Notes

B1: \(A =\) awrt 150

Note \(A = 150.0436\ldots\)

Notes

You may mark (b) (c) and (d) together so do not be concerned about the labelling.

M1: Finds an integrating factor of the form \(\mathrm{e}^{\pm\int \frac{\cos t}{\sin t}\,\mathrm{d}t} = \mathrm{e}^{\pm\ln\sin t}\) or \(\operatorname{cosec} t\) (or \(\sin t\)) and proceeds to multiply through by their IF oe
Look for
\(I.F. = \mathrm{e}^{\pm\int \frac{\cos t}{\sin t}\,\mathrm{d}t} \Rightarrow x \times \text{`their } I.F.\text{'} = \displaystyle\int A\sin 2t \times \text{`their } I.F.\text{'}\,\mathrm{d}t\)
Where they must be using a numerical \(A\)

dM1: Dependent on the previous method mark. Uses the identity \(\sin 2t = 2\sin t\cos t\) to arrange RHS into an integrable form with a correct integrating factor of \(\operatorname{cosec} t\) and then attempts to integrate using their numerical \(A\). Condone a sign slip on their RHS.
\(x\operatorname{cosec} t = 2A\displaystyle\int \cos t\,\mathrm{d}t \Rightarrow x\operatorname{cosec} t = \pm\alpha\sin t\ \ (+c)\)

A1: Correct general solution, condone missing \(+ c\)

M1: Using the model, \(t = 1,\ x = 213\) to find the value of their constant \(c\). Substitution if not seen is implied by sight of 213 for \(x\) and at least one correct value for \(\sin t\). They must use radians and not degrees; you may need to check.

A1: Correct particular solution of \(x = 300\sin^2 t + 0.687\sin t\)
\(c =\) awrt 0.69 (when using \(A = 150\))

Alternatively:
When using answers for \(A\) which have not been rounded to 3sf, the equation will instead be \(x = 300.0872..\sin^2 t + 0.613\sin t\) in which case allow \(c =\) awrt 0.61

(corrected from the printed mark scheme: the integrating factor exponent is printed with d\(x\) in place of d\(t\) in the scheme and in the first M1 note)

Notes

M1: Sets \(x = 0\) and solves a quadratic equation of the form \(\lambda\sin^2 t + \mu\sin t = 0\), where \(\lambda, \mu \neq 0\) finding at least one value for \(\sin t\)

M1: Solves a quadratic equation of the form \(2A\sin^2 t + c\sin t = 0\) where \(A,\ c \gt 0\) to find a correct value of \(t\). May be implied by a value of \(t\) awrt 3.14
Award for \(t = \pi\) but accept the use of \(\arcsin\left(\dfrac{-\mu}{\lambda}\right) + \pi\) oe, where they must add \(\pi\) to a positive value of \(t\). They must also be working in radians.

A1: Correct time awrt 13:08 or 13:09 or 1.08 (pm), 1.09 (pm), which comes from a correct equation

Notes

M1: Solves their 3TQ equation of the form \(2A\sin^2 t + c\sin t = 300.68\) where \(A,\ c \gt 0\) to find an answer for \(\sin t\). You may need to check the value for \(\sin t\) or \(t\) obtained from their quadratic. A correct equation followed by \(t =\) awrt 1.57 is sufficient.

A1: \(t =\) awrt 1.57 isw. Answer must come from a correct equation. Do not allow an answer of \(\dfrac{\pi}{2}\) which may have been obtained from \(\sin t = 1\) even if it proceeds to 1.57
Here it asks for the value of \(t\), so do not accept a time, for example 11.42am

Alternatively uses
\(300.0872\sin^2 t + 0.613\sin t - 300.68 = 0 \Rightarrow \sin t = \ldots\ t =\) awrt 1.56

Notes

B1ft: They must have stated times or values for \(t\) in (c) and (d) for this mark.
We need a comparison consistent with their times from (c) and (d) and reality.
Their values may not be correct, but their statement should be consistent with their times.
You may ignore incorrect statements once a correct statement is given, provided their incorrect statement does not contradict their correct statement.

e.g.

Accept

• The time to go up and down is the same, which is not realistic
• The times are similar which is unlikely
• It should take longer to go up the hill
• In reality it will take longer to go up the hill than down the hill (because of the slope)

Condone

• It takes longer to go up the hill than down the hill
• It takes the same amount of time to go up and down
• In the model it takes them longer to go down the hill
• The rambler may not walk up and down at the same speed, or rate

Reject

• the rambler may need to take a rest
• the model is inaccurate or invalid for certain values of \(t\)
• the rambler may walk down a different path or route
• the rambler is modelled as a particle
• the rambler walks at a constant velocity
• the rambler may not walk at the same speed/pace for the entire journey

If you are uncertain please send to review

Notes

M1: Begins to solve the problem with a correct process to find the integrating factor for the model. Look for an attempt at \(\mathrm{e}^{\int \frac{A}{t + 2}\,\mathrm{d}t} = \ldots\) where \(A\) is a constant (positive or negative). For attempts at IF by recognition they must achieve form \(A(t + 2)^3\)

M1: Attempts the solution of the differential equation using a correct method with their integrating factor. The d\(t\) may be implied. They must attempt the integration as part of the method, though it need not be correct – any changed function will do. So score for \(v \times \text{their IF} = \text{their attempt at integrating their IF} \times \left(k - \dfrac{3}{t + 2}\right)\) allowing for errors expanding this bracket.

A1: Correct solution. The \(+ c\) may be missing for this mark.

M1: Interprets the initial conditions to find the constant of integration – must have a constant of integration for this mark and the correct initial conditions must be clear.

A1*: Obtains the printed answer with no errors and at least one intermediate line between \(v(t + 2)^3 = \dfrac{k}{4}(t + 2)^4 - (t + 2)^3 + c\) and the stated answer. Allow terms in different order, but the \(8 - 4k\) must be factorised.

Notes

M1: Uses the given conditions to establish the value of \(k\) in the model

dM1: Uses their value of \(k\) and the equation of the model with \(t = 5\) to find a value for \(v\). If a value of \(k\) has been found accept “\(t = 5\ \ v = \ldots\)” with any value for an attempt.

