# Separable Differential Equation Confusion

1. Mar 21, 2009

### jumbogala

1. The problem statement, all variables and given/known data
Solve the equation for b.

db/dx = (e^2x)(e^2b) -- therefore:

db / (e^2b) = (e^2x)dx

Also note that b(0) = 8.

2. Relevant equations

3. The attempt at a solution
Following a list of steps my teacher gave me to solve these:
1) Integrate both sides.

-0.5 e^(-2b) = 0.5 e^(2x) + C

2) Isolate b.

e^(-2b) = -e^(2x) + C

-2b = ln(-e^(2x) + C)

And here we run in to the problem - you can't take ln of a negative value!

If that e^(2x) was positive I would finish the problem like this:

2x + C = -2b

x is 0 and b is 8 so C has to be -16.

And my final answer would be b = -x + 8.

But, I didn't think you could do this if there's a negative inside ln?

Last edited: Mar 22, 2009
2. Mar 22, 2009

### SmashtheVan

Correct that you can't take a ln of a negative value, but what if C > -e^(2x), then the value would be positive and a natural log can be taken. You can find a solution, but there will be a restriction!

You can solve for the constant at the step that I bolded above by implementing your initial condition

e^(-16) = -(e^(0)) + C

since we know that e^0 = 1,
C = e^(-16) + 1 $$\approx$$ 1

(e^(-16) is on the order of 10^-7, and I would say that is negligible compared to the 1)

and your solution would be of the form

e^(-2b) = 1- e^(2x) for x>0 (as 0 or any number less would leave the solution undefined)

and if you wanted to, at this step you could take the natural log of both sides and say that the solution is

b= -.5ln(1-e^2x), x>0

Last edited: Mar 22, 2009