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Differential Equations

Practice exam-style IB Math AA questions for Differential Equations, aligned with the syllabus and grouped by topic.

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Question 1
HL • Paper 1
Easy
Non Calculator
HL • Paper 1
Easy
Non Calculator

A solution of the differential equation dydx=x−y\dfrac{\mathrm{d}y}{\mathrm{d}x}=x-y satisfies y(0)=2y(0)=2. Euler's method is used with step length h=12h=\dfrac12.

A

the recurrence formulae for xn+1x_{n+1} and yn+1y_{n+1}.

[2]
B

the Euler approximation to y(1)y(1).

[3]
Question 2
HL • Paper 1
Easy
Non Calculator
HL • Paper 1
Easy
Non Calculator

Euler's method is used to approximate the solution of dydx=x+y\dfrac{\mathrm{d}y}{\mathrm{d}x}=x+y, with y(0)=1y(0)=1, using step length h=12h=\dfrac12.

A

the recurrence formulae used in Euler's method.

[2]
B

Use the recurrence formulae to the Euler approximation to y(1)y(1).

[2]
Question 3
HL • Paper 2
Easy
Calculator Permitted
HL • Paper 2
Easy
Calculator Permitted

Consider the differential equation

dydx=x−y2\frac{dy}{dx}=x-y^2

with initial condition y(0)=1y(0)=1.

A

Use Euler's method with step length h=0.2h=0.2 to an approximation for y(0.6)y(0.6). Give your answer correct to three significant figures.

[5]
Question 4
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The function yy satisfies dydx=2xy2\dfrac{\mathrm{d}y}{\mathrm{d}x}=2xy^2 and y(0)=1y(0)=1.

A

the general solution in implicit form.

[3]
B

the particular solution in the form y=f(x)y=f(x).

[2]

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Question 5
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

For x>0x>0, consider the differential equation dydx+2xy=x2\dfrac{\mathrm{d}y}{\mathrm{d}x}+\dfrac{2}{x}y=x^2, with y(1)=0y(1)=0.

A

an integrating factor for this differential equation.

[2]
B

the differential equation.

[4]
Question 6
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The function yy satisfies dydx−y=ex\dfrac{\mathrm{d}y}{\mathrm{d}x}-y=e^x and y(0)=2y(0)=2.

A

an integrating factor and write the left hand side as the derivative of a product.

[2]
B

yy in terms of xx.

[3]
Question 7
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

For x>0x>0, a function yy satisfies dydx=1+yx\dfrac{\mathrm{d}y}{\mathrm{d}x}=1+\dfrac{y}{x}, with y(1)=0y(1)=0.

A

Use the substitution y=vxy=vx to vv in terms of xx.

[3]
B

the particular solution for yy.

[2]
Question 8
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The function yy satisfies dydx=1+x1+y\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1+x}{1+y} and y(0)=1y(0)=1.

A

an implicit expression relating xx and yy.

[3]
B

write the solution explicitly in the form y=f(x)y=f(x), choosing the branch consistent with the initial condition.

[2]

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Question 9
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A curve satisfies the differential equation

dydx=xe−y\frac{dy}{dx}=xe^{-y}

and passes through the point (0,ln⁡2)(0,\ln 2).

A

yy in terms of xx, and y(2)y(2).

[5]
Question 10
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

For x>0x>0, a curve satisfies

dydx=1+yx\frac{dy}{dx}=1+\frac{y}{x}

and passes through (1,2)(1,2).

A

Using the substitution y=vxy=vx, the differential equation and y(3)y(3).

[5]
Question 11
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The temperature TT, in degrees Celsius, of a heated object is modelled by

dTdt=−0.08(T−18)\frac{dT}{dt}=-0.08(T-18)

where tt is measured in minutes. Initially T=90T=90.

A

the time taken for the object to cool to 30∘C30^\circ\text{C}.

[5]
Question 12
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A radioactive substance has mass MM grams after tt years. Its decay is modelled by

dMdt=−kM,k>0\frac{dM}{dt}=-kM, \qquad k>0

After 200200 years, 65%65\% of the original mass remains.

