CAIE A-Level Mathematics 3.8 Differential Equations Question Bank

CAIE A-Level Mathematics 3.8 Differential Equations Question Bank
Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2028, 2029 and 20302028–2030

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 equations3.8.2—Differential equations3.8.3—An initial condition to find a3.8.4—Differential equations

Question 8

[Maximum number: 8]

The variables x and y satisfy the differential equation

(x2+1)dy dx=kxe2y,\left(x^{2}+1\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=k x \mathrm{e}^{2 y},

where k is a constant. It is given that y=0 when x=0 and that y=−12y=-\frac{1}{2} when x=1.
Solve the differential equation and find the exact value of y when x=3x=\sqrt{3}.

Question 10

[Maximum number: 10]

The variables x and y satisfy the differential equation

sin⁡4y dy dx=xsin⁡2ysin⁡3x.\sin 4 y \frac{\mathrm{~d} y}{\mathrm{~d} x}=x \sin 2 y \sin 3 x .

It is given that y=112πy=\frac{1}{12} \pi when x=12πx=\frac{1}{2} \pi.

Question 10(a)

(a)

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

[ 8 ]

Question 10(b)

(b)

Given that 0<y<12π0<y<\frac{1}{2} \pi, find the values of y when x=0.

[ 2 ]

Question 10

[Maximum number: 9]

The diagram shows a tank for holding water. The tank is in the shape of a cube of side 50 cm. At time t seconds, the depth of water in the tank is h cmh \mathrm{~cm}. Water is poured into the tank at a rate of 5000 cm3 s−15000 \mathrm{~cm}^{3} \mathrm{~s}^{-1}. Water pours out of the tank through a hole in the bottom at a rate proportional to h2h^{2}.

When h=20, the depth of the water is increasing at a rate of 0.4cms−10.4 \mathrm{cms}^{-1}.

Question 10(a)

(a)

Show that dh dt=500−h2250\frac{\mathrm{d} h}{\mathrm{~d} t}=\frac{500-h^{2}}{250}.

[ 4 ]

Question 10(b)

(b)

Given that h=0 when t=0, find the time taken for the depth of the water in the tank to reach 20 cm.

[ 5 ]

Question 10

[Maximum number: 1]

In a chemical reaction, a compound X is formed from two compounds Y and Z.
The masses in grams of X, Y and Z present at time t seconds after the start of the reaction are x, 10-x and 20-x respectively. At any time the rate of formation of X is proportional to the product of the masses of Y and Z present at the time. When t=0, x=0 and dx dt=2\frac{\mathrm{d} x}{\mathrm{~d} t}=2.

Question 10(a)

(a)

Show that x and t satisfy the differential equation

dx dt=0.01(10−x)(20−x).\frac{\mathrm{d} x}{\mathrm{~d} t}=0.01(10-x)(20-x) .

Question 10(c)

(b)

State what happens to the value of x when t becomes large.

[ 1 ]