# Thread: Applications of Nonlinear Equations

1. ## Applications of Nonlinear Equations

Hello,
I am having trouble with this problem, but I am sure that I am just making a dumb mistake...

Find C(t) from dC/dt = C(1-.0005C), C(0) = 1

First you use partial fractions to get: ((1/C) + (.0005/(1-.0005C))dC = dt

After that, I integrated the fractions to get:

ln (C) + .0005(ln(1-.0005)) = t + d d is a constant

But my books says the second natural log: .0005(ln(1-.0005)), should be negative, and then from there on I am completely stuck... Could anyone tell me how to do this?

2. Originally Posted by collegestudent321
Hello,
I am having trouble with this problem, but I am sure that I am just making a dumb mistake...

Find C(t) from dC/dt = C(1-.0005C), C(0) = 1

First you use partial fractions to get: ((1/C) + (.0005/(1-.0005C))dC = dt

After that, I integrated the fractions to get:

ln (C) + .0005(ln(1-.0005)) = t + d d is a constant

But my books says the second natural log: .0005(ln(1-.0005)), should be negative, and then from there on I am completely stuck... Could anyone tell me how to do this?
We have:

d(1-0,0005c) =-0,0005dc and deviding by (1-0.0005c) we have:

$\displaystyle -\frac{d(1-0,0005c)}{(1-0,0005c)} =\frac{0,0005dc}{(1-0,0005c)}$. Now substitute in the original equation and integrate

3. What do the commas mean? I don't think i understand that

4. oh wait, nevermind, now i know, hahah sorry