If you exponentiate the left side, you getP/(P - 1) = Ce^(2t)

In summary: Instead it is e^(2t) * e^C = Ce^(2t)Then you have P/(P-1) = Ce^(2t)Cross multiply to get P = Ce^(2t) * (P - 1)In summary, the dependent variable P can be expressed explicitly as P = Ce^(2t) * (P - 1).
  • #1
SpartanG345
70
1

Homework Statement


dP/dt = 2P(1-P)
write a explicit expression for the dependent variable


Homework Equations



Seperable equation

The Attempt at a Solution


Is the dependent variable P
does this mean you should write a function of P in terms of t?

seperation of variables then integration

lnP - ln(p-1) = 2t + C
p/(P-1) = e^2t + e^c

this is all i can get i cannot get an explicit relationship for the dependent variable which is P
wolfram gave this

http://www.wolframalpha.com/input/?i=dx/dt+=+2x*(1-x)"

i cannot get that no idea totally stuck :(
 
Last edited by a moderator:
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  • #2
SpartanG345 said:

Homework Statement


dP/dt = 2P(1-P)
write a explicit expression for the dependent variable


Homework Equations



Seperable equation

The Attempt at a Solution


Is the dependent variable P
does this mean you should write a function of P in terms of t?

seperation of variables then integration

lnP - ln(p-1) = 2t + C
p/(P-1) = e^2t + e^c
first it doesn't look like you have taken the exponential of the RHS correctly,

second when you have, multply both sides by (p-1), group p terms & divide by the coefficient you get, hopefully should get you closer

SpartanG345 said:
this is all i can get i cannot get an explicit relationship for the dependent variable which is P
wolfram gave this

http://www.wolframalpha.com/input/?i=dx/dt+=+2x*(1-x)"

i cannot get that no idea totally stuck :(
 
Last edited by a moderator:
  • #3
SpartanG345 said:
lnP - ln(p-1) = 2t + C
p/(P-1) = e^2t + e^c
The step missing in the middle is
ln(P/(P - 1)) = 2t + C

When you exponentiate (make each side the exponent on e), what you ended up with on the right side is incorrect. e^(2t + C) != e^(2t) + e^C
 

What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It involves variables, functions, and their rates of change, and is often used to model real-world phenomena.

What is the difference between an ordinary and a partial differential equation?

An ordinary differential equation involves a single independent variable and its derivatives, while a partial differential equation involves multiple independent variables and their partial derivatives.

What are the applications of differential equations?

Differential equations have numerous applications in various fields such as physics, engineering, economics, biology, and more. They are used to model and analyze dynamic systems and predict their behavior over time.

What are the methods for solving differential equations?

There are several methods for solving differential equations, including separation of variables, substitution, integration, and numerical methods such as Euler's method and the Runge-Kutta method.

What is the role of initial and boundary conditions in solving differential equations?

Initial conditions specify the values of the dependent variable and its derivatives at a specific point, while boundary conditions specify the behavior of the dependent variable at the boundaries of the domain. These conditions are crucial in determining unique solutions to differential equations.

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