A

kk and the half-life of the substance.

[5]

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Question 13
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A population nn is modelled by the logistic differential equation dndt=112n(6−n)\dfrac{\mathrm{d}n}{\mathrm{d}t}=\dfrac1{12}n(6-n), where 0<n<60<n<6. Initially n=2n=2 when t=0t=0.

A

ln⁡(n6−n)=t2+C\ln\left(\dfrac{n}{6-n}\right)=\dfrac{t}{2}+C.

[4]
B

the value of tt when n=3n=3.

[2]
Question 14
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

For x>0x>0, a function yy satisfies dydx=yx+xy\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{y}{x}+\dfrac{x}{y} and y(1)=2y(1)=2.

A

Use the substitution y=vxy=vx to obtain a separable differential equation in vv and xx.

[3]
B

the differential equation, giving yy in terms of xx.

[3]
Question 15
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The temperature TT, in degrees Celsius, of a cup of tea after tt minutes is modelled by dTdt=−k(T−20)\dfrac{\mathrm{d}T}{\mathrm{d}t}=-k(T-20), where k>0k>0. Initially T=80T=80, and after one minute T=50T=50.

A

the differential equation in terms of kk.

[2]
B

the exact value of kk.

[2]
C

the time at which T=35T=35.

[2]
Question 16
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

For x>0x>0, the function yy satisfies

dydx=(yx)2+yx\frac{dy}{dx}=\left(\frac{y}{x}\right)^2+\frac{y}{x}

with y(1)=1y(1)=1.

A

Use the substitution y=vxy=vx to yy in terms of xx. the vertical asymptote of the solution curve.

[6]

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Question 17
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The function yy satisfies the differential equation

dydx+2y=e−x\frac{dy}{dx}+2y=e^{-x}

with y(0)=3y(0)=3.

A

the differential equation, and the value of xx for which y=0.2y=0.2.

[6]
Question 18
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

For x>0x>0, a function yy satisfies

dydx+2xy=sin⁡x\frac{dy}{dx}+\frac{2}{x}y=\sin x

and y(1)=0y(1)=0.

A

an expression for yy involving a definite integral, and y(3)y(3).

[6]
Question 19
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

For x>0x>0, a curve satisfies

dydx=x2+y2xy\frac{dy}{dx}=\frac{x^2+y^2}{xy}

and passes through (1,1)(1,1).

A

Use the substitution y=vxy=vx to the differential equation, and y(4)y(4).

[6]
Question 20
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A quantity SS satisfies the differential equation

dSdt+0.1S=5+2e−0.2t\frac{dS}{dt}+0.1S=5+2e^{-0.2t}

where t≥0t\ge 0, and S(0)=20S(0)=20.

A

SS in terms of tt, and the limiting value of SS as t→∞t\to\infty.

[6]

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Question 21
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A curve satisfies the differential equation

dydx=x−y\frac{dy}{dx}=x-y

with initial condition y(0)=1y(0)=1.

A

Euler's method with step length h=12h=\frac12 is used to approximate y(1)y(1).

I.

the recurrence formulae for xn+1x_{n+1} and yn+1y_{n+1}.

[2]
II.

the Euler approximation to y(1)y(1).

[3]
B

the differential equation exactly, giving yy in terms of xx.

[5]
C

whether the Euler approximation found in part (a) is greater than or less than the exact value of y(1)y(1). your answer.

[1]
Question 22
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A function yy satisfies

dydx=x(1+y2),y(0)=0\frac{dy}{dx}=x(1+y^2), \qquad y(0)=0
A

the differential equation, giving yy explicitly in terms of xx.

[5]
B
I.

the least positive value of xx for which y=1y=1.

[2]
II.

the first positive vertical asymptote of the solution curve.

[2]
III.

the largest interval containing 00 on which this solution is continuous.