A1: Correct value including units (allow awrt 8.06 m s\(^{-1}\)). Accept exact answer 5531/686

Notes

B1: Evaluates the model by making a suitable comment – see scheme for examples. Accept answers relating to the size of velocity increasing beyond what is reasonable or answers that allude to the situation not being able to continue for ever. Accept if symbols are used, e.g. “as \(t \to \infty,\ v \to \infty\) which cannot happen” but must include a comment on validity, not just a statement about what happens (e.g. \(t \to \infty,\ v \to \infty\) with no comment).
Do not accept e.g. “\(v\) does not tend to a value” as this is just a statement about what happens.
If contrary information is given then award B0.

Some examples of borderline answers seen.

• “Not valid as rain is limited to how fast its velocity is therefore can’t be big.” Scores B1
• “Due to gravitational force and not huge height between cloud and ground, the model would be inaccurate for large period of time, because the velocity of raindrops would be a big number, which is unlikely.” B1
• “Not valid for very large value of \(t\) as raindrop will not move that fast.” B0
• “Raindrop has very little mass so effect due to gravity is lower. Unlikely to have a 1000 m s\(^{-1}\) raindrop. Not very valid model for large values of \(t\)” B1
• “The water droplet wouldn’t always follow the model in cases where it hits a surface” B0
• “No valid as raindrop will hit ground soon. Speed can be infinitely big which is not plausible.” B1
• “Would not be valid as the raindrop likely has already hit the ground.” B1

You may need to use your judgement on whether their comments are acceptable.

Notes

M1: Sets up the correct equation for the model using the information in the question.

dM1: Uses the initial conditions to find the constant of proportionality for flow out.
Condone use of \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = +3\). Depends on the first mark.

A1*: Correct equation shown from correct work proceeding via \(10k = 4\) to find \(k\).

Attempts in (a) using verification score no marks:
E.g. \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 3 - \dfrac{4}{1 + \mathrm{e}^{0.8t}} - \dfrac{2}{5}V \Rightarrow -3 = 3 - \dfrac{4}{2} - 0.4V \Rightarrow V = \dfrac{4}{0.4} = 10\)

Notes

Alternative to part (b):

Notes

M1: Differentiates to achieve the form shown. Allow \(k = 1\)

A1: Correct derivative in any form. Need not be simplified.

Alternative:
M1: Takes tan of both sides and differentiates implicitly and reaches \(\dfrac{1}{1 + \left(\mathrm{e}^{0.4t}\right)^2} \times k\mathrm{e}^{0.4t}\). Allow \(k = 1\).
A1: Correct derivative in any form. Need not be simplified.

Notes

B1: Deduces the correct integrating factor for the equation. May be implied by sight of \(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\mathrm{e}^{0.4t}V\right) = \ldots\) or equivalent work.

M1: Fully multiplies through by their integrating factor and integrates the LHS (look for \(I.F. \times V = \displaystyle\int I.F. \times \left(3 - \frac{4}{1 + \mathrm{e}^{0.8t}}\right)\mathrm{d}t\) though condone missing \(\mathrm{d}t\).

M1: Attempts the integral of the RHS.
Award for \(\displaystyle\int \alpha\mathrm{e}^{0.4t}\,\mathrm{d}t = \beta\mathrm{e}^{0.4t}\ \ \alpha \neq \beta\) or \(\displaystyle\int \frac{\alpha\mathrm{e}^{0.4t}}{1 + \mathrm{e}^{0.8t}}\,\mathrm{d}t = \beta\arctan\mathrm{e}^{0.4t},\ \ \beta \neq 0\)

A1: Correct integration, need not be simplified. Allow if the \(+ c\) is missing for this mark.

M1: Attempts to find their constant – which must have been treated correctly from point of integration.
Note that this is not formally dependent but there must have been an attempt to integrate.

A1: Correct answer. The question says “simplest form” but allow equivalent expressions e.g. \(V = 7.5 - \dfrac{10\arctan\left(\mathrm{e}^{0.4t}\right)}{\mathrm{e}^{0.4t}} + \dfrac{5\pi}{2\mathrm{e}^{0.4t}} + \dfrac{5}{2\mathrm{e}^{0.4t}}\) but do not allow inexact values for the constants.

Notes

B1ft: Evaluates \(V\) where \(V \gt 0\) at \(t = 10\) and makes an appropriate comment.
For the evaluation, allow if a value of \(V\) is obtained even if there is no evidence of substitution provided that it is clear that \(t = 10\) has not been substituted into something that is not \(V\). So you do not need to check their value.
For the tolerance you may need to use your own judgement but a general guide is:

Notes

M1: Finds the integrating factor and attempts the solution of the differential equation.
Look for \(\text{I.F.} = \mathrm{e}^{\int \tan x\,\mathrm{d}x} \Rightarrow y \times \text{'their I.F.'} = \int \mathrm{e}^{2x}\cos x \times \text{'their I.F.'}\,\mathrm{d}x\)

A1: Correct solution condone missing \(+\,c\)

A1: Correct general solution, Accept equivalents of the form \(y = \mathrm{f}(x)\), such as \(y = \dfrac{\mathrm{e}^{2x}}{2\sec x} + \dfrac{c}{\sec x}\)

Notes

M1: Uses \(x = 0\) \(y = 3\) to find the constant of integration. Allow if done as part of part (a) and allow for their answer to (a) as long as it has a constant of integration to find.

M1: Sets \(y = 0\) in an equation of the form \(y = \left(A\mathrm{e}^{2x} + c\right)\cos x\) (oe) where \(A\) is 1, 2 or \(\dfrac{1}{2}\), with their \(c\) or constant \(c\) and makes a valid attempt to solve the equation to find a value for \(x\). (Allow even if the constant of integration has not been found).

A1: Depends on both M’s. Awrt 1.57 or \(\dfrac{\pi}{2}\) only. There must have been an attempt to find the constant of integration, but allow from a correct answer to (a) as long as a positive value for \(c\) has been found (can be scored from implicit form).

Notes

M1: Clearly identifies that Rate of paint out \(= 3 \times \dfrac{r}{\text{their volume}}\). It is a “show that” question so there must be clearly reasoning. Just answer with no reasoning scores M0.

A1: Puts all the components together to form the correct differential equation.

