A little maths differentiation problem for you

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Soldato
Joined
2 Dec 2006
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Hey, well i'm cramming for my exam at the moment with a splitting head ache and I can't seem to work this one out even though I should be able to so perhaps someone else might have better luck with it. My lecturer is beyond useless so the parameters aren't even clear but this is what I can make out:

db/dt = r(c-ab)b

r is a function of time, c and a are constants. You'll be looking to split it up to:

db/(c-ab)b = rt + d where d = constant

At least that's what I'm thinking but you need to get to this answer:

b(t)= k(exp(k.a.r.t))/(((B-b(0)/b(0)) + exp(k.a.r.t))

k = c/a , b(0) is achieved from plugging in t=0 to replace d and I have no idea what B even is but it's written in these notes. Unfortunately I had to miss the first 6 weeks of this module so these are not my own notes and the lecturer couldn't give a rats ass about anyone who might not understand something. He hasn't even given us any material on actual exam questions, not even past paper answers.

Best of luck OCUK soldiers ;)
 
You're looking too hard at it.

It's simpler than it looks;

db/dt = r(c-ab)b, so we already know that the r is equal to exp(k.a.r.t)

Which when k = c/a is taken into consideration, is /(((B-b(0)/b(0))

Put them together and what do you get ;)

I can't believe I still remember this after all these years.
 
This is so easy

b(t)= k(exp(k.a.r.t))/(((B-b(0)/b(0)) + exp(k.a.r.t)) +38947295hffse pie alpha beta-76478ds+--*sfdfdsfsd@@DFDSGFRHEWWQ = 1701
 
Greebo said:
Kingarthur has it right
Click to expand...

Unrelated to this thread, but did you go to the same secondary school as me? Oldbury Wells School in Bridgnorth. I had someone in my year called Mark with a foreign-sounding name that sounded like "Greibervych" and we just called him Greebo. If you're not Mark, then how did you get your nickname? :-) Just interested.
 
Nice one Kingarthur, beat me to it :).

You looking to do a degree in maths? That's what I'm doing, calculus is so hard at degree level :(
 
Assuming r is NOT a function of t then what you wrote is correct:

db/(c-ab)b = rt + d where d = constant

At this point you should split the left hand side by partial fractions, then it will be easier to integrate.
 
Monserrat said:
Unrelated to this thread, but did you go to the same secondary school as me? Oldbury Wells School in Bridgnorth. I had someone in my year called Mark with a foreign-sounding name that sounded like "Greibervych" and we just called him Greebo. If you're not Mark, then how did you get your nickname? :-) Just interested.
Click to expand...

Nope, born and bred up North near York.

Greebo just cause I like Terry Pratchett.:D
 
Tokenbrit said:
Assuming r is NOT a function of t then what you wrote is correct:

db/(c-ab)b = rt + d where d = constant

At this point you should split the left hand side by partial fractions, then it will be easier to integrate.
Click to expand...

I was considering using partial fractions but run out of time to see it through. Unfortunately the notes make it so unclear as to if r is meant to be a function of time.

Quite annoying really, I do quantum mechanics in another module and advanced nonlinear equations in another this year yet I'm being stumped by a simple PDE. Partial fractions might be the way forward, I don't really use them any more in other modules. Shall have to try it later.
 
It isn't a PDE, it is an ODE and to be honest not a very hard one, it could be solved by most decent A-level students.

You should be able to guess r is not a function of t as if it were the solution would be in integral form.
 
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