Differential equations — Paper 3 · A Level Mathematics 9709

10 A rectangular reservoir has a horizontal base of area 1000 m2. At time t = 0, it is empty and water begins to flow into it at a constant rate of 30 m3 s−1. At the same time, water begins to flow out at a dh rate proportional to √h, where h m is the depth of the water at time t s. When h = 1, = 0.02. dt (i) Show that h satisfies the differential equation dh = 0.01(3 −√h). [3] dt It is given that, after making the substitution x = 3 −√h, the equation in part (i) becomes (x −3)dx = 0.005x. dt (ii) Using the fact that x = 3 when t = 0, solve this differential equation, obtaining an expression for t in terms of x. [5] (iii) Find the time at which the depth of water reaches 4 m. [2]

10 marks

Mark scheme: d V d h 10 (i) State or imply = 1000 B1 d t d t d V d h State or imply = 30 − k h or = 0 . 03 − m h B1 d t d t Show that k = 10 or m = 0.01 and justify the given equation B1 3 [Allow the first B1 for the statement that 0.03 = 30/1000.] x − 3 (ii) Separate variables and attempt integration of with respect to x M1* x Obtain x −3 ln x, or equivalent A1 Obtain 0.005t, or equivalent A1 Use x = 3, t = 0 in the evaluation of a constant or as limits in an answer involving ln x and kt M1(dep*) Obtain answer in any correct form e.g. t = 200(x −3 −3 ln x + 3 ln 3) A1 5 [To qualify for the first M mark, an attempt to solve the earlier differential equation in h and t must involve correct separation of variables, the use of a substitution such as h = u , and an attempt to integrate the resulting function of u.] (iii) Substitute x = 1 and calculate t M1 Obtain answer t = 259 correctly A1 2

This question in 9709/31 Oct/Nov 2004

8 (i) Using partial fractions, find 1 dy. [4] y(4 −y) (ii) Given that y = 1 when x = 0, solve the differential equation dy = y(4 −y), dx obtaining an expression for y in terms of x. [4] (iii) State what happens to the value of y if x becomes very large and positive. [1]

9 marks

Mark scheme: 8 (i) Attempt to express integrand in partial fractions, A B e.g. obtain A or B in + M1 y 4 − y 1 1 1 Obtain ( + ) , or equivalent A1 4 y 4 − y Integrate and obtain 41 ln y − 41 ln ( 4 − y ) , or equivalent A1√ + A1√ 4 A B (ii) Separate variables correctly, integrate + and obtain further y 4 − y term x, or equivalent M1* Use y = 1 and x = 0 to evaluate a constant, or as limits M1(dep*) Obtain answer in any correct form A1 Obtain final answer y = 4 /( 3 e −x4 + )1 , or equivalent A1 4 (iii) State that y approaches 4 as x becomes very large B1 1

This question in 9709/31 May/June 2005

8 In a certain chemical reaction the amount, x grams, of a substance present is decreasing. The rate of decrease of x is proportional to the product of x and the time, t seconds, since the start of the reaction. Thus x and t satisfy the differential equation dx = −kxt, dt where k is a positive constant. At the start of the reaction, when t = 0, x = 100. (i) Solve this differential equation, obtaining a relation between x, k and t. [5] (ii) 20 seconds after the start of the reaction the amount of substance present is 90 grams. Find the time after the start of the reaction at which the amount of substance present is 50 grams. [3]

8 marks

Mark scheme: 8 (i) Separate variables correctly and attempt to integrate both sides M1 Obtain term ln x, or equivalent A1 Obtain term − 1 kt 2 , or equivalent A1 2 Use t = 0, x = 100 to evaluate a constant, or as limits M1 Obtain solution in any correct form, e.g. ln x = − 1 kt 2 + ln 100 A1 [5] 2 (ii) Use t = 20, x = 90 to obtain an equation in k M1* Substitute x = 50 and attempt to obtain an unsimplified numerical expression for t 2 , such as t 2 = 400(ln 100 – ln 50)/(ln 100 – ln 90) M1(dep*) Obtain answer t = 51.3 A1 [3] A Bx + C

This question in 9709/31 Oct/Nov 2005

7 The number of insects in a population t days after the start of observations is denoted by N. The variation in the number of insects is modelled by a differential equation of the form dN = kN cos(0.02t), dt where k is a constant and N is taken to be a continuous variable. It is given that N = 125 when t = 0. (i) Solve the differential equation, obtaining a relation between N, k and t. [5] (ii) Given also that N = 166 when t = 30, find the value of k. [2] (iii) Obtain an expression for N in terms of t, and find the least value of N predicted by this model. [3]

10 marks

Mark scheme: 7 (i) Separate variables correctly and attempt integration of both sides M1* Obtain term ln N, or equivalent A1 k Obtain term sin(0.02t ) , or equivalent A1 .002 Use t = 0, N = 125 to evaluate a constant, or as limits, in a solution containing terms of the form aln N and bsin(0.02t), or equivalent M1 Obtain any correct form of solution, e.g. ln N = 50ksin(0.02t) + ln 125 A1 [5] (ii) Substituting N = 166 and t = 30, evaluate k M1(dep*) Obtain k = 0.0100479…(accept k = 0.01) A1 [2] (iii) Rearrange and obtain N = 125exp(0.502sin(0.02t)), or equivalent B1 Set sin(0.02t) = −1 in the expression for N, or equivalent M1 Obtain least value 75.6 (accept answers in the interval [75, 76]) A1 [3] [For the B1, accept 0.5 following k = 0.01, and allow 4.8 or better for ln 125.]

This question in 9709/31 Oct/Nov 2007

100 8 (i) Express x2(10 −x) in partial fractions. [4] (ii) Given that x 1 when t 0, solve the differential equation = = dx 1 −x), dt = 100x2(10 obtaining an expression for t in terms of x. [6]

10 marks

Mark scheme: A B C 8 (i) State or imply the form + 2 + B1 x x 10 − x Use any relevant method to determine a constant M1 Obtain one of the values A = 1, B = 10, C = 1 A1 Obtain the remaining two values A1 4 Dx + E C [The form 2 + is acceptable and leads to D = 1, E = 10, C = 1] x 10 − x (ii) Separate variables and attempt integration of both sides M1 Obtain terms ln x, –10/x, –ln (10 – x), or equivalent A1√ + A1√ + A1 √ Evaluate a constant or use limits x = 1, t = 0 with a solution containing 3 of the terms kln x, l/x, mln (10 –x) and t, or equivalent M1  9 x  10 Obtain any correct expression for t, e.g. t = ln   − + 10 A1 6  10 − x  x adx [A separation of the form 2 = bdt is essential for the M1. The f.t. is on A, B, C] x (10 − x ) [If A or B (D or E) omitted from the form of fractions, give B0M1A0A0 in (i); M1A1√ A1√M1A0 in (ii)] GCE A/AS LEVEL – May/June 2009 9709 03

This question in 9709/31 May/June 2009

9 The temperature of a quantity of liquid at time t is θ. The liquid is cooling in an atmosphere whose temperature is constant and equal to A. The rate of decrease of θ is proportional to the temperature difference Thus θ and t satisfy the differential equation (θ −A). dθ dt = −k(θ −A), where k is a positive constant. (i) Find, in any form, the solution of this differential equation, given that θ 4A when t 0. [5] = = (ii) Given also that θ 3A when t 1, show that k ln 32. [1] = = = (iii) Find θ in terms of A when t 2, expressing your answer in its simplest form. [3] =

9 marks

Mark scheme: 9 (i) Separate variables correctly B1 Integrate and obtain term ln(θ – A), or equivalent B1 Integrate and obtain term –kt, or equivalent B1 Use θ = 4A, t = 0 to determine a constant, or as limits M1 Obtain correct answer in any form, e.g. ln(θ – A) = –kt + ln 3A, with no errors seen A1 [5] (ii) Substitute θ = 3A, t = 1 and justify the given statement B1 [1] (iii) Substitute t = 2 and solve for θ in terms of A M1 Remove logarithms M1 Obtain answer θ = 7 A, or equivalent, with no errors seen A1 [3] 3 [The M marks are only available if the solution to part (i) contains terms aln(θ – A) and bt.]

This question in 9709/32 Oct/Nov 2009

5 Given that y 0 when x 1, solve the differential equation = = dy xy y2 4, dx = + obtaining an expression for y2 in terms of x. [6]

6 marks

Mark scheme: 5 Separate variables correctly B1 Integrate and obtain term ln x B1 Integrate and obtain term 12 ln( y 2 + 4 ) B1 Evaluate a constant or use limits y = 0, x = 1 in a solution containing aln x and bln(y2 + 4) M1 Obtain correct solution in any form, e.g. 12 ln( y 2 + 4) = ln x + 12 ln 4 A1 Rearrange as y 2 = 4( x 2 − )1 , or equivalent A1 [6] 2 2

This question in 9709/31 May/June 2010

7 The variables x and t are related by the differential equation dx e2t cos2x, dt = where t When t 0, x 0. ≥0. = = (i) Solve the differential equation, obtaining an expression for x in terms of t. [6] (ii) State what happens to the value of x when t becomes very large. [1] (iii) Explain why x increases as t increases. [1]

8 marks

Mark scheme: 7 (i) Separate variables correctly and attempt integration of both sides B1 Obtain term tan x B1 Obtain term − 12 e−2 t B1 Evaluate a constant or use limits x = 0, t = 0 in a solution containing terms a tan x and be–2t M1 Obtain correct solution in any form, e.g. tan x = 12 − 12 e −2 t A1 Rearrange as x = tan −1 ( 12 − 12 e −2 t ) , or equivalent A1 [6] (ii) State that x approaches tan −1 ( 12 ) B1 [1] (iii) State that 1 − e − 2 t increases and so does the inverse tangent, or state that e−2 t cos 2 x is positive B1 [1] GCE AS/A LEVEL – May/June 2010 9709 32 2 2 2 

This question in 9709/32 May/June 2010

4 Given that x 1 when t 0, solve the differential equation = = dx dt = 1x −x4, obtaining an expression for x2 in terms of t. [7]

7 marks

Mark scheme: 4 Separate variables correctly B1 Obtain term k ln(4 – x2), or terms k1 ln(2 – x) + k2 ln(2 + x) B1 Obtain term –2 ln(4 – x2), or –2 ln(2 – x) –2 ln(2 + x), or equivalent B1 Obtain term t, or equivalent B1 Evaluate a constant or use limits x = 1, t = 0 in a solution containing terms a ln(4 – x2) and bt or terms c ln(2 – x), d ln(2 + x) and bt M1 Obtain correct solution in any form, e.g. –2 ln(4 – x2) = t – 2 ln3 A1 Rearrange and obtain x 2 = 4 − 3exp ( − 12 t ) , or equivalent (allow use of 2 ln 3 = 2.20) A1 [7] x 2 x

This question in 9709/33 May/June 2010

10 A certain substance is formed in a chemical reaction. The mass of substance formed t seconds after the start of the reaction is x grams. At any time the rate of formation of the substance is proportional dx to When t 0, x 0 and 1. dt (20 −x). = = = (i) Show that x and t satisfy the differential equation dx dt = 0.05(20 −x). [2] (ii) Find, in any form, the solution of this differential equation. [5] (iii) Find x when t 10, giving your answer correct to 1 decimal place. [2] = (iv) State what happens to the value of x as t becomes very large. [1]

10 marks

Mark scheme: dx 10 (i) State or imply = k (20 − x ) B1 dt Show that k = 0.05 B1 [2] (ii) Separate variables correctly and integrate both sides B1 Obtain term –ln(20 – x), or equivalent B1 Obtain term 201 t , or equivalent B1 Evaluate a constant or use limits t = 0, x = 0 in a solution containing terms a ln(20 – x) and bt M1* Obtain correct answer in any form, e.g. ln 20 – ln(20 – x) = 201 t A1 [5] (iii) Substitute t = 10 and calculate x M1(dep*) Obtain answer x = 7.9 A1 [2] (iv) State that x approaches 20 B1 [1]