Notes

(In these notes \(a\) stands for the constant \(\alpha\) in the differential equation.)

M1: Identifies as a first order differential equation and finds the integrating factor or separates the variables and integrates. Allow if there are sign slips in rearranging (e.g. to \(\dfrac{\mathrm{d}r}{\mathrm{d}t} - \dfrac{r}{10} = 2\)) or in the integrating factor and allow with their value for \(a\) or with \(a\) as an unknown.

M1: Multiplies through by the IF and attempts to integrate or integrates to the form \(\lambda\ln(2a - r) = \dfrac{1}{a}t + c\) oe

A1ft: Correct integration, including constant of integration. Follow through on their value of \(a\), but not sign slips from rearrangement. So allow for \(r\mathrm{e}^{\frac{t}{a}} = 2a\mathrm{e}^{\frac{t}{a}} + c\) or \(-\ln(2a - r) = \dfrac{1}{a}t + c\) oe with \(a\) or their \(a\).

M1: Uses the initial conditions to find the constant of integration. Must see substitution or can be implied by the correct value for their equation. Allow for finding in terms of \(a\) if separation of variables used.

M1: Sets \(r = 15\), achieves \(\mathrm{e}^{\frac{t}{10}} = \alpha \gt 0\) or \(\mathrm{e}^{-\frac{t}{10}} = \beta \gt 0\) as appropriate and solves to find a value for \(t\). Separates the variable method sets \(r = 15\) and rearranges to find a value for \(t\). Note: For this mark a value of \(a\) is needed, but need not be the correct one.

A1cso: \(t =\) awrt 7 seconds from fully correct work.

Alternative: separating the variables

Notes

B1ft: See scheme, follow through on their answer to part (b). Accept any reasonable comparative comment but must have a reason, not just a statement of good or not good. So e.g. look for finding the difference between their answer and 9, or the percentage difference. If their answer is close to 9, then accept a conclusion of being a good model if a suitable reason is given. May substitute 9 into their equation and obtain a value to compare with 15 and make a similar conclusion.

Notes

(a)
B1: A correct rearrangement (may be implied by subsequent work). Alternatively, recognises the LHS as a derivative and writes \((1 + t)\dfrac{\mathrm{d}P}{\mathrm{d}t} + P = \dfrac{\mathrm{d}}{\mathrm{d}t}\left(P(1 + t)\right)\left(= t^{\frac{1}{2}}(1 + t)\right)\) (may be implied).
M1: Uses the model to find the integrating factor (or recognise the derivative) and attempts the solution of the differential equation to achieve \(P \times \text{their IF} = \displaystyle\int \text{their } t^{\frac{1}{2}} \times \text{their IF}\,\mathrm{d}t = \ldots\) but do not be too concerned with the mechanics of integrating the RHS but it must be attempted.
A1: Correct solution
M1: Interprets the initial conditions to find the constant of integration. Must be using \(t = 0\) and \(P = 5\) in an equation with a constant of integration, but their equation may have come from incorrect work. This is correctly interpreting the initial conditions and attempting to use them.
M1: Uses their solution to the problem to find the population after 8 hours. Must be using their solution, but allow for any equations which arise from an attempt at solving the differential equation.
A1: cso Correct number of bacteria (accept awrt 10 300) from a correct equation

Notes

(b)
M1: Realises the need to differentiate the model and uses an appropriate method to find the derivative. Allow the M for attempts at implicit differentiation with \((1+t)P = \ldots\) Trivialised differentiation from incorrect work is M0.
A1ft: Correct differentiation of the correct answer to (a) up to the constant of integration to obtain \(\mathrm{d}P/\mathrm{d}t\) in terms of \(t\) (if implicit differentiation is used, they must get to a function in terms of \(t\) only, or revert to the Alternative method). Follow through on their \(c\) in an otherwise correct equation from (a).
M1: Uses \(t = 4\) in their \(\mathrm{d}P/\mathrm{d}t\) (allow from any attempts at the derivative) to obtain a value for \(\mathrm{d}P/\mathrm{d}t\).
A1: Correct answer, allow 1075 or answers rounding down to 1070 with correct units. Accept as 1.07 thousand bacteria per hour.
(NB If 5000 is used in (a) instead of 5, the answer here would be \(-198.725\))

(b) Alternative:

Alternative:
M1: Substitutes \(t = 4\) into their \(P\)
A1ft: Correct value for \(P\). Follow through on their constant of integration from part (a), but the rest of the equation must be correct.
M1: Uses \(t = 4\) and their \(P\) to find a value for \(\mathrm{d}P/\mathrm{d}t\)
A1: Correct answer allow 1075 or answers rounding down to 1070 with correct units. Accept as 1.07 thousand bacteria per hour.

(corrected from the printed mark scheme: the subscript in \(\left(\dfrac{\mathrm{d}P}{\mathrm{d}t}\right)_{t=4}\) is printed as \(t = 1\))

Notes

(c)
B1: Suggests a suitable limitation which must refer to the model. Allow for a sensible comment even if they have no equation for the model
Do not allow answers such as “the model does not take account of external factors such as temperature” as we do not know what factors the model does take account of.

Notes

(a)
B1: Differentiates the first equation with respect to \(t\) correctly. May have rearranged to make \(y\) the subject first. The dot notation for derivatives may be used.
M1: Uses the second equation to eliminate \(y\) to achieve an equation in \(x\), \(\dfrac{\mathrm{d}x}{\mathrm{d}t}, \dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}\).
A1*: Achieves the printed answer with no errors.

Notes

(b)
M1: Uses the model to form and attempts to attempts to solve the auxiliary equation (Accept a correct equation followed by two values for \(m\) as an attempt to solve.)
A1: Correct roots of the AE
M1: Uses the model to form the complementary function. Must be in terms of \(t\) only (not \(x\))
A1: Correct CF
M1: Chooses the correct form of the PI according to the model and uses a complete method to find the PI
A1ft: Combines their CF (which need not be correct) with the correct PI to give \(x\) in terms of \(t\) so look for \(x\) = their CF + 10

Notes

(c)
B1ft: Correct differentiation of their \(x\). Follow through their \(\mathrm{e}^{\alpha t}(A\cos\beta t + B\sin\beta t)\)
M1: Uses the model and their answer to part (b) to find an expression for \(y\) in terms of \(t\)
A1: Correct equation. Mark the final answer but there is not need for terms to be gathered but must have \(y = \ldots\).

