- #1
courtrigrad
- 1,236
- 2
Given that [tex] \frac{dx}{dt} = k(a-x)(b-x) [/tex]:
(a) Assuming [tex] a \neq b [/tex], find [tex] x [/tex] as a function of [tex] t [/tex]. Use the fact that the initial concentration of [tex] C [/tex] is 0.
(b) Find [tex] x(t) [/tex] assuming that [tex] a = b [/tex]. How does this expression for [tex] x(t) [/tex] simplify if it is known that [tex] [C] = \frac{a}{2} [/tex] after 20 seconds.
(a): So [tex] \frac{dx}{(a-x)(b-x)} = kdt [/tex]. After integrating by partial fractions and using the initial condition, I got [tex] x(t) = \frac{a-abe^{akt-bkt}}{1-\frac{a}{b}e^{akt-bkt}} [/tex].
(b). When I set [tex] a = b [/tex] I got an undefined expression, leading me to believe that part(a) is incorrect.
What did I do wrong?
Thanks
(a) Assuming [tex] a \neq b [/tex], find [tex] x [/tex] as a function of [tex] t [/tex]. Use the fact that the initial concentration of [tex] C [/tex] is 0.
(b) Find [tex] x(t) [/tex] assuming that [tex] a = b [/tex]. How does this expression for [tex] x(t) [/tex] simplify if it is known that [tex] [C] = \frac{a}{2} [/tex] after 20 seconds.
(a): So [tex] \frac{dx}{(a-x)(b-x)} = kdt [/tex]. After integrating by partial fractions and using the initial condition, I got [tex] x(t) = \frac{a-abe^{akt-bkt}}{1-\frac{a}{b}e^{akt-bkt}} [/tex].
(b). When I set [tex] a = b [/tex] I got an undefined expression, leading me to believe that part(a) is incorrect.
What did I do wrong?
Thanks
Last edited: