Help With a Mess of a Separable Equation

[cowmoo32]

Homework Statement

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

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.)

The Attempt at a Solution

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

\int\frac{K + S}{aS} = \intdt

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

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

 

[christoff]

...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.)...

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...

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

(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...

 

[LCKurtz]

\int\frac{K + S}{aS} = \intdt

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

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

 

[cowmoo32]

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...

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

(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...

I'll give it another go and see what I come up with.

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

You're right. I meant to split them into two fractions, not separate the numerator from the denominator.

 

[cowmoo32]

ok, so I get <br /> -\frac{K}{a}(\ln (S(t))+S(t)=t+C<br />

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.

 

[LCKurtz]

ok, so I get


<br /> -\frac{K}{a}(\ln (S(t))+S(t)=t+C<br />

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.

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.