Notes

M1: Forms the correct auxiliary equation, leading to finding a value for \(m\).
Allow the use of any variable instead of \(m\) for this mark.

A1: Correct complementary function, oe such as \(y = (A + Bx)\mathrm{e}^{2x}\)
Do not need \(y =\) here but must be in terms of \(x\), but may be recovered later.

M1: A complete method to find the particular integral. Uses the correct form \(y = \lambda\mathrm{e}^{3x}\), differentiates twice and substitutes correctly into the differential equation to find a value for \(\lambda\ (= 2)\). May use a different variable for \(x\) here.

dM1: Dependent on the previous method mark. Finds the general solution by adding the particular integral (of the correct form) to their complementary function (which may not be of the correct form). Do not need \(y =\) here. May use a different variable for \(x\) but their variable must be consistent.

A1: Correct general solution. Must have \(y =\) here and must be in terms of \(x\)

Notes

M1: For a general solution of the form \(A\mathrm{e}^{\alpha x} + Bx\mathrm{e}^{\alpha x} + \text{``their PI''}\), uses the initial conditions, \(x = 0,\ y = 5\) to find a value for a constant.
(The correct form will be \(A\mathrm{e}^{mx} + Bx\mathrm{e}^{mx} + \lambda\mathrm{e}^{3x}\))

dM1: Dependent on the previous method mark.
Differentiates the general solution to achieve an expression of the form \(mA\mathrm{e}^{mx} + B\mathrm{e}^{mx} + mBx\mathrm{e}^{mx} + 3\lambda\mathrm{e}^{3x}\) using their \(m\) from part (a) where \(m \neq 1\)
Then uses the initial conditions to find a value for the other constant.

A1: Correct particular solution. Must have \(y =\) here and must be in terms of \(x\)

Notes

Note: Accept the dot notation \(\dot{x}\) etc, as equivalents for the derivatives wrt \(t\) throughout.

M1: Differentiates the first equation and substitutes for \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) from the second equation.

M1: Proceeds to an equation in \(x\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) only by substituting for \(y\).

A1*: Achieves the printed answer with no errors.

Notes

M1: Uses the model to form and solve the Auxiliary Equation.

A1: Correct roots of the AE.

M1: Uses the model to form a Complementary Function appropriate for their roots. Accept complex index form.

A1: Correct General Solution. Accept complex index form. Must include the \(x\)

Notes

M1: Uses the model and their answer to part (b) to give \(y\) in terms of \(t\). Must involve an attempt at the product rule. Alternatively, they may repeat the whole process of parts (a) and (b) again on \(y\) – score for a full process leading to the solution for \(y\) in terms of \(t\) only.

A1: Correct simplified equation though may be with expanded brackets but like terms should be gathered (ie the \(A\cos t\) and \(B\sin t\) terms). If starting over, allow if names of constants are the same as for (b) (ie treat independently) but they will lose later marks. Accept complex index form.

Notes

M1: Realises the need to use the initial conditions to establish the values of their constants. But must have consistent constants (ie four different constants if they did part (c) from scratch). Solutions with complex indices are unlikely to achieve this mark. If unsure, use the Review system to consult your team leader.

dM1: Sets \(x = y\) and collects and reaches \(\tan t = \ldots\). Alternatively may convert into sine or cosine and reach an equivalent equation.

A1: Correct equation – accept equivalent if other trig approaches are used. Must come from correct work. Note \(\sin t = \dfrac{7}{\sqrt{65}}\) and \(\cos t = \dfrac{4}{\sqrt{65}}\) are the other ratios.

M1: Solves their \(\tan t = \ldots\) and multiplies by 24. Must be working in radians. If degrees mode is used score M0.

A1: Correct value. Ignore units.

Notes

B1: Must have at least one correct general equation for \(x\) or \(y\) for this mark to be awarded. Rote answers about generic models without evidence to support any claims should not be given credit.
Suggests a suitable correct limitation of the model. This must focus on the fact the equations produce negative values. Award if a correct answer is given and ignore extra comments.
E.g. “gives negative values” would be a minimal acceptable answer.

