Differential Equation dx/dt = k(a-x)(b-x)

[mjbc]

Homework Statement

dx/dt = k(a-x)(b-x)

x(0)=0

The Attempt at a Solution

dx/((a-x)(b-x)) = kdt

Integral (dx/(ab-bx-ax+x^2)) = kt +C

x/ab - (1/b)ln|x| - (1/a)ln|x| - (x^-1) = kt + C

however, if i try to substitute x(0) = 0, i get the ln 0

please help i have no idea if i am doing this right

 

[Dick]

I don't think the integration dx is correct at all. Use partial fractions.

 

[HallsofIvy]

Homework Statement

dx/dt = k(a-x)(b-x)

x(0)=0

The Attempt at a Solution

dx/((a-x)(b-x)) = kdt

Integral (dx/(ab-bx-ax+x^2)) = kt +C

x/ab - (1/b)ln|x| - (1/a)ln|x| - (x^-1) = kt + C

however, if i try to substitute x(0) = 0, i get the ln 0

please help i have no idea if i am doing this right

NO, 1/(a+ bx+ cx²) is NOT equal to 1/a+ 1/bx+ 1/cx².

Rule of thumb: since it is harder to factor than to multiply, NEVER multiply terms that are already factored! (Well, almost never.)

Integrate dx/((a-x)(b-x))= dx/((x-a)(x-b) using partial fractions.