Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By HallsofIvy

Thread: ODE solution of exponential form

  1. #1
    Newbie
    Joined
    Nov 2013
    From
    Austria
    Posts
    20
    Thanks
    2

    Question ODE solution of exponential form

    Hello everybody

    I'm trying to solve the following differential equation derived from a first order kinetic process:

    $\displaystyle \tau {dm(t) \over dt} = m(\infty) - m(t)$

    where $\displaystyle m(\infty)$ is the steady-state of $\displaystyle m(t)$, that I guess, for the moment could be treated as a constant.

    I know that the solution to the ODE is:

    $\displaystyle m(t) = m(\infty) -\big(m(\infty)-m(0)\big ) \exp({-t/\tau} )$

    being $\displaystyle m(0)$ the initial condition.

    I've tried to solve the equation with a piece of paper, but get stuck in the process. Im principle, this equation should be easily solved by separation of variables...



    What I tried so far:
    First, separation of variables

    $\displaystyle \frac{dm(t)}{m(\infty) - m(t)} = \frac{dt}{\tau}$

    applying the integration limits:

    $\displaystyle \int\limits_{0}^{t} \frac{dm(t)}{m(\infty) - m(t)} =\int\limits_{0}^{t} \frac{dt}{\tau}$

    solving the integral

    $\displaystyle \log\frac{m(\infty) - m(t)}{m(\infty) - m(0)} =\frac{t}{\tau}$

    and applying exponential to both sides of the equation leads to:

    $\displaystyle \frac{m(\infty) - m(t)}{m(\infty) - m(0)} =\exp(t/\tau)$

    organizing...
    $\displaystyle m(\infty) - m(t) = \big(m(\infty) - m(0)\big) \exp(t/\tau)$

    finally, solving for $\displaystyle m(t)$

    $\displaystyle m(t) = m(\infty) - \big(m(\infty) - m(0)\big) \exp(t/\tau)$

    which is almost the solution, but the exponential is raised to $\displaystyle t/\tau$ and not to $\displaystyle -t/\tau$ as it should be.

    I would appreciate if somebody would point out where is the mistake in the calculation

    Thanks

    Jose
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,722
    Thanks
    3007

    Re: ODE solution of exponential form

    Quote Originally Posted by jguzman View Post
    Hello everybody

    I'm trying to solve the following differential equation derived from a first order kinetic process:

    $\displaystyle \tau {dm(t) \over dt} = m(\infty) - m(t)$

    where $\displaystyle m(\infty)$ is the steady-state of $\displaystyle m(t)$, that I guess, for the moment could be treated as a constant.

    I know that the solution to the ODE is:

    $\displaystyle m(t) = m(\infty) -\big(m(\infty)-m(0)\big ) \exp({-t/\tau} )$

    being $\displaystyle m(0)$ the initial condition.

    I've tried to solve the equation with a piece of paper, but get stuck in the process. Im principle, this equation should be easily solved by separation of variables...



    What I tried so far:
    First, separation of variables

    $\displaystyle \frac{dm(t)}{m(\infty) - m(t)} = \frac{dt}{\tau}$

    applying the integration limits:

    $\displaystyle \int\limits_{0}^{t} \frac{dm(t)}{m(\infty) - m(t)} =\int\limits_{0}^{t} \frac{dt}{\tau}$

    solving the integral

    $\displaystyle \log\frac{m(\infty) - m(t)}{m(\infty) - m(0)} =\frac{t}{\tau}$
    This integration is wrong. Letting "$\displaystyle u= m(\infty)- m$" gives $\displaystyle du= -dm$. You have the sign wrong in the integral so the fraction in the logarith m is inverted. You should have:
    $\displaystyle \log\left(\dfrac{m(\infty)- m(0)}{m(\infty)- m(t)}\right)= \dfrac{t}{\tau}$

    and applying exponential to both sides of the equation leads to:

    $\displaystyle \frac{m(\infty) - m(t)}{m(\infty) - m(0)} =\exp(t/\tau)$

    organizing...
    $\displaystyle m(\infty) - m(t) = \big(m(\infty) - m(0)\big) \exp(t/\tau)$

    finally, solving for $\displaystyle m(t)$

    $\displaystyle m(t) = m(\infty) - \big(m(\infty) - m(0)\big) \exp(t/\tau)$

    which is almost the solution, but the exponential is raised to $\displaystyle t/\tau$ and not to $\displaystyle -t/\tau$ as it should be.

    I would appreciate if somebody would point out where is the mistake in the calculation

    Thanks

    Jose
    Thanks from jguzman
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2013
    From
    Austria
    Posts
    20
    Thanks
    2

    Re: ODE solution of exponential form

    Fantastic! Thank you very much for your help!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. logs into exponential form
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Oct 11th 2011, 02:20 PM
  2. Set up appropriate form of solution Yp.
    Posted in the Differential Equations Forum
    Replies: 9
    Last Post: Jul 27th 2009, 04:32 AM
  3. Write 1-i exponential form
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: May 19th 2009, 11:03 AM
  4. Exponential form
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Jan 14th 2009, 05:11 PM
  5. Exponential Form
    Posted in the Algebra Forum
    Replies: 4
    Last Post: May 13th 2007, 10:47 AM

Search Tags


/mathhelpforum @mathhelpforum