[1]
Question 23
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

For x>0x>0, the function yy satisfies xdydx+(x+1)y=x2e−xx\dfrac{\mathrm{d}y}{\mathrm{d}x}+(x+1)y=x^2e^{-x}, with y(1)=0y(1)=0.

A

Write the differential equation in the form y′+P(x)y=Q(x)y'+P(x)y=Q(x) and an integrating factor.

[3]
B

the differential equation.

[4]
Question 24
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A quantity nn is modelled by dndt=1100n(20−n)\dfrac{\mathrm{d}n}{\mathrm{d}t}=\dfrac1{100}n(20-n), where 0<n<200<n<20. Initially n=5n=5 when t=0t=0.

A

the general solution may be written as ln⁡(n20−n)=t5+C\ln\left(\dfrac{n}{20-n}\right)=\dfrac{t}{5}+C.

[3]
B

nn explicitly in terms of tt.

[3]
C

the value of tt when n=10n=10.

[1]

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Question 25
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A curve satisfies

dydx=x2+1y+3,y(0)=1\frac{dy}{dx}=\frac{x^2+1}{y+3}, \qquad y(0)=1
A

the differential equation, giving yy explicitly in terms of xx.

[4]
B

Use Euler's method with step length h=1h=1 to approximate y(3)y(3).

[2]
C
I.

the exact value of y(3)y(3).

[2]
II.

the value of xx for which y=5y=5.

[2]
Question 26
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A population PP is modelled by the logistic differential equation

dPdt=kP(500−P),k>0\frac{dP}{dt}=kP(500-P), \qquad k>0

where tt is measured in days. Initially P=50P=50, and after 1010 days P=150P=150.

A

the value of kk and the time at which P=400P=400.

[7]
Question 27
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The differential equation

dydx=y+x\frac{dy}{dx}=y+x

has initial condition y(0)=1y(0)=1.

A

Use Euler's method with step length h=0.5h=0.5 to approximate y(1)y(1). Then the differential equation exactly and the percentage error in the Euler approximation.

[7]
Question 28
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A population NN, where 0<N<100<N<10, is modelled by

dNdt=120N(10−N),N(0)=2\frac{dN}{dt}=\frac1{20}N(10-N), \qquad N(0)=2
A
I.

Separate variables and to obtain an implicit relation between NN and tt.

[3]
II.

NN explicitly in terms of tt.

[4]
B
I.

The rate of growth is greatest when N=aN=a. aa.

[2]
II.

the exact time at which the rate of growth is greatest.

[2]
C

lim⁡t→∞N(t)\lim_{t\to\infty}N(t) and this value in the context of the model.

[2]

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Question 29
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

For x>0x>0, a function yy satisfies

dydx−2xy=x2cos⁡x,y(π2)=0\frac{dy}{dx}-\frac{2}{x}y=x^2\cos x, \qquad y\left(\frac{\pi}{2}\right)=0
A
I.

an integrating factor for the differential equation.

[2]
II.

the differential equation.

[4]
B

Consider the solution curve on the interval 0<x<π0<x<\pi.

I.

x=π2x=\frac{\pi}{2} is a stationary point of the curve.

[2]
II.

whether this stationary point is a local maximum or a local minimum. your answer.

[2]
Question 30
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A function yy satisfies

dydx+y=exy2,y(0)=1\frac{dy}{dx}+y=e^xy^2, \qquad y(0)=1

Let z=1yz=\frac1y.

A
I.

zz satisfies the linear differential equation z′−z=−exz'-z=-e^x.

[3]
II.

the differential equation for yy in terms of xx.

[4]
B
I.

the exact value of y(ln⁡2)y(\ln2).

[1]
II.

the vertical asymptote of the solution and your answer.

[2]
Question 31
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

The function yy satisfies

dydx=y(2−x),y(0)=1\frac{dy}{dx}=y(2-x), \qquad y(0)=1
A

Use Euler's method with step length h=12h=\frac12 to approximate y(1)y(1).

[4]
B
I.

the differential equation exactly.

