Differential equation partial fractions

[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

 

[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.

 

[HallsofIvy]

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.