Note: Answers about unlimited growth that imply the populations increase indefinitely or about scarcity of resource without an accompanying correct statement are B0.

Notes

M1: Attempts to solve \(2m^2 + 5m + 2 = 0\), usual rules apply for solving a quadratic equation.

A1: Correct CF. Do not need “\(x =\)” here, but must be in terms of \(t\)

B1: Correct form for the PI i.e. \(x = pt + q\) or \(x = at^2 + bt + c\)

M1: Differentiates their PI (of the forms \(x = pt + q\) or \(x = at^2 + bt + c\)) twice and substitutes into the given differential equation finding values for their constants to obtain a PI of the form \(pt + q\), \(p, q \neq 0\)

A1: Correct PI

A1ft: Correct GS or correct ft GS, which is the sum of their CF and PI. This is dependent on achieving both previous M marks. Must have \(x =\) and their GS must be in terms of t.

Notes

M1: Substitutes \(x = 3\) when \(t = 0\) into their GS to establish an equation in \(A\) and \(B\), allow minor slips if the intention is clear.

M1: Differentiates their answer to part (a) which must be in terms of \(t\) only, and sets \(= -2\) with \(t = 0\) to establish another equation in \(A\) and \(B\) and solves simultaneously to find \(A\) and \(B\).
Do not be concerned about how their simultaneous equations are solved; award this mark if they then go on to write values for \(A\) and \(B\).
Functions which require the use of product rule or trigonometric functions must be differentiated appropriately.

A1: Correct PS. Need “\(x =\)” here, their answer must be in terms of \(t\) and no other variable.

Notes

(c) Mark (i) and (ii) together but only award for work done in (c)

(i)
M1: Differentiates their Particular Solution of the form \(x = a\mathrm{e}^{-kt} + bt + c\) where c may be 0 to obtain an expression of the form \(C\mathrm{e}^{-kt} + D\)

dM1: Solves an equation of the form \(C\mathrm{e}^{-kt} + D = 0,\ \ C \times D \lt 0\), to obtain \(t = -\dfrac{1}{k}\ln\left(\dfrac{-D}{C}\right)\)

A1*: Substitutes \(t = \dfrac{1}{2}\ln 2\) oe to obtain the printed answer with no errors.
Condone going from \(2\mathrm{e}^{-2\left(\frac{1}{2}\ln 2\right)} + 2\left(\dfrac{1}{2}\ln 2\right) + 1 = 2 + \ln 2\)

(ii)
B1ft: Obtains a second derivative of the form \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = \lambda\mathrm{e}^{-\mu t},\ \ \lambda, \mu \gt 0\) and makes a conclusion e.g.

• \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} \gt 0\) (for all values of \(t\)) hence minimum.
• or substitutes their value of \(t\) (even if incorrect) and states \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} \gt 0\) hence minimum

Notes

B1ft: Dependent on having obtained a Particular Solution of the form \(\mathrm{f}(t) + bt + c\) where \(\mathrm{f}(t)\) only has terms in \(a\mathrm{e}^{-kt}\) where \(k \gt 0\) and \(a, b \neq 0\)
This mark is awarded for the candidate demonstrating that in the model for “large values” or “as \(t \to \infty\)”, the value of their \(\mathrm{e}^{-kt} \to 0\), so there is constant speed and states that the model is suitable, or equivalent statement. (See scheme for examples)

Notes

Alternative

Notes

B1: Differentiates the second equation with respect to \(t\) correctly. May have rearranged to make \(x\) the subject first. The dot notation for derivatives may be used.

M1: Uses the second equation to eliminate \(x\) to achieve an equation in \(y\), \(\dfrac{\mathrm{d}y}{\mathrm{d}t}, \dfrac{\mathrm{d}^2y}{\mathrm{d}t^2}\).

A1*: Achieves the printed answer with no errors, allow dot notation

Alternative

B1: Differentiates the second equation with respect to \(t\) correctly. May have rearranged to make \(x\) the subject first. The dot notation for derivatives may be used.