[3]
II.

the exact value of y(1)y(1).

[1]
C

Without using a calculator, whether the Euler approximation is greater than or less than the exact value.

[3]
Question 32
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A quantity PP satisfies

dPdt=112(P−3)(9−P),3<P<9\frac{dP}{dt}=\frac1{12}(P-3)(9-P), \qquad 3<P<9

and P=5P=5 when t=0t=0.

A
I.

an implicit relation between PP and tt.

[4]
II.

PP explicitly in terms of tt.

[3]
B
I.

the exact time at which P=7P=7.

[2]
II.

the two equilibrium values and which one is approached by P(t)P(t) as t→∞t\to\infty.

[3]

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Question 33
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

For x>0x>0, a curve satisfies

dydx=yx+(yx)3,y(1)=1\frac{dy}{dx}=\frac{y}{x}+\left(\frac{y}{x}\right)^3, \qquad y(1)=1

Use the substitution y=vxy=vx.

A
I.

a separable differential equation in vv and xx.

[3]
II.

yy in terms of xx.

[5]
B
I.

the largest interval containing 11 on which this solution is defined.

[2]
II.

the gradient of the curve at (1,1)(1,1).

[1]
Question 34
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A function yy satisfies

dydx=2x(1−y),y(0)=0\frac{dy}{dx}=2x(1-y), \qquad y(0)=0
A

Use Euler's method with step length h=12h=\frac12 to approximate y(1)y(1).

[4]
B
I.

the differential equation exactly.

[4]
II.

the exact value of y(1)y(1).

[1]
C

whether the Euler approximation is greater than or less than the exact value. your answer without using a calculator.

[2]
Question 35
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A colony of bacteria is modelled by the logistic differential equation

dPdt=0.4P(5−P),0<P<5\frac{dP}{dt}=0.4P(5-P), \qquad 0<P<5

where PP is measured in thousands and tt is measured in hours. Initially P=1.2P=1.2 when t=0t=0.

A

Use Euler's method with step length h=1h=1 to an approximation for P(4)P(4).

[3]
B
I.

the differential equation to

P=51+196e−2tP=\frac{5}{1+\frac{19}{6}e^{-2t}}
[4]
II.

the exact-model value of P(4)P(4).

[2]
C

Using your solution in part (b), the time at which the population first reaches 4.54.5 thousand.

[3]
Question 36
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The temperature TT, in degrees Celsius, of an industrial component after tt minutes is modelled by

dTdt+0.12T=2.4+4e−0.05t,T(0)=70\frac{dT}{dt}+0.12T=2.4+4e^{-0.05t}, \qquad T(0)=70
A

Part (a)

I.

the Euler recurrence formulae for step length h=5h=5.

[2]
II.

Use these formulae to approximate T(15)T(15).

[2]
B

the differential equation exactly.

[4]
C

Part (c)

I.

Use your exact solution to T(15)T(15).

[2]
II.

the time at which T=30T=30.

[2]

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Question 37
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
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A population NN is modelled by

dNdt=kN(800−N),k>0\frac{dN}{dt}=kN(800-N), \qquad k>0

where tt is measured in days. Initially N=40N=40, and after 66 days N=160N=160.

A
I.

the differential equation and the value of kk.

[4]
II.

Write NN explicitly as a function of tt.

[2]
B

the time at which N=600N=600.

[3]
C

Using the unrounded value of kk found in part (a), use Euler's method with step length h=2h=2 to approximate N(6)N(6), and on the approximation.

[2]
Question 38
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For x>0x>0, the function yy satisfies

xdydx+(1−2x)y=x2,y(1)=3x\frac{dy}{dx}+(1-2x)y=x^2, \qquad y(1)=3
A

Write the differential equation in standard linear form and an integrating factor.

[4]
B

the differential equation.

[3]
C
I.

y(1.5)y(1.5).

[2]
II.

the values of xx for which y=10y=10.

