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# Hard differential equation OCR watch

1. Hey guys,

Really stuck on this question....

When a plague of locusts attack a wheat crop, the proportion of the crop destroyed after t hours is denoted by x. In a model, it is assumed that the rate at which the crop is destroyed is proportional to x(1-x).
A plague of locusts is discovered in a wheat crop when 1/4 of the crop has been destroyed.
Given that the rate of destruction at this instant is such that if it remained constant, the crop would be completley destroyed in a further 6 hours.

i) show that dx/dt= 2/3x(1-x)
ii) find the percentage of the crop destroyed 3 hours after the plague of locusts is discovered.

Okay so dx/dt= kx(1-x)

conditions, t=o when X=1/4
and dx/dt=k, t=0?

2. For part i)

When x=0.25, the rate of destruction is such that the remaining 0.75 of the crops would be destroyed in 6 hours, so dx/dt=0.75/6=3/24 where x=0.25

As you rightly stated dx/dt=kx(1-x)

So when x=0.25, kx(1-x)=3/24
3. (Original post by Dj.Clay)
For part i)

When x=0.25, the rate of destruction is such that the remaining 0.75 of the crops would be destroyed in 6 hours, so dx/dt=0.75/6=3/24 where x=0.25

As you rightly stated dx/dt=kx(1-x)

So when x=0.25, kx(1-x)=3/24
4. (Original post by J DOT A)
Substitute 1/4 in to get k
5. If a quarter of the field has been destroyed, then remains, that is then destroyed in 6 hours so the hourly rate

@ x = 0.25

Therefore:

To integrate:

Use partial fractions to simplify the integral:

@ x = 0

@ x = 1

Therefore

And then integrate:

Put in known values to find c:
@ t = 0, x = 0.25

And now put the value of t in to find x:
@ t = 3

Which gives:

And then rearrange to make x the subject:

Which means 71% of the crop has been destroyed 3 hours after discovering the locusts.

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Updated: December 1, 2011
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