This question in 9709/31 Oct/Nov 2010

10 A certain substance is formed in a chemical reaction. The mass of substance formed t seconds after the start of the reaction is x grams. At any time the rate of formation of the substance is proportional dx to When t 0, x 0 and 1. dt (20 −x). = = = (i) Show that x and t satisfy the differential equation dx dt = 0.05(20 −x). [2] (ii) Find, in any form, the solution of this differential equation. [5] (iii) Find x when t 10, giving your answer correct to 1 decimal place. [2] = (iv) State what happens to the value of x as t becomes very large. [1]

10 marks

Mark scheme: dx 10 (i) State or imply = k (20 − x ) B1 dt Show that k = 0.05 B1 [2] (ii) Separate variables correctly and integrate both sides B1 Obtain term –ln(20 – x), or equivalent B1 Obtain term 201 t , or equivalent B1 Evaluate a constant or use limits t = 0, x = 0 in a solution containing terms a ln(20 – x) and bt M1* Obtain correct answer in any form, e.g. ln 20 – ln(20 – x) = 201 t A1 [5] (iii) Substitute t = 10 and calculate x M1(dep*) Obtain answer x = 7.9 A1 [2] (iv) State that x approaches 20 B1 [1]

This question in 9709/32 Oct/Nov 2010

9 A biologist is investigating the spread of a weed in a particular region. At time t weeks after the start of the investigation, the area covered by the weed is A m2. The biologist claims that the rate of increase of A is proportional to √(2A −5). (i) Write down a differential equation representing the biologist’s claim. [1] (ii) At the start of the investigation, the area covered by the weed was 7 m2 and, 10 weeks later, the area covered was 27 m2 . Assuming that the biologist’s claim is correct, find the area covered 20 weeks after the start of the investigation. [9]

10 marks

Mark scheme: dA 9 (i) State = k 2 A − 5 B1 [1] dt (ii) Separate variables correctly and attempt integration of each side M1 1 2 = … or equivalent A1 Obtain (2A − 5 ) Obtain = kt or equivalent A1 Use t = 0 and A = 7 to find value of arbitrary constant M1 Obtain C = 3 or equivalent A1 Use t = 10 and A = 27 to find k M1 Obtain k = 0.4 or equivalent A1 Substitute t = 20 and values for C and k to find value of A M1 Obtain 63 cwo A1 [9]

This question in 9709/33 Oct/Nov 2010

10 The number of birds of a certain species in a forested region is recorded over several years. At time t years, the number of birds is N, where N is treated as a continuous variable. The variation in the number of birds is modelled by dN N(1800 −N) . dt = 3600 It is given that N 300 when t 0. = = (i) Find an expression for N in terms of t. [9] (ii) According to the model, how many birds will there be after a long time? [1]

10 marks

Mark scheme: 10 (i) Separate variables correctly and integrate of at least one side M1 1 A B Carry out an attempt to find A and B such that ≡ + , or equivalent M1 N (1800 − N ) N 1800 − N 2 2 Obtain + or equivalent A1 N 1800 − N Integrates to produce two terms involving natural logarithms M1 Obtain 2 ln N – 2 ln (1800 – N) = t or equivalent A1 Evaluate a constant, or use N = 300 and t = 0 in a solution involving a ln N, b ln(1800) and ct M1 Obtain 2 ln N – 2 ln (1800 – N) = t – 2 ln 5 or equivalent A1 Use laws of logarithms to remove logarithms M1 1 2 t 1800e or equivalent A1 [9] Obtain N = 1 2 t 5 + e (ii) State or imply that N approaches 1800 B1 [1]

This question in 9709/31 May/June 2011

6 A certain curve is such that its gradient at a point (x, y) is proportional to xy. At the point (1, 2) the gradient is 4. (i) By setting up and solving a differential equation, show that the equation of the curve is y = 2ex2−1. [7] (ii) State the gradient of the curve at the point (−1, 2) and sketch the curve. [2]

9 marks

Mark scheme: dy 6 (i) Show that the differential equation is = 2 xy B1 dx Separate variables correctly and attempt integration of both sides M1 Obtain term ln y, or equivalent A1 Obtain term x2, or equivalent A1 Evaluate a constant, or use limits x = 1, y = 2, in a solution containing terms aln y and bx2 M1 Obtain correct solution in any form A1 Obtain the given answer correctly A1 [7] (ii) State that the gradient at (–1, 2) is –4 B1 Show the sketch of curve with correct concavity, positive y-intercept and axis of symmetry x = 0 B1 [2] [SR: A solution with k≠ 2, or not evaluated, can earn B0M1A1A1M1A1A0 in part (i).] dy [SR: If given answer is assumed valid, give B1 if is shown correctly to be equal to dx 2xy, is stated to be proportional to xy, and shown to be equal to 4 at (1, 2).] GCE AS/A LEVEL – May/June 2011 9709 32

This question in 9709/32 May/June 2011

9 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 and 20 −x −x respectively. At any time the rate of formation of X is proportional to the product of the masses of Y dx and Z present at the time. When t 0, x 0 and 2. = = dt = (i) Show that x and t satisfy the differential equation dx dt = 0.01(10 −x)(20 −x). [1] (ii) Solve this differential equation and obtain an expression for x in terms of t. [9] (iii) State what happens to the value of x when t becomes large. [1]

11 marks

Mark scheme: dx 9 (i) State or imply = k (10 − x )(20 − x ) and show k = 0.01 B1 [1] dt (ii) Separate variables correctly and attempt integration of at least one side M1 1 A B Carry out an attempt to find A and B such that ≡ + , or (10 − x )(20 − x ) 10 − x 20 − x equivalent M1 1 1 Obtain A = and B = − , or equivalent A1 10 10 1 1 Integrate and obtain − ln (10 − x ) + ln (20 − x ) , or equivalent A1√ 10 10 Integrate and obtain term 0.01t, or equivalent A1 Evaluate a constant, or use limits t = 0, x = 0, in a solution containing terms of the form a ln (10 − x ) , b ln (20 − x ) and ct M1 1 1 1 Obtain answer in any form, e.g. − ln (10 − x ) + ln (20 − x ) = .001t + ln 2 A1√ 10 10 10 Use laws of logarithms to correctly remove logarithms M1 Rearrange and obtain x = 20(exp (1.0t ) − 1) / (2 exp (1.0t ) − )1 , or equivalent A1 [9] (iii) State that x approaches 10 B1 [1] GCE AS/A LEVEL – May/June 2011 9709 33

This question in 9709/33 May/June 2011

4 The variables x and θ are related by the differential equation dx sin 2θ cos 2θ, dθ = (x + 1) where 0 θ 12π. When θ 12π,1 x 0. Solve the differential equation, obtaining an expression for < < = = x in terms of θ, and simplifying your answer as far as possible. [7]

7 marks

Mark scheme: 4 Separate variables and attempt integration of at least one side M1 Obtain term ln(x + 1) A1 Obtain term k ln sin 2θ, where k = ±1, ±2, or ± 1 M1 2 Obtain correct term 1 ln sin 2θ A1 2 Evaluate a constant, or use limits θ = 1 π, x = 0 in a solution containing terms a ln(x + 1) and 12 b ln sin 2θ M1 Obtain solution in any form, e.g. ln(x + 1) = 1 ln sin 2θ − 1 ln 1 (f.t. on k = ±1, ±2, or ± 1 ) A1√ 2 2 2 2 Rearrange and obtain x = ( 2 sin 2θ ) − 1 , or simple equivalent A1 [7] GCE AS/A LEVEL – October/November 2011 9709 31

This question in 9709/31 Oct/Nov 2011

4 The variables x and θ are related by the differential equation dx sin 2θ cos 2θ, dθ = (x + 1) where 0 θ 12π. When θ 12π,1 x 0. Solve the differential equation, obtaining an expression for < < = = x in terms of θ, and simplifying your answer as far as possible. [7]

7 marks

Mark scheme: 4 Separate variables and attempt integration of at least one side M1 Obtain term ln(x + 1) A1 Obtain term k ln sin 2θ, where k = ±1, ±2, or ± 1 M1 2 Obtain correct term 1 ln sin 2θ A1 2 Evaluate a constant, or use limits θ = 1 π, x = 0 in a solution containing terms a ln(x + 1) and 12 b ln sin 2θ M1 Obtain solution in any form, e.g. ln(x + 1) = 1 ln sin 2θ − 1 ln 1 (f.t. on k = ±1, ±2, or ± 1 ) A1√ 2 2 2 2 Rearrange and obtain x = ( 2 sin 2θ ) − 1 , or simple equivalent A1 [7] GCE AS/A LEVEL – October/November 2011 9709 32

This question in 9709/32 Oct/Nov 2011

7 The variables x and y are related by the differential equation dy 6xe3x . dx = y2 It is given that y 2 when x 0. Solve the differential equation and hence find the value of y when x 0.5, giving your= answer correct= to 2 decimal places. [8] =

8 marks

Mark scheme: 7 Separate variables correctly and attempt integration on at least one side M1 1 3 Obtain y or equivalent on left-hand side A1 3 Use integration by parts on right-hand side (as far as axe 3 x + ∫ b e 3 x d x ) M1 Obtain or imply 2 xe 3 x + ∫ 2 e 3 x d x or equivalent A1 2 Obtain 2 xe 3 x − e 3 x A1 3 Substitute x = 0, y = 2 in an expression containing terms Ay3, Bxe3x, Ce3x, where ABC ≠ 0, and find the value of c M1 1 3 3 x 2 3 x 10 Obtain y = 2 xe − e + or equivalent A1 3 3 3 Substitute x = 0.5 to obtain y = 2.44 A1 [8] GCE AS/A LEVEL – May/June 2012 9709 31 2 

This question in 9709/31 May/June 2012

5 The variables x and y satisfy the differential equation dy e2x+y, dx = and y 0 when x 0. Solve the differential equation, obtaining an expression for y in terms of x. [6] = =

6 marks

Mark scheme: 5 Separate variables correctly and attempt integration of both sides B1 Obtain term −e − y , or equivalent B1 1 2 x Obtain term e , or equivalent B1 2 Evaluate a constant, or use limits x = 0, y = 0 in a solution containing terms a e − y and b e 2 x M1 − y 1 2 x 3 Obtain correct solution in any form, e.g. − e = e − A1 2 2 Rearrange and obtain y = ln( 2 /(3 − e 2x )) , or equivalent A1 [6] GCE AS/A LEVEL – May/June 2012 9709 32 2

This question in 9709/32 May/June 2012

5 In a certain chemical process a substance A reacts with another substance B. The masses in grams of A and B present at time t seconds after the start of the process are x and y respectively. It is given that dy 70. and x When t 0, y = 5e−3t. = = dt = −0.6xy (i) Form a differential equation in y and t. Solve this differential equation and obtain an expression for y in terms of t. [6] (ii) The percentage of the initial mass of B remaining at time t is denoted by p. Find the exact value approached by p as t becomes large. [2]

8 marks

Mark scheme: 5 (i) Substitute for x, separate variables correctly and attempt integration of both sides M1 Obtain term ln y, or equivalent A1 Obtain term e−,t3 or equivalent A1 Evaluate a constant, or use t = 0, y = 70 as limits in a solution containing terms a ln y and be−3 t M1 Obtain correct solution in any form, e.g. ln y − ln 70 = e −t3 − 1 A1 Rearrange and obtain y = 70exp(e −t3 − )1 , or equivalent A1 [6] (ii) Using answer to part (i), either express p in terms of t or use e −t3 → 0 to find the limiting value of y M1 100 Obtain answer from correct exact work A1 [2] e

This question in 9709/33 May/June 2012

6 The variables x and y are related by the differential equation dy x 1 dx = −y2. When x 2, y 0. Solve the differential equation, obtaining an expression for y in terms of x. [8] = =