[2]
Question 39
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The amount MM, in milligrams, of a medicine in the bloodstream after tt hours is modelled by

dMdt+0.4M=20e−0.1t,M(0)=0\frac{dM}{dt}+0.4M=20e^{-0.1t}, \qquad M(0)=0
A

the differential equation.

[4]
B
I.

the time at which MM is a maximum.

[2]
II.

the maximum amount of medicine in the bloodstream.

[3]
C

Use Euler's method with step length h=1h=1 to approximate M(3)M(3), and this with the exact value.

[2]
Question 40
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For x≥0x\ge 0, a function yy satisfies

dydx+2x1+x2y=cos⁡x1+x2,y(0)=4\frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{\cos x}{1+x^2}, \qquad y(0)=4
A

the exact solution for yy in terms of xx.

[3]
B
I.

Use Euler's method with step length h=0.25h=0.25 to approximate y(1)y(1).

[3]
II.

the exact value of y(1)y(1) and the percentage error in the Euler approximation.

[2]
C
I.

the xx-coordinate of the maximum point of the exact solution for 0≤x≤40\le x\le 4.

[2]
II.

the maximum value of yy on this interval.

[2]

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Question 41
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A model for the height yy metres of a fast-growing plant after xx weeks satisfies

dydx=y(2−x),y(0)=1,0≤x≤1\frac{dy}{dx}=y(2-x), \qquad y(0)=1, \qquad 0\le x\le 1

The diagram shows four representative slope segments for this differential equation on the interval 0≤x≤10\le x\le 1.

A
I.

the recurrence formulae for Euler's method with step length h=0.25h=0.25.

[2]
II.

Use the recurrence formulae to the Euler approximation to y(0.5)y(0.5).

[2]
B

the differential equation exactly, giving yy in terms of xx.

[4]
C
I.

Continue Euler's method to x=1x=1 and the percentage error in the approximation. If you did not obtain an exact solution in part (b), use y=e2x−x22y=e^{2x-\frac{x^2}{2}}.

[2]
II.

, using calculus, why Euler's method with this step length gives an underestimate on 0≤x≤10\le x\le 1.

[2]
Question 42
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The number NN of cells, in thousands, in a culture is modelled by the logistic differential equation

dNdt=kN(1000−N),0<N<1000\frac{dN}{dt}=kN(1000-N), \qquad 0<N<1000

where tt is measured in days. Initially N=100N=100, and after 55 days N=250N=250.

A
I.

the general solution can be written in the form

ln⁡(N1000−N)=1000kt+C\ln\left(\frac{N}{1000-N}\right)=1000kt+C
[3]
II.

the exact value of kk.

[2]
B

NN explicitly as a function of tt.

[3]
C
I.

the time when N=750N=750.

[2]
II.

the maximum growth rate predicted by the model, and when it occurs.

[2]
Question 43
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A tank initially contains 5050 litres of solution with 55 kg of dissolved salt. Brine enters at 33 litres per minute with salt concentration 0.40.4 kg per litre. The well-mixed solution leaves at 22 litres per minute.

Let SS be the mass of salt, in kg, in the tank after tt minutes.

A
I.

why SS satisfies

dSdt=1.2−2S50+t\frac{dS}{dt}=1.2-\frac{2S}{50+t}
[2]
II.

Write the differential equation in linear form and an integrating factor.

[2]
B

for SS in terms of tt.

[4]
C
I.

the time when the concentration in the tank first reaches 0.350.35 kg per litre.

[2]
II.

the limiting concentration predicted by the model.

[2]
Question 44
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A metal probe has temperature TT degrees Celsius at time tt minutes. The surrounding air temperature is rising linearly and is given by

A(t)=18+0.5tA(t)=18+0.5t

The probe is modelled by Newton's law of cooling,

dTdt=−0.1(T−A(t)),T(0)=40\frac{dT}{dt}=-0.1(T-A(t)), \qquad T(0)=40
A
I.

Write the differential equation in the form T′+P(t)T=Q(t)T'+P(t)T=Q(t).

[1]
II.

the integrating factor and for TT in terms of tt.

