# C4 help needed!

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#1
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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#2
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#3
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1 year ago
#4
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.
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1 year ago
#5
Start by diving both sides by (2+x)(2-x) and then bringing the dt to the right hand side.
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#6
(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.
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1 year ago
#7
(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
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1 year ago
#8
(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.
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#9
(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.
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1 year ago
#10
(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?
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#11
(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.
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1 year ago
#12
(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
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