Separable differential equation

AI Thread Summary
The discussion revolves around solving the separable differential equation dP/dt = 0.2P(1000 - P) with initial conditions P(0) = 1000 and P(0) = 2000. The correct solutions are P(t) = 1000 for the first case and P(t) = 1000[1 - (1/2)exp(-200t)]^(-1) for the second. The user struggles with the integration process and encounters a division by zero error early on. Clarifications indicate that the general solution is correct, but the initial condition application needs careful handling to avoid errors. The discussion concludes with a resolution that helps the user understand their mistake.
Luminous Blob
Messages
50
Reaction score
0
I'm doing some revision on differential equations, and am getting stuck on what should be a simple problem.

The question is:

Solve: dP/dt = 0.2P(1000 - P)

Find particular solutions when (i) P(0) = 1000 and (ii) P(0) = 2000

The answers are supposed to be:

i) P(t) = 1000
ii) P(t) = 1000[1 - (1/2)exp(-200t)]^(-1)

I keep getting the wrong answers. The general solution I came up with (after doing the partial fraction decomposition and integrating) is:

t + C = 1/200[ ln|P| - ln|1000 - P| ]
exp(200t + C) = exp(ln|P| - ln|1000 - P|)
Aexp(200t) = P/(1000 - P) {where A = exp(C)}
P = 1000Aexp(200t) - (P)Aexp(200t)
P[ 1 + Aexp(200t) ] = 1000Aexp(200t)

P = 1000Aexp(200t)/[ 1 + Aexp(200t)]

But I don't get the right answers when I solve for A with the initial condition.

I'd appreciate it if anyone can tell me where I'm going wrong here.
 
Physics news on Phys.org
i) YOU HAVE PERFORMED DIVISION BY ZERO HERE!
(Right when you started to solve the diff. eq.)

ii) Nothing wrong with your answer here, you'll get A=-2 (right?)

Remember:
\frac{1000Ae^{200t}}{1+Ae^{200t}}=\frac{1000}{1+\frac{1}{A}e^{-200t}}, A\neq0
 
arildno said:
i) YOU HAVE PERFORMED DIVISION BY ZERO HERE!
(Right when you started to solve the diff. eq.)

ii) Nothing wrong with your answer here, you'll get A=-2 (right?)

Remember:
\frac{1000Ae^{200t}}{1+Ae^{200t}}=\frac{1000}{1+\frac{1}{A}e^{-200t}}, A\neq0

Aaahh...of course. Thanks, that clears things up a bit.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top