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Separable Differential Equation Confusion

  • Thread starter jumbogala
  • Start date
1. Homework Statement
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. Homework 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:
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!
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 [tex]\approx[/tex] 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:

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