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C4 help needed!

I am currently doing the June 2016 OCR MEI C4 Maths Paper and I am struggling with question 8(v). It goes as follows:

In another chemical reaction, the mass x (mg) at time t (minutes) is modelled by the differential equation

dx/dt = k(2+x)(2-x)e^-t

where k is a positive constant, and x = 0 when t = 0.

8(v): Show by integration that, for this reaction, ln((2+x)/(2-x)) = 4k(1-e^-t)

The mark scheme makes 0 sense, I was hoping someone could help me out.
The link to the actual paper and mark scheme:http://mei.org.uk/files/papers/C4_paper_AandB_2016.pdf
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aaabattery
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Re-arrange the equation to get all of your x terms on the left hand side of the equation and all the ones containing t on the right hand side

1/(4-x^2) dx = ke^-t dt

Then integrate both sides to get the result they want.
Start by diving both sides by (2+x)(2-x) and then bringing the dt to the right hand side.
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Original post by FP5
Re-arrange the equation to get all of your x terms on the left hand side of the equation and all the ones containing t on the right hand side

1/(4-x^2) dx = ke^-t dt

Then integrate both sides to get the result they want.


Original post by BTAnonymous
Start by diving both sides by (2+x)(2-x) and then bringing the dt to the right hand side.


The integrating is the very part im stuck on :/

How do i actually integrate 1/(2-x)(2+x), the mark scheme introduces new variables, alpha, beta and gamma. Thanks for your input anyway

I used the online integration calculator and even they got a different answer, idk.
(edited 5 years ago)
Original post by aaabattery
The integrating is the very part im stuck on :/


split it into partial fractions, you should be able to integrate then :smile:
Original post by aaabattery
The integrating is the very part im stuck on :/

How do i actually integrate 1/(2-x)(2+x), the mark scheme introduces new variables, alpha, beta and gamma. Thanks for your input anyway

I used the online integration calculator and even they got a different answer, idk.


partial fractions.
Reply 8
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aaabattery
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Original post by BTAnonymous
partial fractions.


dx/dt = k(2+x)(2-x)e^-t

∫ 1/(2+x)(2-x) dx = ∫ ke^-t dt

∫ (4-x^2)^-1 dx = -(ke^-t)/t +c

ln |?| = -(ke^-t)/t +c

What I mean is I physically cant integrate the left side. The mark scheme ( page 43 on link provided) introduces some kind of new variables.
Original post by aaabattery
dx/dt = k(2+x)(2-x)e^-t

∫ 1/(2+x)(2-x) dx = ∫ ke^-t dt

∫ (4-x^2)^-1 dx = -(ke^-t)/t +c

ln |?| = -(ke^-t)/t +c

What I mean is I physically cant integrate the left side. The mark scheme ( page 43 on link provided) introduces some kind of new variables.


do you know how to do partial fractions?
Original post by BTAnonymous
do you know how to do partial fractions?


I only just realised what u were talking about with partial fractions, lmao I’m slow in the head.You are correct, thanks so much.
Original post by aaabattery
I only just realised what u were talking about with partial fractions, lmao I’m slow in the head.You are correct, thanks so much.


no problem :wink:

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