[3]
B

the time when the probe and the surrounding air have the same temperature.

[3]
C
I.

, after the time found in part (b), the probe remains cooler than the surrounding air.

[2]
II.

the limiting value of T−AT-A as t→∞t\to\infty and it in context.

[2]

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Question 45
HL • Paper 3
Hard
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HL • Paper 3
Hard
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For x>0x>0, a curve satisfies

dydx=2yx−xy,y(1)=2\frac{dy}{dx}=2\frac{y}{x}-\frac{x}{y}, \qquad y(1)=2

The curve lies above the xx-axis.

A
I.

Use the substitution y=vxy=vx to form a separable differential equation.

[2]
II.

v2−1=Cx2v^2-1=Cx^2, where CC is a constant.

[2]
B

yy explicitly in terms of xx.

[3]
C

the value of xx at which the gradient of the curve is 33.

[3]
Question 46
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A family of solution curves is defined by

dydx=x(1−y2),y(0)=a\frac{dy}{dx}=x(1-y^2), \qquad y(0)=a

where 0<a<10<a<1.

Student-facing visual


None

A
I.

Separate the variables and to obtain an implicit solution.

[3]
II.

write yy explicitly in terms of xx and aa.

[2]
B

For a=0.2a=0.2, the positive value of xx when y=0.8y=0.8.

[3]
C

why no solution with 0<a<10<a<1 crosses the line y=1y=1.

[3]
Question 47
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The differential equation

dydx=x2−y,y(0)=1\frac{dy}{dx}=x^2-y, \qquad y(0)=1

is solved exactly and also approximated using Euler's method.

Step length hh

Steps to x=1x=1

Step indices

Abscissas xnx_n

Euler update

0.50.5

2

n=0,1n=0,1

0, h, 2h=10,\,h,\,2h=1

xn+1=xn+h,yn+1=yn+h(xn2−yn)x_{n+1}=x_n+h,\quad y_{n+1}=y_n+h(x_n^2-y_n)

0.250.25

4

n=0,1,2,3n=0,1,2,3

0, h, 2h, 3h, 4h=10,\,h,\,2h,\,3h,\,4h=1

xn+1=xn+h,yn+1=yn+h(xn2−yn)x_{n+1}=x_n+h,\quad y_{n+1}=y_n+h(x_n^2-y_n)

A
I.

Use Euler's method with h=0.5h=0.5 to approximate y(1)y(1).

[2]
II.

Use Euler's method with h=0.25h=0.25 to approximate y(1)y(1).

[2]
B

the differential equation exactly and the exact value of y(1)y(1).

[5]
C

Let E0.5E_{0.5} and E0.25E_{0.25} denote the absolute errors in the Euler approximations with step lengths 0.50.5 and 0.250.25, respectively. the ratio E0.5E0.25\frac{E_{0.5}}{E_{0.25}}. on what this suggests about halving the step length.

[3]
Question 48
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
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A falling object has velocity vv metres per second after tt seconds. A simplified air resistance model is

dvdt=9−v225,v(0)=0,0≤v<15\frac{dv}{dt}=9-\frac{v^2}{25}, \qquad v(0)=0, \qquad 0\le v<15
A
I.

the terminal velocity predicted by the model is 1515 m s−1^{-1}.

[1]
II.

the differential equation exactly.

[4]
B

Use the exact solution to the time when v=12v=12.

[2]
C
I.

Use Euler's method with h=0.5h=0.5 to approximate v(1)v(1).

[2]
II.

whether this Euler approximation is an overestimate or an underestimate. your answer.

[3]

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Question 49
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A pollutant enters a lake after a spill. The amount of pollutant in the lake, AA kg, after tt days is modelled by

dAdt+0.15A=60e−0.05t,A(0)=0\frac{dA}{dt}+0.15A=60e^{-0.05t}, \qquad A(0)=0

The term 60e−0.05t60e^{-0.05t} models the decreasing rate at which pollutant enters the lake.

A
I.

an integrating factor for the differential equation.

