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Help With a Mess of a Separable Equation

  • Thread starter cowmoo32
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  • #1
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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 [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.)


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
 
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Answers and Replies

  • #2
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...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...

[tex]
\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
[/tex]

(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]...
 
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  • #3
LCKurtz
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[itex]\int[/itex][itex]\frac{K + S}{aS}[/itex] = [itex]\int[/itex]dt

ln(aS)([itex]\frac{S^2}{2}+KS[/itex]) = 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
 
  • #4
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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...

[tex]
\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
[/tex]

(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]...
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.
 
  • #5
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ok, so I get


[tex]
-\frac{K}{a}(\ln (S(t))+S(t)=t+C
[/tex]

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
LCKurtz
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ok, so I get


[tex]
-\frac{K}{a}(\ln (S(t))+S(t)=t+C
[/tex]

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.
 

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