M1: Substitutes in the printed differential equation uses both equations to remove all derivative, form an expression involving \(x\)’s and \(y\)’s which simplifies to a constant.

A1*: Achieves 2560 with no errors seen

Notes

M1: Uses the model to form and attempt to solve the auxiliary equation the PI (Accept a correct equation followed by two values for \(m\) as an attempt to solve.)

M1: Forms the complementary function correct for their roots (so if repeated or complex roots found, award for appropriate form for CF). 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 \(y\) in terms of \(t\) so look for \(y\) = their CF + \(\dfrac{2560}{13}\), accepting awrt 197

Notes

M1: Uses the initial conditions of the model to set up an equation in \(A\) and \(B\) from their general solution.

B1: Uses the initial conditions of the model to find the value of \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) when \(t = 0\). This can be implied.
Alternatively uses \(x = \dfrac{1}{0.032}\left(\dfrac{\mathrm{d}y}{\mathrm{d}t} + 0.025y\right)\) to find an equation in \(A\) and \(B\) from

M1: Differentiates their general solution, substitutes \(t = 0\) and sets equal to their initial value of \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) to form another equation in \(A\) and \(B\) and proceed at least as far as finding \(A\) in terms of \(B\) oe.
Alternative substitutes \(y\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) into their \(x\) equation and uses \(x = 0\) when \(t = 0\) to find an equation in \(A\) and \(B\)

A1: Correct particular solution, accepting awrt 197

Notes

M1: Uses the limit of the exponential terms is zero to find the long term limit of the concentration

A1ft: Follow through on their constant term and draws a relevant conclusion.

Notes

Note: mark (a) as a whole

M1: Differentiates the given PI twice using the product rule to achieve the required form.
Alternatively, uses a correct form for the PI and differentiates twice using the product rule to achieve the required form. A correct form may involve other terms with coefficients that will be zero, e.g. \(\theta = \lambda t\sin 3t + \mu t\cos 3t\) is fine. Also allow e.g \(\theta = \lambda t\sin\omega t\)

A1: Correct derivatives.

dM1: Depends on first M, substitutes into the given differential equation and attempts to simplify. In the Alt they must go on to find value for \(\lambda\).

A1*: Achieves \(\dfrac{1}{2}\cos 3t\) and makes a minimal conclusion (e.g //). Alternatively reaches the correct PI.

Alternative

Notes

M1: Uses the model to form and solve the auxiliary equation. Accept \(m^2 + 9 = 0 \to m = \pm 3\mathrm{i}\) or \(\pm 3\)

A1: Correct complementary function. Must be in terms of \(t\) but allow recovery if initially in terms of \(x\) but changed later.

dM1: Dependent on the previous method mark. Finds the general solution by adding the particular integral to the complementary function.

A1: Correct general solution including "\(\theta =\)", which may be recovered in part (b).

Notes

M1: Uses the initial conditions of the model, \(t = 0,\ \theta = \dfrac{\pi}{3}\) to find a value for a constant.

M1: Differentiates the general solution and uses the initial conditions of the model \(t = 0,\ \dfrac{\mathrm{d}\theta}{\mathrm{d}t} = 0\) to find a value for the other constant.

ddM1: Dependent on both previous method marks. Substitutes \(t = 10\) into their particular solution. If not substitution is seen, accept any value as the attempt as long as they have found all relevant constants.

A1: Accept awrt \(\pm 0.662\)

Notes

B1ft: Makes a quantitative comparison of the size of their answer to part (b) with 0.62 and makes conclusion (e.g. good model). Follow through on their answer to (b) and draws an appropriate conclusion about the model. Accept “not reasonable” as long as it is supported with evidence but there must be some instructive comparison and a conclusion about the model - not just stating how much it is out. The reason given must be correct.

Accept e.g. a correct percentage error with reasonable conclusion, or statement approximately equal with conclusion.