8 marks

Mark scheme: 6 Separate variables correctly and attempt integration of one side B1 Obtain term ln x B1 1 B State or imply 1 – ≡ 1 – 1 and use a relevant method to find A or B M1 1 1 Obtain A = 2 , B = 2 1 1 Integrate and obtain – 2 ln (1 – y) + 2 ln (1 + y), or equivalent A1 1 + y 1 – y [If the integral is directly stated as k1 ln 1 – y or k2 ln 1 + y give M1, and then A2 for 1 1 k1 = 2 or k2 = – 2] Evaluate a constant, or use limits x = 2, y = 0 in a solution containing terms a ln x, b ln (1 – y) and c ln (1 + y), where abc 0 M1 [This M mark is not available if the integral of 1/(1 – y2) is initially taken to be of the form k ln (1 – y2)] 1 1 + y Obtain solution in any correct form, e.g. 2 ln 1 – y = ln x – ln 2 A1 x2 – 4 Rearrange and obtain y = x2 4, or equivalent, free of logarithms A1 [8] 1 1 dy

This question in 9709/31 Oct/Nov 2012

6 The variables x and y are related by the differential equation dy x 1 dx = −y2. When x 2, y 0. Solve the differential equation, obtaining an expression for y in terms of x. [8] = =

8 marks

Mark scheme: 6 Separate variables correctly and attempt integration of one side B1 Obtain term ln x B1 1 B State or imply 1 – ≡ 1 – 1 and use a relevant method to find A or B M1 1 1 Obtain A = 2 , B = 2 1 1 Integrate and obtain – 2 ln (1 – y) + 2 ln (1 + y), or equivalent A1 1 + y 1 – y [If the integral is directly stated as k1 ln 1 – y or k2 ln 1 + y give M1, and then A2 for 1 1 k1 = 2 or k2 = – 2] Evaluate a constant, or use limits x = 2, y = 0 in a solution containing terms a ln x, b ln (1 – y) and c ln (1 + y), where abc 0 M1 [This M mark is not available if the integral of 1/(1 – y2) is initially taken to be of the form k ln (1 – y2)] 1 1 + y Obtain solution in any correct form, e.g. 2 ln 1 – y = ln x – ln 2 A1 x2 – 4 Rearrange and obtain y = x2 4, or equivalent, free of logarithms A1 [8] 1 1 dy

This question in 9709/32 Oct/Nov 2012

4 The variables x and y are related by the differential equation 6xy. (x2 + 4)dydx = It is given that y 32 when x 0. Find an expression for y in terms of x. [6] = =

6 marks

Mark scheme: 4 Separate variables correctly and integrate one side M1 Obtain ln y = ... or equivalent A1 Obtain = 31n ( x 2 + 4) or equivalent A1 Evaluate a constant or use x = 0, y = 32 as limits in a solution M1 containing terms a ln y and b ln ( x 2 + 4 ) Obtain ln y = 31n ( x 2 + 4) + ln 32 − 31n 4 or equivalent A1 1 2 Obtain y = (x + 4 ) or equivalent A1 [6] 2

This question in 9709/33 Oct/Nov 2012

8 The variables x and t satisfy the differential equation tdx k , −x3 dt = 2x2 for t 0, where k is a constant. When t 1, x 1 and when t 4, x 2. > = = = = (i) Solve the differential equation, finding the value of k and obtaining an expression for x in terms of t. [9] (ii) State what happens to the value of x as t becomes large. [1]

10 marks

Mark scheme: 8 (i) Separate variables correctly and integrate at least one side M1 Obtain term ln t, or equivalent B1 Obtain term of the form a ln(k – x3) M1 2 Obtain term − ln(k − 3x ) , or equivalent A1 3 EITHER: Evaluate a constant or use limits t = 1, x =1 in a solution containing a ln t and b ln(k – x3) M1* 2 3 2 Obtain correct answer in any form e.g. ln t = − ln(k − x ) + ln(k − )1 A1 3 3 Use limits t = 4, x =2, and solve for k M1(dep*) Obtain k = 9 A1 OR: Using limits t = 1, x = 1 and t = 4, x = 2 in a solution containing a ln t and b ln (k – x3) obtain an equation in k M1* 2 2 Obtain a correct equation in any form, e.g. ln 4 = − ln( k − 8) + ln( k − )1 A1 3 3 Solve for k M1(dep*) Obtain k = 9 A1 3 1 − Substitute k = 9 and obtain x = (9 − 8t 2 ) 3 A1 [9] 1 (ii) State that x approaches 9 3 , or equivalent B1 [1]

This question in 9709/33 May/June 2013

10 h 60° C A tank containing water is in the form of a cone with vertex C. The axis is vertical and the semi- vertical angle is as shown in the diagram. At time t 0, the tank is full and the depth of water is H. At this instant,60Å, a tap at C is opened and water begins= to flow out. The volume of water in the tank decreases at a rate proportional to where h is the depth of water at time t. The tank becomes empty when t 60. ïh, = (i) Show that h and t satisfy a differential equation of the form dh −32, dt = −Ah where A is a positive constant. [4] (ii) Solve the differential equation given in part (i) and obtain an expression for t in terms of h and H. [6] (iii) Find the time at which the depth reaches 2H.1 [1] [The volume V of a cone of vertical height h and base radius r is given by V 1 = 30r2h.]

11 marks

Mark scheme: 10 (i) State or imply V = πh 3 B1 d V State or imply = − k h B1 d t d V d V d h Use = . , or equivalent M1 d t d h d t Obtain the given equation A1 [4] d V [The M1 is only available if is in terms of h and has been obtained by a d h correct method.] dV [Allow B1 for = k h but withhold the final A1 until the polarity of the constant dt k has been justified.] 3π GCE A LEVEL – October/November 2013 9709 31 (ii) Separate variables and integrate at least one side M1 5 2 2 Obtain terms h and –At, or equivalent A1 5 5 Use t = ,0 h = H in a solution containing terms of the form ah 2 and bt + c M1 5 Use t = 60, h = 0 in a solution containing terms of the form ah 2 and bt + c M1 5 5 5 2 2 1 2 2 2 Obtain a correct solution in any form, e.g. h = H t + H A1 5 150 5  5    h  2  (ii) Obtain final answer t = 60 1 −   , or equivalent A1 [6]   H     1 (iii) Substitute h = H and obtain answer t = 49.4 B1 [1] 2

This question in 9709/31 Oct/Nov 2013

10 h 60° C A tank containing water is in the form of a cone with vertex C. The axis is vertical and the semi- vertical angle is 60 , as shown in the diagram. At time t = 0, the tank is full and the depth of water is H. At this instant, a tap at C is opened and water begins to flow out. The volume of water in the tank decreases at a rate proportional to h, where h is the depth of water at time t. The tank becomes empty when t = 60. (i) Show that h and t satisfy a differential equation of the form dh −3 = −Ah 2, dt where A is a positive constant. [4] (ii) Solve the differential equation given in part (i) and obtain an expression for t in terms of h and H. [6] [1] (iii) Find the time at which the depth reaches 12H. [The volume V of a cone of vertical height h and base radius r is given by V = 1 r2h.] 3

11 marks

Mark scheme: 10 (i) State or imply V = πh 3 B1 d V State or imply = − k h B1 d t d V d V d h Use = . , or equivalent M1 d t d h d t Obtain the given equation A1 [4] d V [The M1 is only available if is in terms of h and has been obtained by a d h correct method.] d V [Allow B1 for = k h but withhold the final A1 until the polarity of the constant d t k has been justified.] 3π GCE A LEVEL – October/November 2013 9709 32 (ii) Separate variables and integrate at least one side M1 5 2 2 Obtain terms h and –At, or equivalent A1 5 5 Use t = ,0 h = H in a solution containing terms of the form ah 2 and bt + c M1 5 Use t = 60, h = 0 in a solution containing terms of the form ah 2 and bt + c M1 5 5 5 2 2 1 2 2 2 Obtain a correct solution in any form, e.g. h = H t + H A1 5 150 5  5    h  2  (ii) Obtain final answer t = 60 1 −   , or equivalent A1 [6]   H     1 (iii) Substitute h = H and obtain answer t = 49.4 B1 [1] 2

This question in 9709/32 Oct/Nov 2013

4 The variables x and y are related by the differential equation dy 6ye3x . dx = 2 e3x + Given that y 36 when x 0, find an expression for y in terms of x. [6] = =

6 marks

Mark scheme: 4 Separate variables correctly and recognisable attempt at integration of at least one side M1 Obtain ln y, or equivalent B1 3 x Obtain k l n (2 + e ) B1 x k 3 x Use y(0) = 36 to find constant in y = A(2 + 3e ) or l ny = k l n (2 + e ) + c or equivalent M1*  k  3 x *M1 Obtain equation correctly without logarithms from l ny = l n  A( 2 + e )    3 x 2 Obtain y = 4 (2 + e ) A1 [6] GCE A LEVEL – May/June 2014 9709 31 2

This question in 9709/31 May/June 2014

7 In a certain country the government charges tax on each litre of petrol sold to motorists. The revenue per year is R million dollars when the rate of tax is x dollars per litre. The variation of R with x is modelled by the differential equation dR @1 A R , dx = x −0.57 where R and x are taken to be continuous variables. When x 0.5, R 16.8. = = (i) Solve the differential equation and obtain an expression for R in terms of x. [6] (ii) This model predicts that R cannot exceed a certain amount. Find this maximum value of R. [3]

9 marks

Mark scheme: 7 (i) Separate variables correctly and attempt to integrate at least one side B1 Obtain term lnR B1 Obtain ln x − 0.57 x B1 Evaluate a constant or use limits x = 0.5, R = 16.8, in a solution containing terms of the form alnR and blnx M1 Obtain correct solution in any form A1 (3.80 − 0.57 x ) 0.57 x Obtain a correct expression for R, e.g. R = xe , R = 44.7 xe− or (0.285 − 0.57 x ) R = 33.6 xe A1 [6] d R (ii) Equate to zero and solve for x M1 d x State or imply x = 0.57 −1 , or equivalent, e.g. 1.75 A1 Obtain R = 28.8 (allow 28.9) A1 [3]

This question in 9709/31 Oct/Nov 2014

7 In a certain country the government charges tax on each litre of petrol sold to motorists. The revenue per year is R million dollars when the rate of tax is x dollars per litre. The variation of R with x is modelled by the differential equation dR @1 A R , dx = x −0.57 where R and x are taken to be continuous variables. When x 0.5, R 16.8. = = (i) Solve the differential equation and obtain an expression for R in terms of x. [6] (ii) This model predicts that R cannot exceed a certain amount. Find this maximum value of R. [3]

9 marks

Mark scheme: 7 (i) Separate variables correctly and attempt to integrate at least one side B1 Obtain term lnR B1 Obtain ln x − 0.57 x B1 Evaluate a constant or use limits x = 0.5, R = 16.8, in a solution containing terms of the form alnR and blnx M1 Obtain correct solution in any form A1 (3.80 − 0.57 x ) 0.57 x Obtain a correct expression for R, e.g. R = xe , R = 44.7 xe− or (0.285 − 0.57 x ) R = 33.6 xe A1 [6] d R (ii) Equate to zero and solve for x M1 d x State or imply x = 0.57 −1 , or equivalent, e.g. 1.75 A1 Obtain R = 28.8 (allow 28.9) A1 [3]

This question in 9709/32 Oct/Nov 2014

9 The number of organisms in a population at time t is denoted by x. Treating x as a continuous variable, the differential equation satisfied by x and t is dx xe−t , dt = k + e−t where k is a positive constant. (i) Given that x 10 when t 0, solve the differential equation, obtaining a relation between x, k and t. = = [6] (ii) Given also that x 20 when t 1, show that k 1 . [2] e = = = −2 (iii) Show that the number of organisms never reaches 48, however large t becomes. [2]