Alt for (c) and (d) restarting:
B1: Correct second order equation for \(y\) formed: \(\dfrac{\mathrm{d}^2y}{\mathrm{d}t^2} + 2\dfrac{\mathrm{d}y}{\mathrm{d}t} + 5y = 40\)
M1: Full method to obtain the general solution: they may recognise the similarity to the equation in \(x\) and jump straight to finding the PI, or may form the aux equation etc again, but look for an attempt that combines a (correctly formed) CF and a PI. For this mark allow if the constants used are the same as those for the equation in \(x\).
A1: Correct solution for \(y\) with different constants than those for \(x\), though allow recovery if they realise in (d) that they need different constants.

Notes

(d)
M1: Realises the need to use the initial conditions in the equation for \(x\)
M1: Realises the need to use the initial conditions in the equation for \(y\) to find both unknown constants - must have equations from which both unknowns can be found. Alternatively, a complete method using \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) to find the second constant is made.
A1: Deduces the correct equation for \(x\)
A1: Deduces the correct equation for \(y\). For this equation constants should have been gathered.

For (d) (Alt for (c) and (d) restarting):
M1: As main scheme, allow for using the initial conditions in one equation to make a start finding the constants.
M1: For a full method to obtain all four constants – if the same constant were used for both equations in (c) (inconsistently) then this mark cannot be scored. A full method here would, for instance, require finding \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) and using this along with the given initial equations and initial conditions to find the second constants for each equation.
A1: One correct equation with SC of being qualified by the first M only if a full method to find both constants for just one equation is made (so M1M0A1A0 is possible in this case).
A1: Both equations correct.

Notes

(e)
B1: Allow for any appropriate comment with valid supporting reason. They must have equations of the correct form from (d). The coefficients may be incorrect, but they must have positive limits for each of \(x\) and \(y\).
Both \(x\) and \(y\) should be considered (see below for exception), and a reason and some comment about the suitability of the model made (though you may allow implicit conclusions). E.g.

• for values of \(t \gt 8\), the amounts of compounds \(X\) and \(Y\) present settle at 10 and 8 without really varying, which supports the claim.
• \(\dfrac{\mathrm{d}x}{\mathrm{d}t} \approx 0\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t} \approx 0\) when \(t = 8\), so neither are changing, which supports the claim.
• As \(t\) gets large \(x\) and \(y\) tend to limits to 10 and 8 neither will be zero, hence the claim is not supported.
• \(x = 10.0\) (awrt) and \(y = 8.00\) (awrt) when \(t = 8\), since neither is zero it is likely the reaction is still continuing so the claim is not supported.

Exception: Allow a reason that states the model assumes that the reaction continues indefinitely, so the claim is not supported. (The reaction stopping would require a change in the model.)

Do NOT allow an answer that only considers \(x\) or \(y\). E.g. \(x = 10\) when \(t = 8\) so the model is not supported is B0 since there is no consideration that \(y\) may be zero and hence end the reaction.

Notes

M1: A suitable explanation for the “\(100 + t\)” e.g. as a minimum \((v) = 100 + 3t - 2t = 100 + t\)

M1: A suitable explanation for the \(\dfrac{2S}{100 + t}\)

There need to be some explanation (words) for this part of the formula.

e.g. the concentration of (salt) \(= \dfrac{S}{100 + t}\) therefore (salt) out \(= 2 \times \dfrac{S}{100 + t} = \dfrac{2S}{100 + t}\)

e.g. salt out \(= \dfrac{2S}{\text{volume of water}} = \dfrac{2S}{100 + t}\)

Note: M0 for \(2 \times \dfrac{S}{100 + t} = \dfrac{2S}{100 + t}\) only with no explanation

B1: Correct interpretation for the “3” e.g. salt in = 3 or \(\dfrac{\mathrm{d}S}{\mathrm{d}t}\) in \(= 3\)

Note: Salt water in = 3 is B0

A1*: Puts all the components together to form the given differential equation cso

Notes

M1: Uses the model to find the integrating factor and attempts the solution of the differential equation. Look for \(I.F. = \mathrm{e}^{\int \frac{2}{100 + t}\,\mathrm{d}t} \Rightarrow S \times \text{‘their } I.F.\text{’} = \displaystyle\int 3 \times \text{‘their } I.F.\text{’}\,\mathrm{d}t\)

A1: Correct solution condone missing \(+ c\)

For the next three mark there must be a constant of integration

M1: Interprets the initial conditions, \(t = 0\ \ S = 0\), and uses in their equation to find the constant of integration.

dM1: Dependent on having a constant of integration. Uses their solution to the problem to find the amount of salt after 10 minutes.

A1: Awrt 27 or \(\dfrac{3310}{121}\). (If the units are stated they must be correct)

Note: If achieves \(S(100 + t)^2 = 30\,000t + 300t^2 + t^3 + c\) the constant of integration \(c = 0\) and the correct amount of salt can be achieved. If there is no \(+ c\) the maximum they can score is M1A1M0M0A0

Notes

Note: Look out for setting \(S = 0.9\) in this part, which scores no marks.

M1: Uses their solution to the model and divides by \(100 + t\) as an interpretation of the concentration and sets \(= 0.9\).

Alternatively recognises that the amount of salt \(= 0.9(100 + t)\) and substitutes for \(S\) in their solution to the model.

dM1: Dependent on previous method mark. Solves their equation to obtain a value for \(t\). May use a calculator.

A1: Awrt 115 (If the units are stated they must be correct) or 1hr 45 mins with units

(The second line of the scheme is corrected from the printed mark scheme: it is printed with \((100 + 10)^2\) in the denominator, which should be \((100 + t)^2\).)

Notes

B1: Evaluates the model by making a suitable comment – see scheme for examples.

Typical solution

Initially

Typical solution

When \(t = 0\), \(y = 38 \Rightarrow c = -30\)

Typical solution

7 minutes 45 seconds

Typical solution

The model would not work for large values of \(t\) since the temperature of the freezer would not decrease indefinitely.

