# Homework Help: Help With a Mess of a Separable Equation

1. Sep 3, 2012

### cowmoo32

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
$\frac{dS}{dt}$ = $\frac{aS}{K + S}$

$\int$$\frac{K + S}{aS}$ = $\int$dt

ln(aS)($\frac{S^2}{2}+KS$) = t

I'm not even sure how to begin to solve for S

Last edited by a moderator: May 6, 2017
2. Sep 3, 2012

### christoff

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 $S(0)=S_0$...

Last edited: Sep 3, 2012
3. Sep 3, 2012

### LCKurtz

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

4

4. Sep 4, 2012

### cowmoo32

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.

5. Sep 4, 2012

### cowmoo32

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.

6. Sep 4, 2012

### LCKurtz

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.