Do not accept e.g. “does not agree to 1 s.f.” or “out by 0.6” as these lacks context. Do not accept arguments based solely on a difference in sign, they must be referring to the relative size of angle.

Notes

B1: Refines the model, accept any constant on the right hand side.

Notes

M1: Substitutes \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = 13.6\), \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0\) and \(x = -20\) into the differential equation to find a value for \(k\). Allow if there are sign slips but must be attempting the values in the correct places.

A1: Correct value \(k = 5\)

Notes

M1: Forms and solves the auxiliary equation.

A1: Correct solution to the auxiliary equation (not follow through).

A1ft: Correct complementary function for their solutions to their auxiliary equation. (Follow through on distinct real, repeated or complex roots.)

M1: Uses the information from the model \(t = 0\ x = -20\) to find a constant or equation linking two constants in their equation.

M1: Differentiates an expression of the form \(\mathrm{e}^{kt}(A\cos\lambda_1 t + B\sin\lambda_2 t)\) using the product rule to find an expression for the velocity.

dM1: Uses the information from the model, \(t = 0\ \dfrac{\mathrm{d}x}{\mathrm{d}t} = 0\) to find and solve another equation for the constants.

A1: Correct equation for displacement.

Notes

M1: Finds the height above the river by finding the displacement after 15 seconds and adding 30

A1: Vertical height = awrt 29.3 m

Notes

B1: Any suitable comment relating to the given model or the outcomes of it. See scheme for examples. Do not accept just “air resistance has not been considered” as the question does not say this was ignored. However, if a valid consequence of what including air resistance would mean to the model, then the mark may be awarded.

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

(a)
M1: Uses the model to form and solve the auxiliary equation
A1: Correct CF, does not need \(x =\)
B1: Correct PI
A1ft: Deduces the correct GS (follow through their CF + PI). Must have \(x = \mathrm{f}(t)\) and PI not 0

Notes

(b)
M1: Uses the model and the initial conditions to establish the value of “\(A\)”
M1: Differentiates their model using the product rule and uses the initial conditions to establish the value of “\(B\)”. Must be using \(x = 0\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 10\)
A1: Correct particular solution. This can be implied by the correct constants found following a correct answer to part (a).
M1: Uses their solution to the model with a correct strategy to obtain the required value of \(t\) e.g. differentiates, sets equal to zero and solves for \(t\)
M1: Uses a correct trigonometric approach that leads to a value for \(t\)
A1: Correct value for \(t\)
M1: Uses the model and their value for \(t\) to find the maximum concentration.
A1: Correct value

Notes

(c)
M1: Uses the model to find the concentration when \(t = 10\)
A1ft: Makes a suitable comment that is consistent with their calculated value
Special case: If the candidate’s maximum value is less than 5 then
M1: never reaches 5 as maximum is…. or max is less than 5
A1: yes, it is safe

Notes

B1: Differentiates the first equation with respect to \(t\) correctly.

M1: Substitutes \(\dfrac{\mathrm{d}s}{\mathrm{d}t}\) into their derivative.

A1*: Achieves the printed answer with no errors.

Notes

Note: All the mark except the final A1 are available if they use other variables.

M1: Uses the model to form and solve the Auxiliary Equation.

A1: Correct roots of the AE.

M1: Uses the model to form the Complementary Function for their roots (they may be complex roots)

A1: Correct CF

M1: Chooses the correct form of the PI according to the model and uses a complete method to find the PI. Uses \(w = k\mathrm{e}^{-t}\) finds both \(\dfrac{\mathrm{d}w}{\mathrm{d}t}\) and \(\dfrac{\mathrm{d}^2w}{\mathrm{d}t^2}\) substitutes into the differential equation and find the value of \(k\).

A1ft: Dependent on all three of the previous method marks. Following through on their CF only to give \(w = \text{‘their CF’} + 50\mathrm{e}^{-t}\)

Notes

M1: Substitutes into the first equation the answer for part (b) in place of \(w\) and the derivative of their (b) in place of \(\dfrac{\mathrm{d}w}{\mathrm{d}t}\). If they rearrange to make \(S\) the subject first and make a slip but still substitutes for \(w\) and \(\dfrac{\mathrm{d}w}{\mathrm{d}t}\) allow this mark.