Typical solution

Typical solution

Integrating factor \(= \cosh x\)

When \(x = \ln 2\), \(\sinh x = \dfrac{3}{4}\) and \(\cosh x = \dfrac{5}{4}\)

Typical solution

Typical solution

Typical solution

(i)

Integrating factor \(= \mathrm{e}^{\int P\,\mathrm{d}x} = \dfrac{1}{x^2 - 9}\)

(ii)

When \(x = 5.2\)

\(y = 14.7434\) (6 sig. fig.)

Typical solution

The value from part (a) is equal to the answer to part (b)(ii) to four significant figures, meaning that the estimate in part (a) is very accurate.

Typical solution

So \(y = y_0\mathrm{e}^{-kt}\)

Typical solution

Typical solution

CF: \(y = A\mathrm{e}^{-0.174t}\)

PI: \(y = p + qt\)

General solution:

So \(A = 742\)

Typical solution

Births would occur at a particular time of year, not at a steady rate

Over a long period of time the population would increase indefinitely according to the model

Typical solution

Velocity after 71 seconds = 56.30 m s⁻¹

Typical solution

Typical solution

IF: \(\mathrm{e}^{\int 1\,\mathrm{d}t} = \mathrm{e}^t\)

When \(t = 0\), \(v = 0\) \(\therefore c = 1\)

Speed \(= \mathrm{e}^{2\pi} - \mathrm{e}^{-2\pi}\)

Typical solution

Integrating factor \(= \mathrm{e}^{\int P\,\mathrm{d}x} = x^2\)

or

Notes

(corrected from the printed mark scheme: the final A1 guidance prints \(xy = \ldots\); the left-hand side is \(x^2y\), as in the typical solution)

Typical solution

Typical solution

Integrating factor

To find

Notes

M1: Using \(F = 3a\) and \(a = \frac{\mathrm{d}v}{\mathrm{d}t}\) to form a differential equation - allow minor slips or sign errors but intention must be clear. Their \(F\) must be two terms only.

A1: The correct differential equation in the form \(\frac{\mathrm{d}v}{\mathrm{d}t} + P(t)v = Q(t)\) can imply this mark. \(P(t)\) and \(Q(t)\) do not need to be explicitly stated. ISW once correct form seen. If not written in the form, \(\frac{\mathrm{d}v}{\mathrm{d}t} + P(t)v = Q(t)\) then \(P(t)\) and \(Q(t)\) must be explicitly stated.

Notes

M1*: For \(\mathrm{I}(t) = \mathrm{e}^{\int P(t)\,\mathrm{d}t}\) for their \(P(t)\)

M1: For \(\mathrm{I}(t) = \mathrm{e}^{\pm k\ln(\cos 2t)}\) or \(\mathrm{e}^{\pm k\ln(\sec 2t)}\) or \(\pm k\cos 2t\) or \(\pm k\sec 2t\) or \(\pm a\cos^k 2t\) or \(\pm a\sec^k 2t\) for \(a, k \neq 0\).

M1dep*: For \(v \times \mathrm{I}(t) = k_1\displaystyle\int \sin 4t \times \mathrm{I}(t)\,\mathrm{d}t\) with their \(\mathrm{I}(t)\) (in any form) and \(k_1 \neq 0\).

M1: For simplifying \(RHS\) to \(k_2\displaystyle\int \sin 2t\,\mathrm{d}t\) for any \(k_2 \neq 0\) - dependent on all previous M marks.

A1: For correct general solution (any equivalent form). Condone lack of \(+c\).

M1: Using \(v(0) = 4.5\) to find constant term – dependent on first three M marks and an attempt at integration.

M1dep*: For a two-term equation of the form \(\cos 2t = k_3\) where \(|k_3| \lt 1\) and \(k_3 \neq 0\) - dependent on all previous M marks.

A1: Ignore if any other solution(s) found. Allow awrt 0.735 – for reference 0.7353144…
42.1 seconds (calculated in degrees) scores A0.

Notes

*M1: Rearranging to the form \(\frac{\mathrm{d}I}{\mathrm{d}t} + P(t)I = Q(t)\)

A1: (1st) Finding correct integrating factor as an expression not involving exponentials and logs. Ignore unnecessary constant multiplier here

*dep*M1: Multiplying both sides by IF and recognising LHS as exact derivative of \(\left(2t - t^2\right)^{-1}I\). Can be implied by next M1. Can be awarded from slip in initial rearrangement if their IF works for their rearrangement

dep*M1: Taking integral of both sides and attempt to complete square in order to express RHS in a standard form (or using the substitution \(u = t - 1\) including \(\mathrm{d}u = \mathrm{d}t\)), \(= \int\frac{1}{\sqrt{1 - u^2}}\,\mathrm{d}u\). Condone \(\pm 1 \pm (t - 1)^2\) for M1

*A1: For correctly integrating RHS to a function of \(t\). “\(+\,c\)” not necessary here.

dep*A1: AG so use of relevant condition must be explicit. Verification of AG by substitution is not sufficient; value of \(c\) must be derived. Ignore workings using other condition (\(t = 0\), \(I = 0\))

Notes

B1: (1st) If no other later comment or statement to the contrary then accept just \(t = 2\)

B1: (2nd) Do not accept incorrect explanation, e.g. \(-1 \leqslant \sin^{-1}(t - 1) \leqslant 1\). If B0B0 then Sc1 if length of surge determined to be awrt 1.96 s from \(\sin(-5\text{ rads}) + 1\) after \(t = 2\) found

Notes

*M1: Finding an expression for \(\frac{\mathrm{d}I}{\mathrm{d}t}\) in terms of \(t\) by either eliminating given \(I\) from given DE or differentiating given \(I(t)\) using product rule.
\(\left(I = \left(2t - t^2\right)\left(\sin^{-1}(t - 1) + 5\right) \Rightarrow\right) \dfrac{\mathrm{d}I}{\mathrm{d}t} = 2(1 - t)\left(\sin^{-1}(t - 1) + 5\right) + \dfrac{\left(2t - t^2\right)}{\sqrt{1 - (t - 1)^2}}\)
(The printed mark scheme also has a stray “\(\pm 1 \pm (t - 1)^2\)” after the expression for \(I\) in the first line of the answer; it has been left out.)

