CAIE A-Level Mathematics A2 3.8 Differential Equations Questions

Practise formulating rate models, separating variables and integrating first-order differential equations before using initial conditions to find constants, times and limiting…

Exam points

translate proportional or geometric rate information into a differential equation
separate x and t or y and x terms before integrating both sides with a constant
apply initial data to determine constants and evaluate a requested time, value or limit

Covered syllabus points

3.8.1—Differential equations 3.8.2—Differential equations 3.8.3—An initial condition to find a 3.8.4—Differential equations

[Maximum number: 8]

A large plantation of area $20 \mathrm{~km}^{2}$ is becoming infected with a plant disease. At time t years the area infected is $x \mathrm{~km}^{2}$ and the rate of increase of x is proportional to the ratio of the area infected to the area not yet infected.

When t=0, x=1 and $\frac{\mathrm{d} x}{\mathrm{~d} t}=1$ .

(a)

Show that x and t satisfy the differential equation

\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{19 x}{20-x} .

[ 2 ]

State or imply equation of the form $\frac{\mathrm{d} x}{\mathrm{~d} t}=k \frac{x}{20-x}$
Obtain k=19
AG

(b)

Solve the differential equation and show that when t=1 the value of x satisfies the equation $x=\mathrm{e}^{0.9+0.05 x}$ .

[ 5 ]

Separate variables and integrate at least one side
Obtain terms $20 \ln x-x$ and 19t, or equivalent
Evaluate a constant or use t=0 and x=1 as limits in a solution containing
terms $a \ln x$ and b t
Substitute t=1 and rearrange the equation in the given form
AG

(c)

Calculate the value of t at which the entire plantation becomes infected.

[ 1 ]

[Maximum number: 8]

In a certain chemical reaction the amount, x grams, of a substance is increasing. The differential equation satisfied by x and t, the time in seconds since the reaction began, is

\frac{\mathrm{d} x}{\mathrm{~d} t}=k x \mathrm{e}^{-0.1 t},

where k is a positive constant. It is given that x=20 at the start of the reaction.

(a)

Solve the differential equation, obtaining a relation between x, t and k.

[ 5 ]

Separate variables correctly
$\int \frac{1}{x} \mathrm{~d} x=\int k \mathrm{e}^{-0.1 t} \mathrm{~d} t$

Obtain term $\ln x$
Obtain term $-10 k \mathrm{e}^{-0.1 t}$
Not from $\int x \mathrm{e}^{-0.1 t} \mathrm{~d} t$

Use x=20, t=0 to evaluate a constant or as limits in a solution containing
terms $a \ln x, b e^{-0.1 t}$ where $a b \neq 0$
Obtain $\ln x=10 k\left(1-\mathrm{e}^{-0.1 t}\right)+\ln 20$
or equivalent ISW

(b)

Given that x=40 when t=10, find the value of k and find the value approached by x as t becomes large.

[ 3 ]

Use x=40, t=10 to find k or 10 k
Available for their function of the correct structure even if they found no constant in (a).

Obtain 10 k=1.09654
or equivalent e.g. $10 k=\frac{\ln 2}{1-e^{-1}}$

State that x tends to 59.9
Need a number, not an expression for that value
3 sf or better 59.87595.....

[Maximum number: 8]

The variables x and y satisfy the differential equation
$\frac{dy}{dx}=\frac{1+4y^2}{e^x}.$
It is given that y=0 when x=1.

(a)

Solve the differential equation, obtaining an expression for y in terms of x.

[ 7 ]

Separate variables and attempt integration:
$\int\frac{1}{1+4y^2}\,dy=\int e^{-x}\,dx.$
Obtain a term of the form $a\tan^{-1}(2y)$ .
$\frac12\tan^{-1}(2y)$
Obtain
$-e^{-x}.$
Use x=1, y=0 to evaluate the constant.
Obtain a correct answer in any form.
Final answer:
$y=\frac12\tan\left(2e^{-1}-2e^{-x}\right).$

(b)

State what happens to the value of y as x tends to infinity.

[ 1 ]

As $x\to\infty$ , $e^{-x}\to0$ , so
$y\to\frac12\tan\left(2e^{-1}\right).$
B1FT

The follow-through is on correct work on a solution containing $e^{-x}$ .

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (b)