A1: Correct simplified equation.

Notes

M1: Uses the initial conditions \(t = 0,\ w = 65\) and \(s = 85\) to form simulations equations and solves to find the values of their constants

dM1: Dependent on the previous method mark. Sets \(w = 0\)

A1: Processes the indices correctly to obtain a 3-term quadratic equation in terms of \(\mathrm{e}^{1.5t}\). It does not need to all be on one side and condone missing \(= 0\).

M1: Solves their three-term quadratic (3TQ) to reach \(\mathrm{e}^{pt} = q\)

M1: Correct use of logarithms to reach \(pt = \ln q\) where \(q \gt 0\) and rejects the other solution

A1: awrt 1.165

Note: the final 3 marks only can be implied by a correct answer following the correct 3-term quadratic equation in terms of \(\mathrm{e}^{1.5t}\)

Notes

B1: Suggests a suitable limitation of the model, not valid when negative population

Any mention of other factors such as does not take into account e.g. other predictors, fishing, disease, lack of food etc is B0

Notes

M1: Uses the model to form and solve the auxiliary equation \(4m^2 + 4m + 37 = 0\)

See General Guidance for awarding this mark. This can be implied by correct values for \(m\) (from calculator)

A1: Correct general solution including “\(h =\)”

Notes

M1: Uses the model and the initial conditions to establish the value of “\(A\)”. Need to see \(t = 0\) and \(h = \pm 20\) leading to a value for “\(A\)”. This may be implied by \(A = -20\) or \(A = 20\).

M1: Differentiates their model using the product rule and uses the initial conditions, \(t = 0\) with \(\mathrm{d}h/\mathrm{d}t = \pm 55\), to establish the value of “\(B\)”

A1: Correct particular solution or correct values for \(A\) and \(B\)

M1: Uses their solution to the model with a correct strategy to obtain a value for \(t\) e.g. differentiates or uses their derivative from earlier, sets equal to zero and solves for \(t\)

A1(M1 on ePEN): Correct equation for \(t\)

A1: Correct value for \(t\) (allow awrt 0.778 if necessary) but this value may be implied.

dM1: Uses the model and their positive value for \(t\) to find the maximum displacement - if their \(t\) is incorrect there must be some indication that they are using their \(h\) and not just a number written down. E.g. must see substitution into their \(h\) or they re-state their \(h\) and obtain a value for \(h\).

Dependent on all the previous method marks

A1: Correct value (awrt 16.7 (units not needed))

Notes

M1: Considers the model for large values of \(t\) either by substituting values or by considering the expression and commenting on its behaviour for large values of \(t\). E.g. as \(t \to \infty,\ h \to 0\) or as \(t \to \infty,\ e^{-0.5t} \to 0\) or as \(t \to \infty\) the oscillations become smaller etc.

A1: Makes a suitable comment – see scheme for examples

Typical solution

In equilibrium position

After release

Typical solution

Typical solution

Complementary Function

Particular Integral

General Solution

\(x = 0\)

Typical solution

Typical solution

(i)

(ii)

Typical solution

The resistance force in a thin tube might not be the same as in a large bath.

Typical solution

45 m s⁻¹

Typical solution

(i)

CF:

CF:

PI:

So

(ii)

When \(t = 0\), \(x = 0\)

When \(t = 0\), \(\dot{x} = 14\)

To the nearest integer, \(A = -38\) and \(B = 16\)

Typical solution

Setting \(R = 0\)

When \(t = 0\), \(x = 0 \Rightarrow A = -38.3\)

When \(t = 0\), \(\dot{x} = 14\)

Typical solution

Typical solution

\(\mathrm{f}(t) = 7\mathrm{e}^{-t} + 2\mathrm{e}^{-2t}\)

Typical solution

Typical solution

Sub into (2)

CF: \(5m^2 - 2m + 0.65 = 0\)

PI: \(x = 195\)

When \(t = 0\), \(x = 1755\) and \(y = 30\) so

Typical solution

Using calculator:

When \(t = 5.38\), \(x = 0\)

At this time \(y \approx 108\) so there are still birds of prey

The rabbits die out first.