10 marks

Mark scheme: 9 (i) Separate variables correctly and attempt integration of one side B1 Obtain term ln x B1 Obtain term of the form a ln( k + e − t ) M1 Obtain term − ln( k + e − t ) A1 Evaluate a constant or use limits x = 10, t = 0 in a solution containing terms a ln( k + e − t ) and bln x M1* Obtain correct solution in any form, e.g. ln x − ln 10 = − ln( k + e − t ) + ln( k + )1 A1 [6] (ii) Substitute x = 20, t = 1 and solve for k M1(dep*) Obtain the given answer A1 [2] (iii) Using e −t → 0 and the given value of k, find the limiting value of x M1 Justify the given answer A1 [2]

This question in 9709/32 May/June 2015

7 The number of micro-organisms in a population at time t is denoted by M. At any time the variation in M is assumed to satisfy the differential equation dM k cos 0.02t , dt = ïM where k is a constant and M is taken to be a continuous variable. It is given that when t 0, M 100. = = (i) Solve the differential equation, obtaining a relation between M, k and t. [5] (ii) Given also that M 196 when t 50, find the value of k. [2] = = (iii) Obtain an expression for M in terms of t and find the least possible number of micro-organisms. [2]

9 marks

Mark scheme: 7 (i) Separate variables correctly and integrate one side B1 Obtain term 2 M , or equivalent B1 Obtain term 50 k sin( .0 02t ) , or equivalent B1 Evaluate a constant of integration, or use limits M = 100, t =0 in a solution with terms of the form a M and b sin( .0 02t ) M1* Obtain correct solution in any form, e.g. 2 M = 50 k sin( .002t ) + 20 A1 5 (ii) Use values M =196, t =50 and calculate k M1(dep*) Obtain answer k = 0.190 A1 2 (iii) State an expression for M in terms of t, e.g. M = ( .4 75 sin( .0 02t ) + 10 2) M1(dep*) State that the least possible number of micro-organisms is 28 or 27.5 or 27.6 (27.5625) A1 2 i

This question in 9709/33 May/June 2015

7 The variables x and y satisfy the differential equation dy xex+y, dx = and it is given that y 0 when x 0. = = (i) Solve the differential equation and obtain an expression for y in terms of x. [7] (ii) Explain briefly why x can only take values less than 1. [1] ` a ` a ` a

8 marks

Mark scheme: 7 (i) Separate variables and attempt integration of one side M1 Obtain term − e − y A1 Integrate x ex by parts reaching x e x ±∫ e x d x M1 Obtain integral xe x − e x A1 Evaluate a constant, or use limits x = 0, y = 0 M1 Obtain correct solution in any form A1 Obtain final answer y = − ln(e x (1 − x )) , or equivalent A1 [7] (ii) Justify the given statement B1 [1]

This question in 9709/32 Feb/March 2016

6 The variables x and satisfy the differential equation 1 dx 3 cos x sin + 21 = 21, d1 and it is given that x 3 when 1 = 1 = 40. (i) Solve the differential equation and obtain an expression for x in terms of [7] 1. (ii) State the least value taken by x. [1]

8 marks

Mark scheme: 6 (i) Separate variables correctly and attempt integration of at least one side B1 Obtain term ln x B1 Obtain term of the form k ln(3 + cos2θ ) , or equivalent M1 Obtain term − 12 ln(3 + cos2θ ) , or equivalent A1 Use x = 3, θ = 14 π to evaluate a constant or as limits in a solution with terms a ln x and b ln(3 + cos2θ ) ,where ab ≠ 0 M1 State correct solution in any form, e.g. ln x = − 12 ln(3 + cos2θ ) + 32 ln3 A1  27  Rearrange in a correct form, e.g. x =   A1 [7]  3 + cos2θ  (ii) State answer x = 3 3 / 2 , or exact equivalent (accept decimal answer in [2.59, 2.60]) B1 [1] B C

This question in 9709/32 May/June 2016

5 The variables x and y satisfy the differential equation dy e−2y tan2x, dx = for 0 1 and it is given that y 0 when x 0. Solve the differential equation and calculate the ≤x < 20, 1 = = value of y when x [8] = 40.

8 marks

Mark scheme: 5 Separate variables and make reasonable attempt at integration of either integral M1 Obtain term 12 e 2 y B1 Use Pythagoras M1 Obtain terms tan x − x A1 Evaluate a constant or use x = 0, y = 0 as limits in a solution containing terms a e ± 2 y and b tan x ,( ab ≠ 0) M1 Obtain correct solution in any form, e.g. 12 e 2 y = tan x − x + 12 A1 Set x = 14 π and use correct method to solve an equation of the form e ± 2 y = a or e ± y = a , where a > 0 M1 Obtain answer y = 0.179 A1 [8]

This question in 9709/33 May/June 2016

10 A large field of area 4 km2 is becoming infected with a soil disease. At time t years the area infected is dx x km2 and the rate of growth of the infected area is given by the differential equation kx 4 , dt = −x where k is a positive constant. It is given that when t 0, x 0.4 and that when t 2, x 2. = = = = (i) Solve the differential equation and show that k 1 ln 3. [9] = 4 (ii) Find the value of t when 90% of the area of the field is infected. [2]

11 marks

Mark scheme: 10 (i) Separate variables correctly and integrate at least one side M1 Integrate and obtain term kt, or equivalent A1 1 A B Carry out a relevant method to obtain A and B such that ≡ + , or equivalent M1* x (4 − x ) x 4 − x 1 Obtain A = B = , or equivalent A1 4 1 1 Integrate and obtain terms ln x − ln(4 − x ) , or equivalent A1 4 4 EITHER: Use a pair of limits in an expression containing plnx, qln(4 – x) and rt and evaluate a constant DM1 Obtain correct answer in any form, e.g.ln x − ln(4 − x ) = 4 kt − ln9 ,  x = or ln   4 kt − 8k A1  4 − x  Use a second pair of limits and determine k DM1 Obtain the given exact answer correctly A1 OR: Use both pairs of limits in a definite integral M1* Obtain the given exact answer correctly A1 Substitute k and either pair of limits in an expression containing plnx, qln(4 – x) and rt and evaluate a constant DM1 x Obtain ln = t ln 3 − ln 9 or equivalent A1 [9] 4 − x (ii) Substitute x = 3.6 and solve for t M1 Obtain answer t = 4 A1 [2]

This question in 9709/31 Oct/Nov 2016

10 A large field of area 4 km2 is becoming infected with a soil disease. At time t years the area infected is dx x km2 and the rate of growth of the infected area is given by the differential equation kx 4 , dt = −x where k is a positive constant. It is given that when t 0, x 0.4 and that when t 2, x 2. = = = = (i) Solve the differential equation and show that k 1 ln 3. [9] = 4 (ii) Find the value of t when 90% of the area of the field is infected. [2]

11 marks

Mark scheme: 10 (i) Separate variables correctly and integrate at least one side M1 Integrate and obtain term kt, or equivalent A1 1 A B Carry out a relevant method to obtain A and B such that ≡ + , or equivalent M1* x (4 − x ) x 4 − x 1 Obtain A = B = , or equivalent A1 4 1 1 Integrate and obtain terms ln x − ln(4 − x ) , or equivalent A1 4 4 EITHER: Use a pair of limits in an expression containing plnx, qln(4 – x) and rt and evaluate a constant DM1 Obtain correct answer in any form, e.g.ln x − ln(4 − x ) = 4 kt − ln9 ,  x = or ln   4 kt − 8k A1  4 − x  Use a second pair of limits and determine k DM1 Obtain the given exact answer correctly A1 OR: Use both pairs of limits in a definite integral M1* Obtain the given exact answer correctly A1 Substitute k and either pair of limits in an expression containing plnx, qln(4 – x) and rt and evaluate a constant DM1 x Obtain ln = t ln 3 − ln 9 or equivalent A1 [9] 4 − x (ii) Substitute x = 3.6 and solve for t M1 Obtain answer t = 4 A1 [2]

This question in 9709/32 Oct/Nov 2016

5 In a certain chemical process a substance A reacts with and reduces a substance B. The masses of A dy and B at time t after the start of the process are x and y respectively. It is given that and dt = −0.2xy 10 x . At the beginning of the process y 100. 2 = 1 t = + (i) Form a differential equation in y and t, and solve this differential equation. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (ii) Find the exact value approached by the mass of B as t becomes large. State what happens to the mass of A as t becomes large. [2] … … … … … … … … … … …

8 marks

Mark scheme: 5(i) d y 2 y B1 State = − , or equivalent d t (1 + t ) 2 Separate variables correctly and attempt integration of one side M1 Obtain term ln y , or equivalent A1 2 A1 Obtain term , or equivalent (1 + )t Use y = 100 and t = 0 to evaluate a constant, or as limits in an expression containing terms of M1 b the form a ln y and 1 + t 2 A1 Obtain correct solution in any form, e.g. ln y = − 2 + ln100 1 + t Total: 6 5(ii) 100 B1 State that the mass of B approaches , or exact equivalent e 2 State or imply that the mass of A tends to zero B1 Total: 2

This question in 9709/32 May/June 2017

6 The variables x and y satisfy the differential equation dy 4 cos2y tan x, dx = for 0 1 and x 0 when y 1 Solve this differential equation and find the value of x when ≤x < 20, = = 40. y 1 [8] = 30. … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6 Separate variables correctly and attempt integration of one side B1 Obtain term tan y , or equivalent B1 Obtain term of the form k lncos x , or equivalent M1 Obtain term − 4lncos x , or equivalent A1 Use x = 0 and y = 14π in solution containing a tan y and b lncos x to evaluate a M1 constant, or as limits Obtain correct solution in any form, e.g. tan y = 4lnsec x + 1 A1 Substitute y = 13πin solution containing terms a tan y and b lncos x , and use correct M1 method to find x Obtain answer x = 0.587 A1 8

This question in 9709/31 Oct/Nov 2017

6 In a certain chemical reaction the amount, x grams, of a substance is decreasing. The differential equation relating x and t, the time in seconds since the reaction started, is dx t, dt = −kx where k is a positive constant. It is given that x 100 at the start of the reaction. = (i) Solve the differential equation, obtaining a relation between x, t and k. [5] … … … … … … … … … … … … … … … … … … … … … … (ii) Given that t 25 when x 80, find the value of t when x 40. [3] = = = … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(i) Separate variables correctly and integrate at least one side B1 Obtain term ln x B1 2 B1 Obtain term − kt t , or equivalent 3 Evaluate a constant, or use limits x = 100 and t = 0, in a solution containing M1 terms a ln x and bt t 2 A1 Obtain correct solution in any form, e.g. 1n x = − kt t + 1n 100 3 5 6(ii) Substitute x = 80 and t = 25 to form equation in k M1 Substitute x = 40 and eliminate k M1 Obtain answer t = 64.1 A1 3

This question in 9709/31 May/June 2018

1 6 (i) Express in partial fractions. [2] 4 −y2 … … … … … … … … … … (ii) The variables x and y satisfy the differential equation xdy 4 dx = −y2, and y 1 when x 1. Solve the differential equation, obtaining an expression for y in terms of x. = = [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(i) Carry out relevant method to find A and B such that M1 1 A B ≡ + 4 − y 2 2 + y 2 − y 1 A1 Obtain A = B = 4 Total: 2 6(ii) Separate variables correctly and integrate at least one side to obtain one of the terms M1 a ln x, b ln (2 + y) or c ln (2 – y) Obtain term ln x B1 1 1 A1FT Integrate and obtain terms ln ( 2 + y ) − ln ( 2 − y ) 4 4 Use x = 1 and y = 1 to evaluate a constant, or as limits, in a solution containing at M1 least two terms of the form a ln x, b ln (2 + y) and c ln (2 – y) Obtain a correct solution in any form, e.g. A1 1 1 1 ln x = ln ( 2 + y ) − ln ( 2 − y ) − ln3 4 4 4 4 A1 2 ( 3 x − 1) Rearrange as , or equivalent ( 3 x 4 + 1) Total: 6

