# Differential equation partial fractions

1. Sep 26, 2006

### courtrigrad

Given that $$\frac{dx}{dt} = k(a-x)(b-x)$$:

(a) Assuming $$a \neq b$$, find $$x$$ as a function of $$t$$. Use the fact that the initial concentration of $$C$$ is 0.
(b) Find $$x(t)$$ assuming that $$a = b$$. How does this expression for $$x(t)$$ simplify if it is known that $$[C] = \frac{a}{2}$$ after 20 seconds.

(a): So $$\frac{dx}{(a-x)(b-x)} = kdt$$. After integrating by partial fractions and using the initial condition, I got $$x(t) = \frac{a-abe^{akt-bkt}}{1-\frac{a}{b}e^{akt-bkt}}$$.

(b). When I set $$a = b$$ I got an undefined expression, leading me to believe that part(a) is incorrect.

What did I do wrong?

Thanks

Last edited: Sep 26, 2006
2. Sep 26, 2006

### StatusX

You can't just set a=b, you have to take the limit. Alternatively, go back and integrate again with the knowledge that a=b. This will change how the partial fractions expansion goes.

3. Sep 27, 2006

### HallsofIvy

Staff Emeritus
If a= b the method you used to solve the equation assuming a $\ne$ b does not work. Go back to the original equation, set a= b, and solve again.

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