So the conservationists’ plan succeeds.

Typical solution

In equilibrium position

After release

as required

Typical solution

(i)

Critical Damping so:

(ii)

(iii)

At Max \(\dot{x}\)

Max Speed = 0.5 m s⁻¹

Typical solution

Typical solution

Tension in \(AP = 24m(0.5) = 12m\)
Tension in \(BP = 10m(1.2) = 12m\)
So tensions are equal

Typical solution

Typical solution

But \(v = -\dot{x}\)

So

CF:

PI: \(x = 0.196\)

General Solution:

Typical solution

Of form \(\ddot{x} = -\omega^2x\), therefore SHM

Typical solution

Typical solution

Continuing Charlotte’s method gives

This would make \(10\mathrm{e}^{-2x}\) equal to zero, which is impossible, so the method fails.

Typical solution

Charlotte needs to find the complementary function first.

CF: \(y = A\mathrm{e}^x + B\mathrm{e}^{-2x}\)

The RHS has a similar form to the CF and so we need to introduce a factor of \(x\) into the particular integral.

PI: \(y = \lambda x\mathrm{e}^{-2x}\)

General solution:

Typical solution

Typical solution

Flow into A = flow out of A (to keep volume of water constant)

Typical solution

\(x = 50\dot{y} + y\) and

Sub in (1):

CF: \(50m^2 + 2m + 0.015 = 0\)

PI: \(y = \dfrac{36}{0.015} = 2400\)

\(x = 50\dot{y} + y\) so

When \(t = 0\), \(x = 0\) and \(y = 0\) so

\(A + B + 2400 = 0\) and

\(-0.5A + 0.5B + 2400 = 0\)

\(\Longrightarrow A = 1200\) and \(B = -3600\)

Alternative

\(y = 200\dot{x} + 4x - 7200\) and

Sub in (2):

CF: \(200m^2 + 8m + 0.06 = 0\)

PI: \(x = \dfrac{144}{0.06} = 2400\)

When \(t = 0\), \(x = 0\) and \(y = 0\) so

\(A + B + 2400 = 0\) and

\(-2A + 2B + 2400 = 0\)

\(\Longrightarrow A = -600\) and \(B = -1800\)

(corrected from the printed mark scheme: the “Sub in (2)” line is printed as \(200\ddot{x} + 4\dot{x} = 0.02(200\dot{x} + 4x - 7200)\), and the PI as \(x = \dfrac{144}{0.006} = 2400\).)

Typical solution

CF:

PI:

General Solution:

Typical solution

Critical damping, because the Auxiliary Equation has equal roots

Notes

B1: States \(\sin\theta \approx \theta\) (or in words), and this is only valid for small angles (oe e.g. ‘close to zero’)

Notes

B1: Correct general solution. Allow exact equivalents e.g. \(\theta = A\cos\left(\sqrt{1.4}t\right) + B\sin\left(\sqrt{1.4}t\right)\), \(\theta = A\cos\left(\sqrt{1.4}t\right)\).
Allow 1.18 or better (1.183215…) for \(\sqrt{1.4}\)

M1*: Use correct initial conditions \(\theta = \frac{1}{15}\pi, \dot{\theta} = 0\) when \(t = 0\) to find constant(s) from a GS of the correct form. Allow \(\theta_0\) for \(\frac{1}{15}\pi\), and allow stating \(B = 0\) without corresponding working. If differentiation seen, then must see \(\sin \to \pm\cos\) and \(\cos \to \pm\sin\) together with a change in coefficients.

A1: cao www – allow \(\theta = \theta_0\cos\left(\frac{\sqrt{35}}{5}t\right)\) provided \(\theta_0 = \frac{1}{15}\pi\) stated explicitly in their working. The correct particular solution (with no incorrect working seen) scores the first three marks. Allow \(\theta = 0.209\cos(1.18t)\) or better.