This question in 9709/33 May/June 2018

6 The variables x and y satisfy the differential equation dy = ky3e−x, dx where k is a constant. It is given that y 1 when x 0, and that y e when x 1. Solve the = = = = differential equation, obtaining an expression for y in terms of x. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 6 Separate variables correctly and attempt integration of at least one side B1 Obtain term 2 1 2 − y , or equivalent B1 Obtain term – k e−x B1 Use a pair of limits, e.g. x = 0, y = 1 to obtain an equation in k and an arbitrary constant c M1 Use a second pair of limits, e.g. x = 1, y = e , to obtain a second equation and solve for k or for c M1 Obtain k = 1 2 and c = 0 A1 Obtain final answer 1 2 e = x y , or equivalent A1 7

This question in 9709/32 Feb/March 2019

1 5 (i) Differentiate with respect to [2] 1. sin21 … … … … … … … … … … … … (ii) The variables x and satisfy the differential equation 1 dx x tan 0, 1 + cosec21 = d1 for 0 1 and x 0. It is given that x 4 when 1 Solve the differential equation, < 1 < 20 > = 1 = 60. obtaining an expression for x in terms of [6] 1. … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(i) Use chain rule M1 3 2 cos sin cosec cot θ θ θ θ − = − k k Allow M1 for 1 2cos sin θ θ − − Obtain correct answer in any form A1 e.g. 2 2cosec cot θ θ − , 3 2cos sin θ θ − Accept 4 2sin cos sin θ θ θ − 2 5(ii) Separate variables correctly and integrate at least one side B1 2 d cosec cot d θ θ θ = − ∫ ∫ x x Obtain term 2 1 2 x B1 Obtain term of the form 2 sin k θ M1* or equivalent Obtain term 2 1 2sin θ A1 or equivalent Use x = 4, 1 6 θ π = to evaluate a constant, or as limits, in a solution with terms 2 ax and 2 sin b θ , where ab ≠ 0 DM1 Dependent on the preceding M1 Obtain solution ( ) 2 cosec 12 x θ = + A1 or equivalent 6

This question in 9709/31 May/June 2019

dy 0.7 The variables x and y satisfy the differential equation It is given that y 0 when x = = dx = xex+y. (i) Solve the differential equation, obtaining y in terms of x. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (ii) Explain why x can only take values that are less than 1. [1] … … … … … … … … …

8 marks

Mark scheme: 7(i) Separate variables correctly and attempt integration of at least one side B1 e d e d − = ∫ ∫ y x y x x Obtain term e− − y B1 B0B1 is possible Commence integration by parts and reach e e d ± ∫ x x x x M1 B0B0M1A1 is possible Obtain e e − x x x A1 or equivalent B1B1M1A1 is available if there is no constant of integration Use x = 0, y = 0 to evaluate a constant, or as limits in a definite integral, in a solution with terms e−y a , ex bx and ex c , where abc ≠ 0 M1 Must see this step Obtain correct solution in any form A1 e.g. e e e −= − y x x x Rearrange as ( ) ln 1 = − − − y x x A1 or equivalent e.g. ( ) 1 ln e 1 = − x y x ISW 7 7(ii) Justify the given statement B1 e.g. require 1 0 − > x for the ln term to exist, hence 1 < x Must be considering the range of values of x, and must be relevant to their y involving ( ) ln 1−x 1

This question in 9709/32 May/June 2019

4 The number of insects in a population t weeks after the start of observations is denoted by N. The population is decreasing at a rate proportional to Ne−0.02t. The variables N and t are treated as dN continuous, and it is given that when t 0, N 1000 and = = dt = −10. (i) Show that N and t satisfy the differential equation dN [1] dt = −0.01e−0.02tN. … … … … … … … … (ii) Solve the differential equation and find the value of t when N 800. [6] = … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (iii) State what happens to the value of N as t becomes large. [1] … … … … … …

8 marks

Mark scheme: 4(i) State d d N t = 0.02 e t k N − and show k = – 0.01 B1 OE ( ) 10 1 1000 k − = × × 1 4(ii) Separate variables correctly and integrate at least one side B1 0.02 1 d 0.01e d t N t N − = − ∫ ∫ Obtain term ln N B1 OE Obtain term 0.02 0.5e t − B1 OE Use N = 1000, t = 0 to evaluate a constant, or as limits, in a solution with terms ln a N and 0.02 e t b − , where ab ≠ 0 M1 Obtain correct solution in any form e.g. ( ) 0.02 ln ln1000 0.5 e 1 t N − − = − A1 1 ln1000 6.41 2 − = Substitute N = 800 and obtain t = 29.6 A1 6 4(iii) State that N approaches 1000 e B1 Accept 606 or 607 or 606.5 1

This question in 9709/31 Oct/Nov 2019

20 40 109 The variables x and t satisfy the differential equation 5dx . It is given that x −x −x = dt = when t 0. = (i) Using partial fractions, solve the differential equation, obtaining an expression for x in terms of t. [9] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (ii) State what happens to the value of x when t becomes large. [1] … … … … …

10 marks

Mark scheme: 9(i) Separate variables correctly and integrate one side B1 Obtain term 0.2t, or equivalent B1 Carry out a relevant method to obtain A and B such that ( )( ) 1 20 40 − − x x ≡ 20 − A x + 40 − B x *M1 OE Obtain A = 1 20 and B = 1 20 − A1 Integrate and obtain terms ( ) ( ) 1 1 ln 20 ln 40 20 20 − − + − x x OE A1FT +A1FT The FT is on A and B Use x = 10, t = 0 to evaluate a constant, or as limits DM1 Obtain correct answer in any form A1 Obtain final answer = x 4 4 60e 40 3e 1 − − t t A1 OE 9 9(ii) State that x approaches 20 B1 1

This question in 9709/33 Oct/Nov 2019

6 The variables x and y satisfy the differential equation dy 1 4y2 + . dx = ex It is given that y 0 when x 1. = = (a) Solve the differential equation, obtaining an expression for y in terms of x. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) State what happens to the value of y as x tends to infinity. [1] … … … … … … … …

8 marks

Mark scheme: 6(a) Separate variables correctly and attempt integration of at least one side B1 Obtain term of the form 1 tan (2 ) − a y M1 Obtain term ( ) 1 1 tan 2 2 − y A1 Obtain term e− − x B1 Use x = 1, y = 0 to evaluate a constant or as limits in a solution containing terms of the form ( ) 1 tan− a by and e±x c M1 Obtain correct answer in any form A1 Obtain final answer ( ) 1 1 tan 2e 2e 2 − − = − x y , or equivalent A1 7 Question Answer Marks Guidance 6(b) State that y approaches ( ) 1 1 tan 2e 2 − , or equivalent B1FT The FT is on correct work on a solution containing e−x . 1

This question in 9709/32 Feb/March 2020

y 8 A certain curve is such that its gradient at a point x, y is proportional to The curve passes x x. through the points with coordinates 1, 1 and 4, e . (a) By setting up and solving a differential equation, find the equation of the curve, expressing y in terms of x. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Describe what happens to y as x tends to infinity. [1] … … … … …

9 marks

Mark scheme: 8(a) State d d = y y k x x x , or equivalent B1 Separate variables correctly and attempt integration of at least one side M1 Obtain term ln y, or equivalent A1 Obtain term 1 2 −k x , or equivalent A1 Use given coordinates to find k or a constant of integration c in a solution containing terms of the form a ln y and b x , where ab ≠ 0 M1 Obtain k = 1 and c = 2 A1 + A1 Obtain final answer y = exp 2 2   − +     x , or equivalent A1 8 Question Answer Marks 8(b) State that y approaches eଶ (FT their c in part (a) of the correct form) B1FT 1 ( )   ( )  

This question in 9709/31 May/June 2020

7 The variables x and t satisfy the differential equation dx e3t cos22x, dt = for t It is given that x 0 when t 0. ≥0. = = (a) Solve the differential equation and obtain an expression for x in terms of t. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) State what happens to the value of x when t tends to infinity. [1] … … … … … …

8 marks

Mark scheme: 7(a) Correct separation of variables B1 2 3 sec 2 d e d t x x t − =   Needs correct structure Obtain term 3 1e 3 t − − B1 Obtain term of the form tan 2 k x M1 From correct working Obtain term 1 tan2 2 x A1 Use x = 0, t = 0 to evaluate a constant, or as limits in a solution containing terms of the form tan 2 a x and 3 e t b −, where ab ≠ 0 M1 Obtain correct solution in any form A1 e.g. 3 1 1 1 tan2 e 2 3 3 t x − = − + Obtain final answer ( ) 1 3 1 2 tan 1 e 2 3 t x − −   = −     A1 7 7(b) State that x approaches 1 1 2 tan 2 3 −      B1 FT Correct value. Accept 0.294 x → The FT is dependent on letting 3 e 0 t − → in a solution containing 3 e t −. 1

This question in 9709/32 Oct/Nov 2020

4 The variables x and y satisfy the differential equation dy 1 y sin x. −cosx dx = It is given that y 4 when x = = π. (a) Solve the differential equation, obtaining an expression for y in terms of x. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Sketch the graph of y against x for 0 x [1] < < 2π.

7 marks

Mark scheme: 4(a) Separate variables correctly and attempt integration of at least one side M1 Obtain term ln y A1 Obtain term of the form ln(1 cos ) ± − x M1 Obtain term ( ) ln 1 cos − x A1 Use π = x , y = 4 to evaluate a constant, or as limits, in a solution containing terms of the form ln a y and ln(1 cos ) − b x M1 Obtain final answer 2(1 cos ) = − y x A1 OE 6 Question Answer Marks Guidance 4(b) Show a correct graph for 0 2π x < < with the maximum at x = π B1 FT The FT is for graphs of the form (1 cos ) = − y a x , where a is positive. 1

This question in 9709/32 Feb/March 2021

7 y P x O M N For the curve shown in the diagram, the normal to the curve at the point P with coordinates x, y meets the x-axis at N. The point M is the foot of the perpendicular from P to the x-axis. The curve is such that for all values of x in the interval 0 1 the area of triangle PMN is equal to tan x. ≤x < 2π, MN dy (a) (i) Show that [1] y dx. = … … … … … 1 dy (ii) Hence show that x and y satisfy the differential equation 2y2 tan x. [2] dx = … … … … … … … (b) Given that y 1 when x 0, solve this differential equation to find the equation of the curve, expressing y in= terms of x.= [6] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(a)(i) Justify the given statement MN y = d d y x B1 1 7(a)(ii) Express the area of PMN in terms of y and d d y x and equate to tan x M1 Obtain the given equation correctly A1 2 7(b) Separate variables and integrate at least one side M1 Obtain term 3 1 6 y A1 Obtain term of the form lncos ± x M1 Evaluate a constant or use x = 0 and y = 1 in a solution containing terms 3 ay and lncos ± x , or equivalent M1 Obtain correct answer in any form, e.g. 3 1 1 lncos 6 6 = − + y x A1 Obtain final answer 3 (1 6ln cos ) = − y x A1 OE 6

This question in 9709/33 May/June 2021

7 (a) Given that y ln ln x , show that = dy 1 [1] dx x ln x. = … … … … … The variables x and t satisfy the differential equation dx x ln x t 0. dt + = It is given that x e when t 2. = = (b) Solve the differential equation obtaining an expression for x in terms of t, simplifying your answer. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (c) Hence state what happens to the value of x as t tends to infinity. [1] … … … …

9 marks

Mark scheme: 7(a) Show sufficient working to justify the given answer B1 1 7(b) Correct separation of variables B1 e.g. 1 1 ln d d t x x t x − =   Obtain term ( ) ln ln x B1 Obtain term ln − t B1 Evaluate a constant or use x = e and t = 2 as limits in an expression involving ( ) ln ln x M1 Obtain correct solution in any form, e.g. ln(ln ) ln ln 2 = − + x t A1 Use log laws to enable removal of logarithms M1 Obtain answer 2 e = t x , or simplified equivalent A1 7 7(c) State that x tends to 1 coming from e = k t x B1 1