[1]
II.

for AA in terms of tt.

[4]
B

the maximum amount of pollutant in the lake.

[4]
C
I.

the limiting amount of pollutant as t→∞t\to\infty.

[1]
II.

why the amount eventually decreases even though pollutant is still entering the lake.

[1]
Question 50
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

For x>0x>0, a curve satisfies

dydx=x+yx−y,y(1)=0\frac{dy}{dx}=\frac{x+y}{x-y}, \qquad y(1)=0

Use the substitution y=vxy=vx where appropriate.

A
I.

the differential equation may be written as a separable differential equation in vv and xx.

[3]
II.

an implicit equation for the curve in terms of xx and yy.

[5]
B
I.

the equation of the tangent to the curve at (1,0)(1,0).

[1]
II.

Using the implicit relation, the limiting value of xx as the solution approaches the line y=xy=x.

[2]
Question 51
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A tank contains AA grams of salt after tt minutes. Initially A=12A=12. For 0≤t≤40\le t\le4 the amount of salt satisfies

dAdt=8−A6\frac{dA}{dt}=8-\frac{A}{6}

After t=4t=4, fresh water is added so that, for t>4t>4,

dAdt=−A6\frac{dA}{dt}=-\frac{A}{6}
A

AA in terms of tt for 0≤t≤40\le t\le4.

[5]
B
I.

the amount of salt in the tank when t=4t=4.

[2]
II.

AA in terms of tt for t>4t>4.

[3]
C

the elapsed time after t=4t=4 until the amount of salt has returned to 1212 grams.

[2]
Question 52
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A curve satisfies

dydx=xy+xy,y(0)=2\frac{dy}{dx}=xy+\frac{x}{y}, \qquad y(0)=2

Let z=y2z=y^2.

A
I.

zz satisfies the linear differential equation z′−2xz=2xz'-2xz=2x.

[3]
II.

the original differential equation, giving yy in terms of xx.

[5]
B
I.

the stationary point of the curve.

[2]
II.

whether this stationary point is a local maximum or local minimum.

[1]

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Question 53
HL • Paper 2
Hard
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HL • Paper 2
Hard
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For x>0x>0 and y>0y>0, a curve satisfies

dydx=x+2y2x+y\frac{dy}{dx}=\frac{x+2y}{2x+y}

and passes through the point (1,2)(1,2).

A

Use the substitution y=vxy=vx to

xdvdx=1−v2v+2x\frac{dv}{dx}=\frac{1-v^2}{v+2}
[3]
B

the solution may be written in the form

v+1(v−1)3=3x2\frac{v+1}{(v-1)^3}=3x^2

where v=yxv=\frac{y}{x}.

[4]
C
I.

the value of yy when x=2x=2.

[2]
II.

the value of yy when x=0.5x=0.5.

[2]
Question 54
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For x>0x>0, a curve satisfies

dydx=yx+(yx)3,y(1)=1\frac{dy}{dx}=\frac{y}{x}+\left(\frac{y}{x}\right)^3, \qquad y(1)=1
A

Use the substitution y=vxy=vx to obtain a separable differential equation in vv and xx.

[3]
B

y=x1−2ln⁡xy=\frac{x}{\sqrt{1-2\ln x}}
[3]
C
I.

Use Euler's method with step length h=0.1h=0.1 to approximate y(1.3)y(1.3).

[2]
II.

the exact value of y(1.3)y(1.3) and the largest possible value of xx for this solution.

[3]
Question 55
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A tank contains a salt solution. Let AA be the mass of salt, in kg, in the tank after tt minutes. A model for AA is

dAdt+0.05A=0.4+0.1sin⁡(0.1t),A(0)=10\frac{dA}{dt}+0.05A=0.4+0.1\sin(0.1t), \qquad A(0)=10
A

the differential equation.

[4]
B

A(30)A(30).

[2]
C
I.

the first time after t=0t=0 at which A=9A=9.

[2]
II.

the mean value and amplitude of the long-term oscillation of AA.

