# A seperation of variables intergration

1. Nov 24, 2008

### thomas49th

1. The problem statement, all variables and given/known data

$$\frac{dV}{dt} = 20 - kV$$

By solving this show that

$$V = A + Be^{-kt}$$

2. Relevant equations

Well im guessing there is a ln coming into play somewhere during the intergration
if the diff the bottom = the top then you get a ln(bottom)

3. The attempt at a solution

seperate the variables

$$\int{\frac{1}{20-kV}dv} = \int{1dt}$$

i dont know how to form the ln part out of the LHS, cus you cant just take a constant out here?

Thanks :)

2. Nov 24, 2008

### gabbagabbahey

Hint: what is $$\frac{d}{dx} \ln(20-kV)$$?

3. Nov 24, 2008

### Staff: Mentor

Let w = 20 - kV. Then dw = -k dV

4. Nov 24, 2008

### thomas49th

ahh take a constant of -1/k out

so -1/k ln (20-kV) = t
multiply through by -k to give

ln(20-kV) = -tk
then e it man
gives you
20-kV = e^(-tk)
V = 20/k -e^(-tk)/k
but the stupid mark scheme says
V = 20/k -20e^(-tk)/k

who is right?

Thanks :)

5. Nov 24, 2008

### gabbagabbahey

Remember, your integrals are indefinite integrals; so you need to include a constant of integration:

$$\Rightarrow \frac{-1}{k} \ln(20-kV) = t +C$$

6. Nov 24, 2008

### thomas49th

Ooops missed that.
however how does that effect the co-efficient of e being 20?

Thanks :)

7. Nov 24, 2008

### gabbagabbahey

Well, what do you get when you solve it with the constant C?

8. Nov 24, 2008

### thomas49th

using v=0 and t=0
(as im told the container is empty)
so that means c = -1/k ln 20

which means i can combine the ln s!!!

cheers :)