This question in 9709/31 Oct/Nov 2021

7 The variables x and y satisfy the differential equation dy e2x 4xy2, dx = and it is given that y 1 when x 0. = = Solve the differential equation, obtaining an expression for y in terms of x. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 7 Separate variables correctly B1 2 2 1 d 4 e d x y x x y − =   2 1 dy y =  1 y − B1 OE Commence the other integration and reach 2 2 e e d x x ax b x − − +  M1 Obtain 2 2 2 e 2 e d x x x x − − − +  or 2 2 1 1 e e d 2 2 x x x x − − − +  A1 SOI (might have taken out factor of 4) Complete integration and obtain 2 2 2 e e x x x − − − − A1 Evaluate a constant or use x = 0 and y = 1 as limits in a solution containing terms of the form 2 2 , e , e x x p qx r y − − , or equivalent. M1 Obtain y = 2e 2 1 x x + , or equivalent expression for y A1 ISW 7

This question in 9709/32 Oct/Nov 2021

10 A large plantation of area 20 km2 is becoming infected with a plant disease. At time t years the area infected is x km2 and the rate of increase of x is proportional to the ratio of the area infected to the area not yet infected. dx When t 0, x 1 and 1. = = dt = (a) Show that x and t satisfy the differential equation dx 19x [2] dt = 20 −x. … … … … … … … (b) Solve the differential equation and show that when t 1 the value of x satisfies the equation = [5] x = e0.9+0.05x. … … … … … … … … … … … … … … … … … … … … (c) Use an iterative formula based on the equation in part (b), with an initial value of 2, to determine x correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … (d) Calculate the value of t at which the entire plantation becomes infected. [1] … … … …

11 marks

Mark scheme: 10(a) State or imply equation of the form d d x t = k 20 x x − M1 Obtain k = 19 A1 AG 2 10(b) Separate variables and integrate at least one side M1 Obtain terms 20 ln x – x and 19t, or equivalent A1 A1 Evaluate a constant or use t = 0 and x = 1 as limits in a solution containing terms a ln x and bt M1 Substitute t = 1 and rearrange the equation in the given form A1 AG 5 10(c) Use 1 nx + = 0.9 0.05 e n x + correctly at least once M1 Obtain final answer x = 2.83 A1 Show sufficient iterations to 4 decimal places to justify 2.83 to 2 d.p. or show there is a sign change in the interval (2.825, 2.835) A1 3 10(d) Set x = 20 and obtain answer t = 2.15 B1 1

This question in 9709/33 Oct/Nov 2021

6 The variables x and y satisfy the differential equation dy = xey−x, dx and y 0 when x 0. = = (a) Solve the differential equation, obtaining an expression for y in terms of x. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the value of y when x 1, giving your answer in the form a b, where a and b are = −ln integers. [1] … … … … … … … … …

8 marks

Mark scheme: 6(a) Correct separation of variables B1 e d e d y x y x x      Condone missing integral signs. Obtain term e y   B1 Commence integration by parts and reach e e d x x x x     *M1 M0 if clearly using differentiation of a product. Complete integration and obtain e e x x x     A1 Use x = 0 and y = 0 to evaluate a constant or as limits in a solution containing or derived from terms e , e and e y x x a bx c   , where 0 abc  DM1 Must see working for this. In a correct solution they should have e e e y x x C x        or equivalent. If they take logarithms before finding the constant, the constant must be of the right form. Correct solution in any form Must follow from correct working A1 e.g. e e e y x x x       A0 if constant of integration ignored or assumed to be zero. Obtain final answer     ln 1 e x y x    from correct working A1 OE e.g.   ln 1 y x x    , e ln 1 x y x         . A0 if constant of integration ignored or assumed to be zero. 7 6(b) Obtain answer  1 ln2 y   B1 Must follow from at least 6 or7 obtained in part 6(a). 1

This question in 9709/32 May/June 2022

8 At time t days after the start of observations, the number of insects in a population is N. The variation dN 3 in the number of insects is modelled by a differential equation of the form kN 2 cos 0.02t, where dt = k is a constant and N is a continuous variable. It is given that when t 0, N 100. = = (a) Solve the differential equation, obtaining a relation between N, k and t. [5] … … … … … … … … … … … … … … … … … … … … … … … (b) Given also that N 625 when t 50, find the value of k. [2] = = … … … … … … … … … … … … (c) Obtain an expression for N in terms of t, and find the greatest value of N predicted by this model. [2] … … … … … … … … … … …

9 marks

Mark scheme: 8(a) Separate variables correctly B1 3 2 dN N = (k cos 0.02t) dt Allow without integral signs. Obtain term – 2 N B1 OE Ignore position of k . Obtain term 50sin0.02t B1 OE Ignore position of k . Use t = 0, N = 100 to evaluate a constant, or as limits, in a solution containing terms a N and sin0.02 b t , where 0  ab M1 0.2 e.g. 0.2 or          c c k Obtain correct solution in any form, e.g. – 2 N = 50 sin0.02 0.2  k t A1 OE ISW e.g.   2 1 25 sin0.02 0.1   N k t 1 2 1 2 sin0.02 0.02 5     k N t 2 1 50 sin0.02 5   k t N   1 1 1 50sin0.02 2 10   k t N 1 2 20 50sin 50 100        N t kN k 5 8(b) Use the substitution N = 625 and t = 50 in expression of appropriate form to evaluate k M1 Expression must contain a + bksin 0.02t , ( N )±n, where n =  1, 1, 3 or 5 and a and b are constants 0  ab or (a + bksin 0.02t) ±2 and (N) ±n. Allow with k replaced by 1 k , error due to k(N 3/2) when separating variables in 8(a). If invert term by term when 3 terms shown then M0. Obtain k = 0.00285[2148] A1 Must evaluate sin1. Degrees k = 0.138 M1 A0. 2 Question Answer Marks Guidance 8(c) Rearrange and obtain N =   2 4(0.2 0.142 607 sin 0.02 )  t Substitution for k required M1 Anything of the form N = c(d  ek sin 0.02t)2, where c, d and e are constants 0  cde and value of k substituted. Allow with k replaced by 1/k, error due to k(N 3/2) when separating variables in 8(a). OE ISW e.g.   2 2 10 1 0.7125sin0.02 1 0.0713sin0.02 0.1            N N t t 2 100 0.6 sin0.02 1 sin1               N t 2 1 3 1 sin0.02 50sin1 10           N t 2 0.06 sin0.02 0.1 sin1          N t 2 800 80 57 0.02        N sin t Do not need to substitute for sin(0.02t) = 1, but must substitute for k. Accept answers between 1209 and 1215 A1 ISW Substitute sin 0.02t = 1 or t = 50 sin1 1 or 78.5 or 25π. Answer with no working (rubric) 0/2. SC N = … not seen but correct numerical answer B1 1/2. 2

This question in 9709/33 May/June 2022

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 dx = kxe−0.1t, dt 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] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (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] … … … … … … … … … … … …

8 marks

Mark scheme: 8(a) Separate variables correctly B1 1 −0.1t dx = ke dt x   Obtain term ln x B1 0.1t B1 −0.1t xe dt Obtain term −10 ke− Not from  Use x = 20, t = 0 to evaluate a constant or as limits in a solution containing M1 terms alnx , be −0.1t where ab  0 −0.1t A1 or equivalent ISW Obtain ln x = 10k 1 − e + ln20 ( ) 5 8(b) Use x = 40, t = 10 to find k or 10k M1 Available for their function of the correct structure even if they found no constant in (a). Obtain 10k = 1.09654 A1 ln 2 or equivalent e.g. 10 k = −1 1 − e State that x tends to 59.9 A1 Need a number, not an expression for that value 3 sf or better 59.87595….. 3

This question in 9709/31 Oct/Nov 2022

10 A gardener is filling an ornamental pool with water, using a hose that delivers 30 litres of water per minute. Initially the pool is empty. At time t minutes after filling begins the volume of water in the pool is V litres. The pool has a small leak and loses water at a rate of 0.01V litres per minute. dV The differential equation satisfied by V and t is of the form a dt = −bV. (a) Write down the values of the constants a and b. [1] … … … … (b) Solve the differential equation and find the value of t when V 1000. [6] = … … … … … … … … … … … … … … … … … … … … … … … … … … … … (c) Obtain an expression for V in terms of t and hence state what happens to V as t becomes large. [2] … … … … … … … … … … … …

9 marks

Mark scheme: 10(a) a = 30 and b = 0.01 B1 1 10(b) Separate variables and integrate one side M1 Obtain terms −100ln ( 30 − 0.01V ) and t, or equivalent A1 FT FT their a and b. + A1 FT Evaluate a constant, or use t = 0, V = 0 as limits, in a solution containing terms c M1 ln ( 30 − 0.01V ) and dt where cd ≠ 0 Obtain solution 100ln30 − 100ln ( 30 − 0.01V ) = t , or equivalent A1 Substitute V = 1000 and obtain answer t = 40.5 A1 6 10(c) Obtain V = 3000 1 − e−0.01t B1 OE ( ) State that V approaches 3000 B1 2

This question in 9709/33 Oct/Nov 2022

9 The variables x and y satisfy the differential equation dy e3y sin2 2x. dx = It is given that y 0 when x 0. = = Solve the differential equation and find the value of y when x 12. [7] = … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 9 Separate variables correctly and obtain e−y3 and sin 2 2x on the opposite B1 sides 1 −3 y B1 Obtain term − e 3 Use correct double angle formula for sin 2 2x = (1/2)[1 – cos 4x] M1 1  1  A1 Obtain terms x − sin4 x oe   2  4  Use x = 0, y = 0 to evaluate a constant or as limits in a solution containing M1 terms of the form ax and b sin4 x and cey3 Obtain correct answer in any form A1 1 −3 y 1  1  1 e.g. − e = x − sin4 x −   3 2  4  3 1 Substitute x = 12 and obtain y = 0.175 or − 13 ln ( 4 + 83 sin2 ) A1 OE ISW 7

This question in 9709/32 Feb/March 2023

8 (a) The variables x and y satisfy the differential equation dy 4 9y2 + . dx = e2x+1 It is given that y 0 when x 1. = = Solve the differential equation, obtaining an expression for y in terms of x. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) State what happens to the value of y as x tends to infinity. Give your answer in an exact form. [1] … … … … …

8 marks

Mark scheme: 8(a) Separate variables correctly B1   2 1 2 1 d e d 4 9 x y x y       . Condone missing integral signs or dx and dy missing. Obtain term 2 1 1 e 2 x    B1 OE e.g. 2 1 e 2e x   . Obtain term of the form 1 3 tan 2 y a       M1 Obtain term 1 1 3 tan 6 2 y       A1 OE e.g. 1 1 3 3 tan 9 2 2 y   . Use x = 1, y = 0 to evaluate a constant or as limits in a solution containing or derived from terms of the form   1 tan a by  and   2 1 e x c   M1 If they rearrange before evaluating the constant, the constant must be of the correct form. Obtain correct answer in any form A1 e.g.   2 1 1 3 1 3 1 1 tan e e 6 2 2 2 x y       . Obtain final answer   3 2 1 2 tan 3e 3e 3 x y      A1 OE Allow with 3 3e 0.149 …  7 8(b) State that y approaches   3 2 tan 3e 3  B1 FT Or exact equivalent. The FT is on correct work on a solution containing e–2x–1. Condone y = … Accept correct answer stated with minimal wording. 0.10032… is not exact so B0. 1

This question in 9709/32 May/June 2023

7 The variables x and 1 satisfy the differential equation x dx = x2 + 3. tan 1 d1 It is given that x = 1 when 1 = 0. Solve the differential equation, obtaining an expression for x2 in terms of 1. [7] … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 7 Separate variables correctly B1 x  tan=d  x dx . 2 + 3 Condone missing integral signs or missing dx, d. Can be implied by later work. Obtain term –ln(cos ) B1 Or equivalent e.g. ln(sec ). 2 M1 Obtain term of the form a ln x + 3 ( ) 1 2 A1 Obtain term ln x + 3 ( ) 2 Use x = 1, 𝜃 = 0 to evaluate a constant or as limits in a solution containing M1 If they have rearranged then the constant must be of the 2 correct form. terms of the form a ln x + 3 and b ln(cos) ( ) Obtain correct answer in any form A1 1 2 x + 3 = − lncos+ ln2 . 2 ln ( ) 2 4 A1 Or equivalent e.g. x 2 = 4sec 2 x − 3 . Obtain final answer x = − 3 cos 2 lns removed. 7