[2]
Question 56
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The velocity vv, in metres per second, of an object moving vertically through a fluid is modelled by

dvdt=12−0.08v2,v(0)=0\frac{dv}{dt}=12-0.08v^2, \qquad v(0)=0

where tt is measured in seconds.

A
I.

the terminal velocity predicted by the model.

[2]
II.

Use Euler's method with step length h=0.5h=0.5 to approximate v(1.5)v(1.5).

[2]
B

the differential equation, giving vv in terms of tt.

[4]
C
I.

the time at which the object first reaches 90%90\% of its terminal velocity.

[2]
II.

The displacement from the starting point is denoted by ss. Given that s(0)=0s(0)=0 and dsdt=v\frac{ds}{dt}=v, the displacement when the object first reaches 90%90\% of its terminal velocity.

[2]

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Question 57
HL • Paper 3
Hard
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HL • Paper 3
Hard
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A curve which enters the first quadrant through (1,0)(1,0) satisfies

dydx=x+2y2x+y,y(1)=0\frac{dy}{dx}=\frac{x+2y}{2x+y}, \qquad y(1)=0

The gradient at a point depends only on the ratio yx\frac{y}{x}.

A
I.

Use the substitution y=vxy=vx to obtain a separable differential equation in vv and xx.

[2]
II.

2+v1−v2=32(1−v)+12(1+v)\frac{2+v}{1-v^2}=\frac{3}{2(1-v)}+\frac{1}{2(1+v)}
[2]
B

the differential equation, giving an implicit equation in xx and yy.

[4]
C
I.

the point on the curve where y=x2y=\frac{x}{2}. If you did not obtain an implicit equation, use 1+v(1−v)3=x2\frac{1+v}{(1-v)^3}=x^2.

[2]
II.

the limiting value of yx\frac{y}{x} as xx increases along this branch of the curve.

[2]
Question 58
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
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For a positive constant aa, the function yy satisfies

y′+ay=e−x,y(0)=0y'+ay=e^{-x}, \qquad y(0)=0

The behaviour changes when a=1a=1.

A
I.

For a≠1a\ne 1, the differential equation in terms of aa and xx.

[3]
II.

the differential equation for the special case a=1a=1.

[2]
B

the expression found in part (a)(i) tends to the result in part (a)(ii) as a→1a\to 1.

[2]
C
I.

For a=2a=2, the maximum value of yy.

[3]
II.

why y>0y>0 for all x>0x>0 when a>0a>0.

[1]
Question 59
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For a real constant cc, consider the family of solutions of

dydx+2x1+x2y=4x,y(0)=c\frac{dy}{dx}+\frac{2x}{1+x^2}y=4x, \qquad y(0)=c

where x≥0x\ge 0.

A
I.

an integrating factor for the differential equation.

[2]
II.

the solution is

y=x4+2x2+c1+x2y=\frac{x^4+2x^2+c}{1+x^2}
[3]
B

the value of cc for which the solution has a stationary point at x=1x=1.

[3]
C
I.

For c=5c=5, the stationary point at x=1x=1 is a local minimum.

[3]
II.

the coordinates of this local minimum.

[1]
Question 60
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For x>0x>0, a curve satisfies

dydx=x+yx−y,y(1)=0\frac{dy}{dx}=\frac{x+y}{x-y}, \qquad y(1)=0
A

Use the substitution y=vxy=vx to

xdvdx=1+v21−vx\frac{dv}{dx}=\frac{1+v^2}{1-v}
[3]
B

the solution satisfies

arctan⁡(yx)−12ln⁡(1+y2x2)=ln⁡x\arctan\left(\frac{y}{x}\right)-\frac12\ln\left(1+\frac{y^2}{x^2}\right)=\ln x
[3]
C
I.

the value of yy when x=1.2x=1.2.

[2]
II.

the limiting upper bound (supremum) of xx for this solution curve as it approaches the singular line y=xy=x.

[3]

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Differential Calculus

Integral Calculus