e^(i*pi) View Post
    \displaystyle kv^2 \cdot \frac{dv}{dt} = m(g-1)
    This is not the same DE as \displaystyle m\,\frac{dv}{dt} = m\,g - k\,v^2...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Quote Originally Posted by AlanC877 View Post
    Hi, can anyone help me with the following question:


    Find the solution of the ODE:

    m(dv/dt)=mg-kv^2

    with the initial condition v=v0 at t=0

    Is it integration by parts that should be used?
    Thanks in advance for any help!
    Several years ago, during an interesting discussion in an argument [apparently...] 'far from math', I imposted the problem of the fall of an object of mass m with initial speed 0 and initial position L under action of gravity and 'aerodinamic resistance'. The DE that describes the problem is...

    v^{'}= -g + k\ v^{2}\ , \\ v(0)=0 (1)

    In (1) You can separate the variables obtaining...

    \displaystyle t= \frac{1}{k}\ \int \frac{dv}{v^{2}-\frac{g}{k}} + \ln c = \frac{1}{2\ \sqrt{k\ g}}\ \ln \frac{v-\sqrt{\frac{g}{k}}} {v+\sqrt{\frac{g}{k}}} + \ln c \implies

    \displaystyle \implies e^{2\ \sqrt{k\ g}\ t}= c\ \frac{v-\sqrt{\frac{g}{k}}} {v+\sqrt{\frac{g}{k}}} \implies v= -\sqrt{\frac{g}{k}}\ \frac{e^{2\ \sqrt{k\ g}\ t}+c}{e^{2\ \sqrt{k\ g}\ t}-c} (2)

    If we set in (2) v(0)=0 we obtain c=-1 so that is...

    \displaystyle v= - \sqrt{\frac{g}{k}}\ (1-\frac{2}{1+e^{2\ \sqrt{k\ g}\ t}}) (3)

    If we integrate (3) with the condition y(0)= L we obtain...

    \displaystyle y(t)=L-\sqrt{\frac{g}{k}}\ \int_{0}^{t} (1-\frac{2}{1+e^{2\ \sqrt{k\ g}\ \tau}})\ d \tau = L + \sqrt{\frac{g}{k}}\ t - \frac{1}{k}\ \ln \frac{1+e^{2\ \sqrt{k\ g}\ t}}{2} (4)

    May be that, sooner or later', I will explain You what has been the 'scope' amd the 'queue' of the 'discussion' ...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2010
    Posts
    15
    Quote Originally Posted by e^(i*pi) View Post
     kv^2 + m\dfrac{dv}{dt} = mg

    Separate the variables: \dfrac{k}{m}v^2\, dv = g\, dt

    If, as I assume m is mass then it is non-zero and can be cancelled. I have also assumed that v << c so we can ignore relativity
    Thats kinda what I was thinking, rearranging and dividing throught by m gives (k/m)v^2+(dv/dt)=g, but multiplying the dt on each side would then give (k/m)v^2+dv=g dt, can you really just get rid of the plus?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    No, that was a mistake on my part. Please disregard all of my posts in this thread
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,324
    Thanks
    1294
    Quote Originally Posted by AlanC877 View Post
    Hi, can anyone help me with the following question:


    Find the solution of the ODE:

    m(dv/dt)=mg-kv^2
    \frac{m}{mg- kv^2}dv= dt

    \frac{m}{(\sqrt{mg}-\sqrt{k}v()(\sqrt{mg}+ \sqrt{k}v)}= d

    Use "partial fractions" to do the integral on the left.

    with the initial condition v=v0 at t=0

    Is it integration by parts that should be used?
    Thanks in advance for any help![/QUOTE]
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. ordinary differential equation
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: March 7th 2010, 08:08 AM
  2. First order ordinary differential equation
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: February 13th 2010, 01:46 PM
  3. Ordinary Differential equation
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: December 11th 2008, 11:44 AM
  4. Ordinary Differential Equation
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 23rd 2007, 09:24 AM
  5. Ordinary Differential Equation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 1st 2007, 06:20 PM

Search Tags


/mathhelpforum @mathhelpforum