This question in 9709/31 Oct/Nov 2023

11 The variables x and y satisfy the differential equation dy x2 y2 y 0. dx + + = It is given that x 1 when y 1. = = (a) Solve the differential equation to obtain an expression for y in terms of x. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) State what happens to the value of y when x tends to infinity. Give your answer in an exact form. [1] … … … … …

9 marks

Mark scheme: 11(a) Correct separation of variables. B1  1 1 − dx .  2 dy =  2  y + y  x Condone missing integral signs or missing dx, dy, but not both. 1 B1 Obtain x 1 *M1  1 Express in partial fractions Allow for the correct split of . y 2 + y y 2  y ( ) or express the denominator of the fraction as a difference of two squares 1 1 1 A1 Allow if coefficients for the partial fractions are correct Obtain − or 1 1 y y + 1 ( y + 2 ) 2 − ( 2 ) 2 but followed by an error. Obtain ln y − ln ( y + 1) A1 Or equivalent, dependent on where they left the minus sign. Use x = 1, y = 1 to find constant of integration or as limits in a definite DM1 ln 12 = 1 + C integral in an expression containing terms of the form p , q ln y and r ln (1 + y ) If they rearrange the equation before finding the constant x of integration then the constant must be of the correct form. Correct equation in x and y A1 y 1 1 ln = −+1 ln . 1 + y x 2 1 x −1 A1 1 1 e . Or equivalent e.g. y = , y = Obtain y = 1 x x −1 2e1− 1 − 1 e1−+1 x ln2 − 1 2 − e Accept with decimal value for e-1. 8 11(b) 1 B1 FT 1 State that y approaches Or exact equivalent. Condone y = 2e − 1 2e −.1 FT on an expression in 1ex . 1

This question in 9709/32 Oct/Nov 2023

8 The variables x and y satisfy the differential equation dy e4x = cos2 3y. dx It is given that y = 0 when x = 2. Solve the differential equation, obtaining an expression for y in terms of x. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 8 Separate variables correctly and reach asec2 3y or be−4x B1 Condone missing integral signs or dy and dx, but allow if recognisable integrals follow. Not for 1/cos2 3y and 1/e4x. 1 −4 x B1 Can recover the previous B1 if de–4x seen here. Obtain term − e 4 Obtain only a term of the form a tan3 y M1 Can recover the first B1 if a tan3 y seen here. 1 A1 Obtain term tan3 y 3 Use x = 2, y = 0 to evaluate a constant or as limits in a solution containing M1 May see tan by and e  4 x here. terms of the form a tan by and ce 4 x Obtain correct answer in any form A1 1 1 1 e.g. tan3y = − e−4x + e−8 3 4 4 1 1 or tan3y = − e−4x + 8.39  10–5 3 4 1 −1  3 −8 3 −4 x  A1 ISW e − Obtain final answer y = tan  e  3  −1  3 4  4 e −4 x  OE e.g. y = 1 tan  2.52 10 −4 −  3  4  7

This question in 9709/33 Oct/Nov 2023

11 The variables y and i satisfy the differential equation d y 3 y ( 1 + y)( 1 + cos 2 i) = e . d i r . It is given that y = 0 when i = 14 Solve the differential equation and find the exact value of tani when y = 1. [9] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 11 Separate variables correctly B1 −3 y 1 dθ. ∫ (1 + y ) e d y = ∫ 1 + cos2θ Allow 1/e3y and missing integral signs. Integrate to obtain p (1 + y ) e −3 y +∫ q e −3 y d y M1 Allow unless clear evidence that formula used has a + sign. −1 −3 y 1 −3 y A1 Allow unsimplified. Obtain (1 + y ) e +∫ e dy 3 3 −1 −3 y 1 −3 y A1 Condone no constant of integration. Obtain (1 + y ) e − e ( + A ) 3 9 1 B1 dθ Use correct double angle formula to obtain ∫ 2cos 2 θ Obtain ktanθ[ + B ] B1 Condone no constant of integration. π M1* 1 1 1  17  Use y = 0, θ = to evaluate a constant of integration in an expression of the form = − − + C  C =  4 2 3 9  18  αye −3 y , βe −3 y and γtanθ only. Allow αye3y and βe3y. Must have integrated LHS twice. Use y = 1 DM1 − (1 + 1) 3 1 17 = tan θ− . − 1 ( 9e ) 3 3e 2 18 Must have integrated LHS. 17 14 −3 A1 Or exact equivalent . Exact ISW. Obtain tanθ = − e 9 9 −1  17 14 −3  Allow θ = tan  − e  .  9 9  If x instead of θthen withhold final A1. 9

This question in 9709/32 Feb/March 2024

11 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 dx variables x and t are treated as continuous, and it is given that = 0. 2 and x = 1 when t = 0 . dt (a) Show that x and t satisfy the differential equation dx 1495 = x ( 300 - x) . [2] dt … … … … … … … … … … … … … (b) Using partial fractions, solve the differential equation and obtain an expression for t in terms of a single logarithm involving x. [9] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 11(a) State or imply equation of the form d (300 ) d   x kx x t and use d 0.2 and 1 d   x x t Obtain k = 1 1495 and rearrange to the given answer A1 d 1495 (300 ). d x x x t   2 Question Answer Marks Guidance 11(b) Separate variables correctly B1   1 1 d d 300 1495        x t x x Correct integration of t term B1 E.g. obtain t or . 1495 t State or imply partial fractions of the form 300   A B x x B1 Correct method to find A or B M1 A = 1 300 and B = 1 . 300 May see 1495 299. 300 60 A B    Obtain terms 1495 1495 ln ln(300 ) 300 300   x x A1 OE. May see   1 1 ln ln 300 . 300 300 x x   Use t = 0, x = 1 to evaluate a constant or as limits in a solution containing terms of the form ln , ln(300 ) x x  and t. M1 Obtain correct answer in any form A1 E.g.   1495 1495 ln ln(300 ) ln 299. 300 300 x x t    Use law of logarithms twice to obtain an expression for t M1 Obtain final answer 299 299 ln 60 300   x t x or equivalent single logarithm A1 9

This question in 9709/31 May/June 2024

10 (a) By writing y = sec 3i as cos3i, show that di = 3 sin i sec 4 i . [2] … … … … … … … … … … … … (b) The variables x and i satisfy the differential equation 2 d i 4 x + 9 sin i = ( x + 3) cos i . ` j d x r . It is given that x = 3 when i = 13 Solve the differential equation to find the value of cosi when x = 0 . Give your answer correct to 3 significant figures. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(a) Use of correct chain rule (and correct quotient rule) and 3 cos   4 cos sin k      or equivalent.   4 4 d 3 sin cos 3sin sec d y         Must be expressed in the given form A1 Obtain given answer from full and correct working (signs must be shown), but condone   3 d d sec …   and '( ). y  2 Question Answer Marks Guidance 10(b) Separate variables:     4 2 3 sin d d cos 9 x x x            B1 Or 4 2 d d . 3sin 3 9 cos 9 x x x             Condone missing integral signs or missing d or d , x  but not both. Obtain   3 sec p A  B1 Correct form, p any constant but not 0. Use 2 2 2 d d 3 9 3 9 9 9 9 x x x x x x x                  and obtain   2 ln 9 q x  or   1 3 tan . x r C   *M1 Might have one third of both sides. Alt: substitute 3tan x   to obtain 1 tan d ; q    condone if have  in place of  in this method. Obtain   2 ln 9 q x  and   1 3 tan x r C   DM1 Obtain     ln cos q    OE. Obtain     3 2 1 3 2 3 sec ln 9 3tan x x C       or equivalent A1 Or might see a third of both sides. Must have 2 different variables. Use 1 3 , 3 x     in an equation including 3 sec , p    2 ln 9 q x  and 1 3 tan x r  to evaluate the constant of integration M1 Or as limits in a definite integral. Limits for  are 0 and 1 4 .  Obtain constant = 3 3 2 4 8 ln18    A1 OE, e.g. 1.308… to at least 3sf. Obtain cos 0.601  A1 Accept AWRT 0.601. 8

This question in 9709/32 May/June 2024

10 A balloon in the shape of a sphere has volume V and radius r. Air is pumped into the balloon at a constant rate of 40r starting when time t = 0 and r = 0 . At the same time, air begins to flow out of the balloon at a rate of 0.8rr . The balloon remains a sphere at all times. (a) Show that r and t satisfy the differential equation dr 50 - r = . [3] td 5r 2 … … … … … … … … … … … (b) Find the quotient and remainder when 5r 2 is divided by 50- r . [3] … … … … … … … … … … … … (c) Solve the differential equation in part (a), obtaining an expression for t in terms of r. [6] … … … … … … … … … … … … … … … … … … … … … … (d) Find the value of t when the radius of the balloon is 12. [1] … … … …

13 marks

Mark scheme: 10(a) dV B1 Need a complete correct statement seen or Obtain = 40π − 0.8πr or equivalent implied. dt dV 2 dV 2 dr B1 Need a complete correct statement seen or Obtain = 4πr or equivalent e.g. = 4 r implied. dr dt dt Use the chain rule to obtain given answer (including the derivative) B1 dr 50 − r dr 40 − 0.8r Allow if = follows = dt 5r 2 dt 4 r 2 without further explanation (π already cancelled) and no incorrect statements seen. 3 10(b) Commence division and reach quotient of the form M1 Allow M1 if divide by r − 50 to obtain –5r ± 250 5r  250 . or 5r2 = (50 – r)(Ar + B) + C and reach A = –5 and B = ± 250 Obtain quotient –5r – 250 A1 Do not need to state which is quotient and which is remainder. However, if clearly muddled, then M1A1A0 for both expressions correct. Obtain remainder 12 500 A1 Note: 12 500 following division by r – 50 is correct and scores this A1 ISW. SC B1 only for correct use of remainder theorem to obtain correct remainder. 3 10(c) Prepare to integrate e.g. separate variables correctly B1FT 2 5 r 1d t d r =  2  50 − r d t 5 r  12500  Condone missing dr, dt or missing integral Or express in the form =  = − ( 5 r + 250 ) +  dr 50 − r  50 − r  signs, but not both. Follow their division in (b) if substitute before separating. Obtain term t DB1 A 2 M1 C Obtain terms r + Br − Cln ( 50 − r ) From their Ar + B + in (b) where 2 50 − r ABC ≠ 0. Allow a single slip in the coefficients. 5 2 A1FT FT their (b), provided of the correct form. Obtain terms − r − 250r − 12500ln(50 − r ) 2 Use t = 0, r = 0 to evaluate a constant or as limits in a solution containing terms of M1 the form r2, r, ln(50 – r) and t 5 2 A1 OE Obtain final answer t = − r − 250r − 12500ln(50 − r ) + 12500ln50 Must be t = ….. 2 Allow with 12500ln50 = 48900 or better. 6 10(d) Obtain t = 70.5 B1 May be more accurate (70.4605…). 1

This question in 9709/32 Oct/Nov 2024

6 The variables x and i satisfy the differential equation d x 1 2 = b x + 1l sin 2 i , d i 5 and x = 5 when i = 0 . Solve the differential equation and obtain an expression for x in terms of i. [7] … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 6 Use correct double angle formula to express sin2 2θ in terms of cos 4θ B1 1 sin2 2θ = (1 – cos 4θ) 2 1 Separate variables correctly and reasonable attempt at integration of at least one side M1 Position of (x + 5) or ( 5 x + 1) and sin2 2θ sufficient for correct separation.  1  B1 OE Obtain term 5ln  x + 1  May see 5ln ( x + 5 ) .  5  1 1 B1 FT 1  1  1 Obtain term (− sin4) Allow   sin 4 from = ( 1 ± cos 4θ). 2 4 2  4  2 Use x = 5 when θ = 0 to evaluate a constant or as limits in a solution containing M1 OE  1  terms of the form ln  x + 1  ,  and sin4  5  Obtain correct answer in any form A1 1 1 E.g. 5ln ( x + 5 ) = (− sin4) + 5 ln 10 2 4  1  1   A1 FT  1  1   Obtain final answer x = 10exp − sin4 − 5 or equivalent x = 10exp  sin4 − 5         10  4    2  4   with ln removed Must remove ln from 1 1 ln ( x + 5 ) = (  sin4 ) +  ln 10. 2 4 7

