1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Help With a Mess of a Separable Equation

  1. Sep 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Let S(t) represent the amount of a chemical reactant present at time t, where t>= 0. Assume that S(t) can be determined by solving the initial value problem
    http://webwork.math.ncsu.edu/webwork2_files/tmp/equations/21/885ac2eff6f65b363662233870e25e1.png [Broken]

    where a, K, and S0 are positive constants. Obtain an implicit solution of the initial value problem. (The differential equation, often referred to as the Michaelis-Menten equation, arises in the study of biochemical reactions.)

    3. The attempt at a solution
    [itex]\frac{dS}{dt}[/itex] = [itex]\frac{aS}{K + S}[/itex]

    [itex]\int[/itex][itex]\frac{K + S}{aS}[/itex] = [itex]\int[/itex]dt

    ln(aS)([itex]\frac{S^2}{2}+KS[/itex]) = t

    I'm not even sure how to begin to solve for S
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 3, 2012 #2
    Technically, if all you need is an implicit solution, then you're almost done - you never need to solve for S(t)... With an implicit solution, all you need is an equation which relates S and t. S and t are related by the expression you came to at the end: ln(aS)(S22+KS) = t. Moreover, that ODE doesn't have a solution S(t) in terms of elementary functions. You need the Lambert W-function to express it.

    However, I would caution you that your integral seems to be missing a minus sign. Moreover, when I come up with the implicit solution, I get a completely different answer...

    \int -\frac{K+S}{aS}dS=\int dt\\
    \Rightarrow \int -\frac{K}{aS}-\frac{S}{aS}dS=t+c\\
    \Rightarrow \int -\frac{K}{aS}-\frac{1}{a}dS=t+c\\
    \Rightarrow -\frac{1}{a}(K\ln (S(t))+S(t))=t+C

    (I'm not sure where you went wrong, but this is what you should be doing.)
    Now, can you see what you need to do? Your constant of integration C is still there, but you have an initial condition [itex]S(0)=S_0[/itex]...
    Last edited: Sep 3, 2012
  4. Sep 3, 2012 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That antiderivative isn't correct. You can't do
    $$\int \frac{K+S}{aS}dS = \int\frac 1 {aS}dS\int K+S\, dS$$

  5. Sep 4, 2012 #4
    I'll give it another go and see what I come up with.

    You're right. I meant to split them into two fractions, not separate the numerator from the denominator.
  6. Sep 4, 2012 #5
    ok, so I get

    -\frac{K}{a}(\ln (S(t))+S(t)=t+C

    Substituting S(0)=S0. I can't get the tex code correct, but I'm replacing S(t) with S0. When I solve for C, I get the entire equation =0, which isn't correct.
  7. Sep 4, 2012 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That should be$$-\frac{1}{a}(K\ln S(t)+S(t))=t+C$$
    When you put ##t=0## and ##S(0) = S_0## in that equation you get$$
    -\frac{1}{a}(K\ln (S_0)+S_0)=C$$Put that in for ##C## in the top equation.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook