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MEI C4 Qs 7i)

I Just don't understand the mark scheme!

http://www.mei.org.uk/files/papers/c410ja_4754.pdf

Thx :smile:
Reply 1
Avatar for ttoby
ttoby
15
You need to take the equation given, x=11+e−tx=\dfrac{1}{1+e^{-t}} and calculate dx/dt and x(1-x) to check if they give the same answer, as well as checking the initial conditions.


The easiest way of doing it is to calculate (1-x) in terms of t by adding fractions, then work out x(1-x). Then calculate dx/dt using the chain rule.
Original post by ttoby
You need to take the equation given, and calculate dx/dt and x(1-x) to check if they give the same answer, as well as checking the initial conditions.


The easiest way of doing it is to calculate (1-x) in terms of t by adding fractions, then work out x(1-x). Then calculate dx/dt using the chain rule.


How? :frown: ... What fractions?
Original post by Newyork Nagaram...
How? :frown: ... What fractions?

1−x≡1−11+e−t≡1+e−t1+e−t−11+e−t1-x \equiv 1 - \dfrac{1}{1+e^{-t}} \equiv \dfrac{1+e^{-t}}{1+e^{-t}} - \dfrac{1}{1+e^{-t}}.
Reply 4
Avatar for RK92
RK92
2
what you do now is put that solution x=1/(1+e^-t) into the right hand side of dx/dt=x(1-x)

if you then show that the right hand side you obtained by putting x into the equation is the same as just differentiating the x solution you were given, then the left hand side and right hand side are equal so the solution x=1/(1+e^-t) satisfies the differential equation.

lol neg
(edited 12 years ago)
Original post by Farhan.Hanif93
.


This is the Bit i don't Get... How is 1-x= 1-(1/1+e^-t)??
Original post by Newyork Nagaram...
This is the Bit i don't Get... How is 1-x= 1-(1/1+e^-t)??

You're told to verify that the differential equation dxdt=x(1−x)\dfrac{dx}{dt}=x(1-x) is satisfied by x=11−e−tx=\dfrac{1}{1-e^{-t}} so you're gonna have to sub that in, in place of x, in the RHS and also calculate the derivative directly for the LHS and show that they are equal.
Reply 7
Avatar for ttoby
ttoby
15
Original post by Newyork Nagaram...

Original post by Newyork Nagaram...
This is the Bit i don't Get... How is 1-x= 1-(1/1+e^-t)??


You're told that x=11+e−tx=\dfrac{1}{1+e^{-t}} so 1−x=1−11+e−t1-x=1-\dfrac{1}{1+e^{-t}}
Thank you everyone!!! I got IT :biggrin: !!! Rep+ Everyone!

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