This question in 9709/32 Feb/March 2025

10 (a) Find the quotient and remainder when x 3 + 5x 2 - 2x - 15 is divided by x 2 - 3 . [3] … … … … … … … … … … … … … (b) The variables x and y satisfy the differential equation d y x 3 + 5 x 2 - 2 x - 15 = . dx 6y ( x 2 - 3 ) It is given that y = 2 when x = 2 . Solve the differential equation to obtain an expression for y2 in terms of x. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: M1 x + 510(a) Divide by ( x 2 − 3 ) to obtain x + k ( k  0 ) x 2 − 3 3x +5x 2 − 2x −15 3x −3x +5x 2 + x +5x 2 −15 x Obtain quotient x + 5 A1 Obtain remainder x A1 ISW x Allow . x 2 − 3 3 2 10(b) x 3 + 5 x − 2 x − 15 Separate variables correctly and obtain 6 y dy = 3 y 2 B1  6 y dy = dx. OE from  2  x − 3  Obtain 12 x 2 + 5 x B1ft Follow their linear quotient. 1 2 B1ft From the x term in their remainder ax + b. x − 3 Obtain 2 ln ( ) C = 0 Use y = 2, x = 2 to evaluate the constant of integration in an integral containing M1 12 = 2 + 10 + 12 ln1 + C  k ln x 2 − 3 ( ) 2 1 2 5 1 2 A1 OE x − 3 Obtain y = 6 x + 3 x + 6 ln ( ) 5

This question in 9709/31 May/June 2025

8 The variables x and i satisfy the differential equation d x sin 2i = ( 4x + 3) cos 2i , d i and x = 0 when i = 1 r . 12 Solve the differential equation and obtain an expression for x in terms of i. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 8 Separate variables correctly B1  1  cos2  dx =  d  4 x + 3  sin2 Can be implied by obtaining both correct integrals. Obtain term 14 ln ( 4 x + 3 ) B1 OE 3 or 1 ) . 4 ln ( x + 4 Obtain term of the form A ln ( sin2) M1 Or A ln ( k sin2OE,) e.g. P ln a sin+ Q ln b cos from using the tan2 formula. Or expanding cos2as cos 2 − sin 2 . Obtain term 12 ln ( sin 2) A1 OE Correct in any form, e.g. 12 ln a sin+ 12 ln b cos. 1 Use x = 0 when θ = 121 π to evaluate a constant or as limits in a solution M1 E.g. c = 14 ln3 − 12 lnsin ( 6 π ) 1 1 containing terms of the form ln ( sin 2) and ln ( 4 x + 3 ) . ln 1 4 ln ( 4 x + 3 ) − 4 ln3 = 12 ln ( sin 2) − 12 2 c = … seen or implied Note that the constant may be expressed as a logarithm 1 1 Obtain correct answer in any form with the trigonometry evaluated A1 E.g. 14 ln ( 4 x + 3 ) = 2 ln ( sin2) + 4 ln12 1 1 4 ln ( 4 x + 3 ) = 2 ln ( sin2) + 0.621 1 ln 4 x + 3 = 1 ln ( 2sin2) 4 3 2 12sin 2 2− 3 A1 OE Obtain final answer x = 2.48 … 2 e sin 2− 3 4 Allow . 4 Allow 2.48. 1 2ln ( sin2+) 2.48... e − 3 . Allow x = 4 ( ) 7

This question in 9709/32 May/June 2025

10 The variables x and y satisfy the differential equation dy sin 4 y = x sin 2y sin 3x . dx It is given that y = 1 r when x = 1 r . 12 2 (a) Solve the differential equation, obtaining a relation between x and y. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Given that 0 1 y 1 1 r, find the values of y when x = 0 . [2] 2 … … … … … … … … … … … … …

10 marks

Mark scheme: 10(a) Separate variables correctly B1 Use correct double angle formula to simplify integral in y *M1  sin4 y  dy 2cos2 y dy =  sin2 y Obtain sin 2y A1 *M1 cos3 x dx Commence integration by parts and obtain px cos3 x + q  1 1 A1 cos3 x dx Obtain − 3 x cos3 x + 3  Complete integration and obtain − 13 x cos3 x + 19 sin3 x A1 Use y = 121 π when x = 12 π in an expression with sin2 y , x cos3 x and sin3 x to obtain DM1 12 = 0 − 19 + c the constant of integration Obtain sin2 y = − 13 x cos3 x + 19 sin3 x + 1811 A1 OE ISW 8 10(b) Solve sin2y = 1811 to obtain one solution, e.g. 0.329 M1 AWRT Allow for their constant from a solution involving sin2 y , x cos3 x and sin3 x. M0 for answers in degrees. Obtain a second solution, e.g. 1.24, and no others in range A1 AWRT 2

This question in 9709/33 May/June 2025

11 The variables x and y satisfy the differential equation 2 d y 3 y ( x + 3 ) = e ( x - 2) . d x It is given that y = 0 when x = 0 . Solve the differential equation, and find the value of y when x = 2 . [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 11 Separate variables correctly B1 1 Sight of sufficient for f(y) in f(y)dy = g(x)dx to obtain e 3 y B1. 1 −3 y B1 Obtain term − e 3 1 2 B1 Separate fractions and obtain term ln ( x + 3 ) 2 Separate fractions and obtain term of the form a tan −1 bx M1 2 −1 x A1 OE Obtain term − tan 3 3 Use x = 0, y = 0 to evaluate a constant or as limits in a solution containing terms M1 −1 mx of the form a e 3 y , b ln ( x 2 + 3 ) and c tan Obtain correct solution in any form relating x and y A1 1 −3 y 1 2 2 −1 x 1 1 E.g. − e = ln ( x + 3 ) − tan − − ln 3. 3 2 3 3 3 2 1 1 Constant − − ln3 may be – 0.883. 3 2 Obtain – 0.331 A1 OE AWRT. E.g. –0.33084… Note A0A1 is possible. 8

This question in 9709/35 May/June 2025

8 The variables x and y satisfy the differential equation 2 dy 2 y `x + 1j = kxe , dx where k is a constant. It is given that y = 0 when x = 0 and that y =- 1 when x = 1. 2 Solve the differential equation and find the exact value of y when x = 3 . [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 8 Separate variables correctly B1 kx −2 y dx = e dy 2  x + 1  Condone missing integral signs or missing dx, dy, but not both. Obtain term Ae−y2 B1 OE − 12 e−2 y 2 B1 OE Obtain term B ln x + 1 ( ) k 2 x + 1 2 ln ( ) 1 −2 y k 2 B1 OE x + 1 +C ) Obtain − 2 e = 2 ln ( )( Use y = 0 when x = 0 to evaluate c or as limits in a solution containing terms of M1 the form e−y2 and ln x 2 + 1 ( ) Use y = − 12 when x = 1 to evaluate k or as limits in a solution containing terms of M1 the form e−y2 and ln x 2 + 1 ( ) 1 1 − e A1 1 −2 y 1 − e 2 1 Obtain c = – and k = OE, e.g. − e = ln x + 1 − . ( ) 2 ln2 2 2ln2 2 1 A1 Or exact equivalent. Obtain final answer y = − ln ( 2e − 1) 2 8

This question in 9709/31 Oct/Nov 2025

2 10 (a) Express in partial fractions. [2] 1 - 9y 2 … … … … … … … … … … … … (b) The variables x and y satisfy the differential equation 2 d y 2 2 cos 3 x = 1 - 9 y , d x and y = 0 when x = 1 r . 12 Solve the differential equation and obtain an expression for y in terms of x. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 10(a) Carry out a relevant method to find A and B such that M1 OE 2 A B A B = + o Allow M1 for finding A and B for + 1 − 9 y 2 1 + 3 y 1 − 3 y 1 + 3 y 3 y − 1 and A1 for A = 1, B = –1 if –2 = A(3y – 1) + B(1 + 3y). But, A0 for A = –1, B = 1 if 2 = A(3y – 1) + B(1 + 3y). Obtain A = 1 and B = 1 A1 If work with x and never see y, award M1A0, but allow M1A1 if y is seen anywhere on right hand side. 2 10(b) Separate variables correctly and attempt integration of at least one side M1 Integrate to obtain at least one log term of the form a + by p ln (a + by) OE, q ln on one side, or a tan a − by term on the other, and disregard the 2 if it appears. 2 1 1 A1FT 1  1 + 3 y  Integrate 2 to obtain ln (1 + 3 y ) − ln (1 − 3 y ) OE, e.g. ln   . 1 −y9 3 3 3  1 − 3 y  A B FT ln (1 + 3 y ) − ln (1 − 3 y ) 3 3 A B or ln (1 + 3 y ) + ln ( 3 y − 1) if their partial 3 3 fractions used. The ‘2’ must have been dealt with correctly for this mark (check right hand side for 2 appearing here) Obtain r tan 3x B1 1 B1 1 Obtain term tan3 x Allow tan3 x if ‘2’ not dealt with correctly 3 6 earlier. 1 M1 0 + 0 +1/3 + C = 0 Use y = 0 when x = π to evaluate a constant or as limits in a solution of the No errors in substitution. 12 form p ln (1 + 3 y ) + q ln (1 − 3 y ) + r tan3x where p,q,r ≠ 0 1 1 1 1 A1 OE ln (1 + 3 y ) − ln (1 − 3 y ) = tan3 x − ISW 3 3 3 3 1 2 e tan3 x −1 − 1 or y = Obtain answer y = − tan3 x −1 + 1) 3 3 (1 tan3 x −1 + e ) 3 ( e 6

This question in 9709/33 Oct/Nov 2025

11 A fungal disease is affecting some of the trees in a forest. The fraction of the trees affected after t years is denoted by x. The rate of increase of x is proportional to the product of the fraction of the trees affected and the fraction of the trees not affected. d x (a) Explain why, after t years, = kx ( 1 - x) , where k is a constant. [1] d t … … … … (b) When the disease is first detected, one quarter of the trees are affected. Two years later, one third of the trees are affected. Solve the differential equation to find the number of years from the time when the disease is first detected until the time when three quarters of the trees are affected. Give your answer correct to the nearest year. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 11(a) d x B1 AG Obtain = kx (1 − x ) , stating where the factors come from Factor 1 – x explained. d t 1 11(b) Separate variables correctly B1 1 k dt dx =   x (1 − x ) Condone missing integral signs or dx and dt but not both. 1 M1 Condone a sign error between the fractions. Correct method to express in partial fractions x (1 − x ) 1 1 A1 Obtain + x 1 − x Obtain general solution ln x − ln (1 − x ) = kt + A A1 OE Use t = 0, x = 14 in an expression containing pt , q ln x and q ln (1 − x ) to obtain the *M1 1 A = ln constant of integration 3 1 3 Use t = 2, x = 13 in an expression containing pt , q ln x and q ln (1 − x ) to calculate DM1 k = ln the value of k 2 2 x 1 3 1 A1 Must be seen, as part of question demand. Obtain ln = ln t + ln x 1 − x 2 2 3 OE, e.g. ln = (0.2027...)t − 1.0986... 1 − x Substitute x = 0.75 and obtain t = 11 years (to the nearest year). A1 10.838… may be seen. 8 SC M1DM1A0A1 for candidates who use t = 1 and t = 3 rather than t = 0 and t = 2 to obtain t = 12 (11.838…) years (to the nearest year).

This question in 9709/35 Oct/Nov 2025