M1dep*: Simplifying their \(\frac{\mathrm{d}I}{\mathrm{d}t}\) so that there is no denominator which is zero at \(t = 0\)
\(\therefore \dfrac{\mathrm{d}I}{\mathrm{d}t} = 2(1 - t)\left(\sin^{-1}(t - 1) + 5\right) + \dfrac{\left(2t - t^2\right)}{\sqrt{2t - t^2}} = (2 - 2t)\left(\sin^{-1}(t - 1) + 5\right) + \left(2t - t^2\right)^{\frac{1}{2}}\)

A1: Must be in a simplified non-trigonometric form. Ignore units. If M1M0 or M0M0, then Sc B1 for correct answer

Notes

M1: Differentiate \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) and substitute \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\)

A1: AG convincingly shown after second substitution.

Notes

M1: Attempt to solve auxiliary equation for their DE

M1: Correct (or recovered) trial function for their CF.

A1: For GS

M1: Using \(t = 0\) and \(x = 0\) in their GS to find an equation in \(A\) and \(B\).

M1: Differentiating and using \(t = 0, x = 0, (y = 0), \dot{x} = k\) to find another equation in (\(A\) and) \(B\). Could also use \(\dot{x} + \dot{y} = k\) and \(x + y = kt\).

A1: N.B. \(\dfrac{3120}{3200} = \dfrac{39}{40}\)

Alternative method for first 4 marks

M1: Integrating \(\ddot{x} + 80\dot{x} = 78k\) wrt \(t\) and then using integrating factor

M1: Integrating wrt \(t\)

M1: Integrating by parts (\(78kt + c\) may be separated)

A1: GS of form \(x = A + B\mathrm{e}^{-80t} + \frac{39}{40}kt\)

Alternative method for this M mark

M1: Finding GS for \(y\) and using \(x = 0, y = 0\) to find another equation for \(A\) and \(B\)

Another alternative method for Q8(a)(ii) for first 4 marks

Mark scheme for Q8(a)(ii) if candidate solves for \(y\), rather than \(x\)

M1: Auxiliary equation for their DE

M1: Correct trial function for their CF
Note the GS for \(x\) in the main mark scheme achieves A1 but here A1 is not awarded until GS for \(x\) found.

A1: GS of form \(x = C + D\mathrm{e}^{-80t} + \frac{39}{40}kt\)

M1 M1: Using two of \(x = y = \dot{y} = 0\) to find simultaneous equations for \(A\) and \(B\) (M1 for each equation).

Alternative method (solving for \(y\))

M1: Integrating \(\ddot{y} + 80\dot{y} = 2k\) wrt \(t\) and then using integrating factor (corrected from the printed mark scheme, which reads \(\ddot{y} + 78\dot{y} = 2k\))

M1: Integrating wrt \(t\)

M1: Integrating by parts

Notes

B1: Use of \(t = 50\) must be seen to give a term in \(k\)
Must reference \(k \gt 6\).
Condone the idea of ignoring exponential term as negligible
Sight of \(292.5 \gt 250\) earns the mark

Notes

M1: Turning function into exponentials. Could see eg \(\mathrm{e}^{-0.0488t}\) and \(\mathrm{e}^{-81.95t}\). Condone error(s) in coefficients.
Allow argument such as \(\mathrm{e}^{-41t}\) dominates

A1: Argument must be complete and correct but could be based on sufficiently large values of \(t\) rather than formal limiting process.

SC B1 argument that \(20k = 240 \lt 250\) if 2^nd term is assumed to tend to 0

Notes

B1: The idea that regularly is not the same as continuously (and may not even mean “at a constant average rate”).
Other sensible answers possible, but must be in context.

Notes

M1: (first) Use of NII with \(m\) and \(a\) replaced and with 2 forces, the given force and \(kv\)
\(F = ma\) can be implicit here

M1: (second) Use of given conditions to derive an equation in \(k\)
Can be done first

A1: AG. Complete argument including \(F = ma\)

Notes

*B1: Or CF

*M1: Multiplying by IF and writing LHS as an exact derivative
Or subst correct PI into DE

A1: (first) “\(+\,c\)” required
Or GS \(v = A\mathrm{e}^{-t} - 2\mathrm{e}^{-2t}\)

dep*M1: Use of initial conditions to derive a value for \(c\)
Or using alternative boundary condition

Notes

M1: Deducing time when \(v\) is maximum
Or by finding expression for \(\dfrac{\mathrm{d}v}{\mathrm{d}t}\) and solving \(\dfrac{\mathrm{d}v}{\mathrm{d}t} = 0\)

Notes

M1: (first) Integrating to find expression for \(x\)

M1: (second) Using initial conditions to find value of (new) constant
Or definite integral with correct lower limit…

M1: (third) Recognising that the model is only valid when \(x\) lies between 0 and 0.9
…and upper limit

A1: Rejecting \(t = \ln\left(\dfrac{10}{10 + 3\sqrt{10}}\right) \lt 0\) (can be implicit)

Notes

M1: Use of Newton II with constants of proportionality. \(F = ma\) must be seen

Notes

M1*: Attempt to find IF

M1dep: Getting DE in correct form

M1dep: Attempt integration by parts

Alternative method

Notes

M1: Consider the behaviour of the exponential function in their equation from (b) soi

A1: or \(v \approx 2t\)

Notes

B1: AG. Intermediate step as shown or both sides halved

Notes

B1: Or \(\mathrm{e}^{\frac{1}{2}t + c}\) or \(A\mathrm{e}^{\frac{1}{2}t}\)

M1: Multiplying both sides by IF and writing new LHS as an exact derivative

M1: Use initial conditions to determine \(c\). Allow if using a solution of correct form with wrong (non-zero) coefficients

Alternate method

M1: Substitutes boundary condition into general solution to find \(A\). Allow if using a solution of correct form with wrong (non-zero) coefficients

Notes

B1: or awrt 0.368

Notes

SC1 \(v\mathrm{e}^{\frac{1}{2}t} = \frac{N}{2}\mathrm{e}^{\frac{1}{2}t} + c\) oe

M1: substitute their boundary condition from (c) into correct GS to find \(c\)

Or \(\dfrac{\mathrm{d}v}{\mathrm{d}t} + \dfrac{1}{2}v = \dfrac{1}{4}\) rearranged to \(\displaystyle\int \frac{2\,\mathrm{d}v}{1 - 2v} = \frac{1}{2}\int \mathrm{d}t = \frac{1}{2}t\) giving \(-\ln(1 - 2v) + c' = \dfrac{1}{2}t\); \(c' = 1 + \ln(1 - 2\mathrm{e}^{-1})\)

Notes

(a)(i)

B1: or any rearrangement. ISW.

