CAIE A-Level Mathematics A2 3.5.3 Integration by Partial Fractions

[Maximum number: 9]

In a field there are 300 plants of a certain species, all of which can be infected by a particular disease. At time t after the first plant is infected there are x infected plants. The rate of change of x is proportional to the product of the number of plants infected and the number of plants that are not yet infected. The variables x and t are treated as continuous, and it is given that $\frac{\mathrm{d} x}{\mathrm{~d} t}=0.2$ and x=1 when t=0.

Using partial fractions, solve the differential equation and obtain an expression for t in terms of a single logarithm involving x.

Separate variables correctly
$\int \frac{1}{x(300-x)} \mathrm{d} x=\int \frac{1}{1495} \mathrm{~d} t$

Correct integration of t term
E.g. obtain t or $\frac{t}{1495}$ .

State or imply partial fractions of the form $\frac{A}{x}+\frac{B}{300-x}$
Correct method to find A or B
$A=\frac{1}{300}$ and $B=\frac{1}{300}$ .

May see $A=B=\frac{1495}{300}=\frac{299}{60}$ .

Obtain terms $\frac{1495}{300} \ln x-\frac{1495}{300} \ln (300-x)$
OE. May see $\frac{1}{300} \ln x-\frac{1}{300} \ln (300-x)$ .

Use t=0, x=1 to evaluate a constant or as limits in a solution containing terms
of the form $\ln x, \ln (300-x)$ and t.
Obtain correct answer in any form
E.g. $\frac{1495}{300}[\ln x-\ln (300-x)]=t-\frac{1495}{300} \ln 299$ .

Use law of logarithms twice to obtain an expression for t

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Obtain final answer $t=\frac{299}{60} \ln \frac{299 x}{300-x}$ or equivalent single logarithm

CAIE A-Level Mathematics A2 3.5.3 Partial Fractions Questions

Exam points

CAIE A-Level Mathematics A2 3.5.3 Partial Fractions Questions question 1