(a)(ii)

B1: or converse. Could be in words. Do not condone \(h \to 20\).
Accept their expression for \(\frac{\mathrm{d}h}{\mathrm{d}t} = 0\) from (a)(i) provided it is correct or of the form \(k(20 - h)\).

(a)(iii)

M1*: correctly separating variables or using integrating factor for their DE, leading to an integral soi. If a CF/PI approach used award for a correct method leading to the CF and a value for the PI.

A1: or \(h\mathrm{e}^{0.2t} = 20\mathrm{e}^{0.2t}[+c]\) or \(h = B\mathrm{e}^{-0.2t} + 20\). Condone missing brackets if recovered.

M1dep: substituting initial conditions into their equation leading to a value for \(A\), \(B\) or \(c\)

A1: AG www, condone missing brackets earlier if recovered but do not condone \(\frac{\mathrm{d}h}{\mathrm{d}t}\) or \(\frac{\mathrm{d}}{\mathrm{d}x}\) used for \(\frac{\mathrm{d}}{\mathrm{d}t}\)

(a)(iv)

M1: substituting \(t = 5\) into given expression for \(h\)

A1: or \(12.64\ldots \neq 10\) soi but 10 oe must be seen, e.g. \(\frac{20}{2}\) or “half the maximum height”. Comment of inconsistency required (must not say that the value is consistent).

Notes

(b)(i)

M1: \(= 0\) must be seen unless implied by correct roots

M1: correct form of particular integral (any polynomial with a constant term) and substituting into DE correctly, soi by correct \(k\)

A1: must see \(h =\)

(b)(ii)

B1: must state that each exponential term (or their sum) tends to 0. Accept general term with negative exponent, e.g. \(\mathrm{e}^{-kt}\), but not \(\mathrm{e}^{-t}\). Condone \(h = 20\).

(b)(iii)

M1FT: substituting \(t = 0\), \(h = 0\) correctly into their formula for \(h\) from part (b)(i) to form an equation in \(A\) and \(B\); must be seen

M1FT: differentiate their \(h\) from part (b)(i) correctly, soi by correct second equation in \(A\) and \(B\)

M1FT: substituting \(t = 0\) and \(\frac{\mathrm{d}h}{\mathrm{d}t} = 2.9\) to form (a second) equation in \(A\) and \(B\); must be seen

A1: or \(10.017\ldots \times 2 = 20.03\ldots\) so does fit the data well. Accept “half the maximum height” oe for 10.

Notes

M1: division by \(x\) seen. Condone errors on RHS.

M1: integrating factor

A1: AG An intermediate step must be seen before the final answer. Condone errors on RHS. Requires M1M1.

Notes

M1: multiplying through by \(x^{-2}\) (allow FT from an incorrect rearrangement in (a) if clear) and writing as an exact derivative or setting up integral

A1: Ignore modulus signs if used

M1: substituting \(x = 1, y = 0\) in leading to a value for \(c\)

A1cao: oe

M1*: differentiating their \(y\), allow a slip

A1: correctly (do not FT), need not be simplified

M1dep: setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and meaningfully attempting to solve; soi by correct final answer

A1: www, do not condone \(\frac{\mathrm{d}y}{\mathrm{d}x}\) used for \(\frac{\mathrm{d}}{\mathrm{d}x}\). Must dismiss \(x = 0\) if previously given
Ignore \(x = -\mathrm{e}^{-\frac{3}{2}}\) if stated

Alternative method for last 4 marks

B1: could be stated anywhere or implied by next line

M1: setting up an equation in \(x\) only, cannot be implied

M1: re-arranging and attempting to solve; soi by correct final answer

A1: www, do not condone \(\frac{\mathrm{d}y}{\mathrm{d}x}\) used for \(\frac{\mathrm{d}}{\mathrm{d}x}\). Must dismiss \(x = 0\) if previously given
Ignore \(x = -\mathrm{e}^{-\frac{3}{2}}\) if stated

Notes

(a)(i)

B1: AG Volume of liquid (allow “volume”, “amount of liquid”, “capacity” or “liquid”) in the container must be given in first step. Do not allow “volume leaving”.

(a)(ii)

Second B1*: Not oe

B1dep: AG successful completion

Notes

M1: Finding integrating factor

A1: Allow \(A\frac{1}{200 - t}\) where \(A\) is clearly an unevaluated constant of integration and not from incorrect integration

M1: Multiplying equation by their IF, soi by next line

M1*: Either, ft their IF (i.e. their IF \(\times\ x\))

A1: First A1 must have been scored. Condone missing \(+c\) and bracketing errors.

B1FT: Correct \(c\) for their integral

M1dep: Combining their log terms correctly, do not condone bracketing errors without clear recovery

A1: AG www

Notes

(c)(i)

M1: May be embedded but must be seen

A1: Accept 690 or better, or \(1000\ln 2\)

(c)(ii)

M1: soi. Implied by \(x = 10(200 - t)\).

M1*: Or finding \(\frac{\mathrm{d}x}{\mathrm{d}t} = 10\left(1 - \ln\left(\frac{200}{200 - t}\right)\right)\) correctly

M1dep: Eliminating logarithms correctly

A1: 130 (2sf) or better. Do not accept exact value. Ignore units.

Notes

B1: Or “the salt does not dissolve/mix/spread/diffuse/disperse/distribute uniformly”. Must be explicit that modelling assumption is unrealistic.
Ignore subsequent comments
Do not accept reference to volume or exogenous factors, e.g. evaporation/temperature

Notes

(a)(i)

M1*: differentiate \(\frac{\mathrm{d}x}{\mathrm{d}t}\) wrt \(t\)

M1dep: (1st) substitution for \(\frac{\mathrm{d}y}{\mathrm{d}t}\)

M1dep: (2nd) substitution for \(y\)

A1: (1st) correct differential equation \(= 0\). Could see \(ab\) instead of \(n^2\).

M1: giving and solving their AE

A1: (2nd) AG www

Alternative method

M1*: differentiate \(y\) wrt \(t\)

M1dep: (1st) substitution for \(\frac{\mathrm{d}y}{\mathrm{d}t}\)

M1dep: (2nd) substitution for \(y\)

A1: (1st) correct differential equation \(= 0\). Could see \(ab\) instead of \(n^2\).

M1: giving and solving their AE

A1: (2nd) AG www

(a)(ii)

M1: differentiation of \(x\) must be seen

A1: may not be factorised but must be simplified

Notes

(b)(i)

M1: (1st) all three values established

(corrected from the printed mark scheme, which has \(n = \sqrt{0.004}\); \(n = \sqrt{ab} = \sqrt{0.04 \times 0.01} = \sqrt{0.0004} = 0.02\). The same correction applies in the alternative method.)

M1: (2nd) a method to find both A and B (ft their \(y\)) from expressions for \(x_0\) and \(y_0\)

A1: AG dependent upon both M1s

Alternative method

M1: (1st) all three values established

M1: (2nd) a method to find both A and B from two expressions for \(\frac{\mathrm{d}x}{\mathrm{d}t}\)

A1: AG dependent upon both M1s

(b)(ii)

B1: AG first step or equivalent substitution must be seen

Notes

(c)(i)

M1: for either
may be implied by awrt 365 or 303

A1: for both correct
allow \(y = 303\)

(c)(ii)

M1: (1st) equating \(x\) and \(y\) with \(x_0\), \(y_0\) substituted

M1: (2nd) collecting like terms

M1: (3rd) oe. Taking logs for a single term in \(t\).

(c)(iii)

M1: for correct conclusion with some explanation

A1: for complete explanation (e.g. \(x = 0\) at \(t = 59.94\ldots\))

B1: reason must be given, e.g. \(y = 0\) when \(t = 25\ln(-11)\) which is undefined or \(\mathrm{e}^{0.04t} = -11\) has no solution. Implied by M1A1 without further working seen unless incorrect conclusion for \(y\).

Notes

(d)(i)

B1: oe. Subscripts must be seen.

(d)(ii)

M1: oe; e.g. \(C\) is \(\frac{1}{2}(x_0 + 2y_0)\) or \(2y_0\) or \(x_0\)

A1: oe, e.g. both species disappear. Must be supported by explanation.

Notes

B1: \(\frac{\mathrm{d}y}{\mathrm{d}x} - 2y\tanh x = 1\)

M1: (1st) or \(\mathrm{e}^{\int -2\tanh x\,\mathrm{d}x}\). Integral must come from an attempt to get \(\frac{\mathrm{d}y}{\mathrm{d}x}\) on its own

M1: (2nd) or multiplying through by their IF

M1: (3rd) substituting \(x = 0\) and \(y = 1\) to lead to a value for \(c\) (corrected from the printed mark scheme, which says “substituting \(x = 1\) and \(y = 1\)”; the condition is \(y = 1\) when \(x = 0\))

A1: (final) oe, e.g. \(\cosh x\,(\sinh x + \cosh x)\) or \(\mathrm{e}^x\cosh x\)

Notes

M1: Integral must come from an attempt to get \(\frac{\mathrm{d}y}{\mathrm{d}x}\) on its own

M1: For integrating

M1: Multiplying both sides by their IF

M1: Rearranging into the form \(y =\), equation must come from an attempt at integration having used IF and include \(c\)

M1: Substituting in \(x = 0\), \(y = 1\) to lead to a value of \(c\)

Notes

A1: (a)(i) AG

A1: (a)(ii) AG

(Corrected from the printed mark scheme: in (a)(ii) the rates are printed in litres/hr; the question uses litres per minute, as typed above.)

Notes

(b)(i)

M1: separating variables, or IF \(\mathrm{e}^{at}\)
\(x\mathrm{e}^{at} = \mathrm{e}^{at} + c\)
\(c = -1\)

Notes

(c)(i)

B1: oe e.g. volume negative when \(t \gt \dfrac{1}{a}\)

(c)(ii)

M1: \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0\): or completing the square

Notes

(a)(ii)

B1: or by integration by separating variables

Notes

M1: attempt at partial fractions

M1: substituting values and/or equating coeffs; need to attempt 3 values of t (or alternative) for mark

M1 A1: \(\displaystyle\int\left(\frac{t}{1 + t^2}\right)\mathrm{d}t = k\ln(1 + t^2)\)

M1: combining lns

E1cao: NB AG

Notes

(c)(i)

E1: NB AG

(c)(ii)

M1: substituting \(P = 0.5A\) and squaring to solve for \(t\)

Notes

E1: NB AG

Notes

B1: Award for either P seen

B1: the other value of P and conclusion

Notes

B1: allow ‘proportional to’, ‘varies with’

Notes

B1: [by Newton’s 2nd Law]

Notes

(c)(i)

M1: must be correct

M1: substituting \(t = 10\), \(v = 0\)

A1cao: or \(54.4\mathrm{e}^{-0.1t} - 20\)

Alternative solution

M1: substituting \(t = 10\), \(v = 0\); M1: anti-logging

Alternative solution

M1: substituting \(t = 10\), \(v = 0\)

(corrected from the printed mark scheme: the printed line reads “When \(t = 0\). \(v = 10\)”; the condition used is \(t = 10\), \(v = 0\).)

(c)(ii)

B1: 13 or better

Notes

B1: [by Newton’s 2nd Law]

Notes

(e)(i)

M1: integrating by parts

M1: substituting \(t = 0\), \(v = 0\)

A1: NB AG

Alternative solution

M1: substituting \(t = 0\), \(v = 0\); A1cao: NB AG

(e)(ii)

M1: subst \(t = 5\) and equating to their \(v\) when \(t = 5\)

A1: 1.2 or better

Notes

M1: integrating \(v\) between 0, 5

B1: \(\left[\frac{1}{2}t^2 - 10t - 100\mathrm{e}^{-0.1t}\right]\)

A1: art 22.5 (soi)

M1: integrating their \(v\) between 5 and 10

A1: art 29.7 (soi)