M1*: Correct method to find \(\theta\) after one complete oscillation or for \(\theta\) when \(t = 6\). Follow through their PS which must be of the form \(\theta = \frac{1}{15}\pi\cos(pt)\) where \(p \gt 0\) and \(p \neq 1\). For reference if correct then \(\theta = 0.1434\ldots\)

M1dep*: For their numerical \(7\left(\frac{4}{15}\pi + \frac{1}{15}\pi - \theta\right)\) or \(7\left(\frac{4}{15}\pi + \theta\right)\) only

A1: awrt 6.33 (6.32603…)

Notes

M1: For consideration of correct discriminant (allow any inequality or equals)

A1: cao – oe but must be exact

Notes

M1: Differentiating given CF twice using product rule. Allow sign errors in either derivative (but coefficients must be correct).
Condone use of e.g. \(x\) instead of \(t\) for this mark but not a mixture.

A1: Cancelling or combining terms need not be seen, but all must be present before \(= 0\).
Condone arguments leading to e.g. \(0 = 0\) provided all terms correct.

Notes

B1: Correct form for PI. Allow more general forms e.g. \(\alpha t^2\mathrm{e}^{-3t} + \cdots\) if subsequent coefficients \(\beta, \gamma\) etc. later found to be 0.
Condone use of e.g. \(x\) instead of \(t\) for this mark and the first 3 M marks (but no further).

M1*: Attempt to differentiate their \(y\) twice (finding \(y'\) and \(y''\)), using the product rule at least once.
Allow this mark if their PI is of the form \(\alpha p(t)\mathrm{e}^{-3t}\) where \(p(t)\) is any polynomial in \(t\) of degree \(\geq 1\).

M1dep*: Correctly differentiating their \(y\) to find \(y'\) and \(y''\). Condone sign errors.
Condone missing 3, extra 3 or 1/3 coming down but not \(t\) coming down and/or \(3t - 1\) as exponent.

M1: Substituting their \(y'\) and \(y''\) into the DE (with RHS seen or \(\mathrm{e}^{-3t}\) cancelled from both sides). Allow this mark following M0M0.
This mark may be implied by the next A1 (i.e. a correct GS).

A1: Full form of GS can be implied by later work.

M1: Substituting \(y = 0\) and \(t = 5\) into their GS and cancelling any exponential part properly.
May be implied by later working including \(\mathrm{e}^{-3t}\) if fully correct.

M1: Correctly interpreting and using the 2^nd condition which, coupled with the 1^st, must mean that the quadratic is a perfect square. May see \(3A + 14B = -325\) oe.
Or \(B + 10t = 0\) when \(t = 5\) (at what must be a minimum point). Or \(5\left(t^2 + \frac{Bt}{5}\right) + A = 5\left(t + \frac{B}{10}\right)^2 + A - \frac{B^2}{20} \Rightarrow A - \frac{B^2}{20} = 0\) since \(y = 0\).

A1: Must be \(y =\) or \(y(t) = \ldots\) May see \(5(t^2 - 10t + 25)\) or \(5(t - 5)^2\)
Allow SCB1 for \(A = 125\) and \(B = -50\) following M0, provided all 6 of the first marks are gained (7/8).

Notes

B1FT: FT their PS of the form \(p(t)\mathrm{e}^{kt}\) where \(p(t)\) is any polynomial of non-zero degree.
Must be a value (i.e. not an expression with \(A, B\), etc.) and must be \(\geq 0\).

Notes

M1: Finding the value of their \(y\) when \(t = 1\). FT their PS of the form \(p(t)\mathrm{e}^{kt}\) where \(p(t)\) is any polynomial.
Must be a value (i.e. not an expression with \(A, B\), etc.)

A1FT: Sensible conclusion in context with correct comparison to 4 for their value (A0 for a value \(\lt 0\)). (i.e. \(\lt 4 \Rightarrow\) operating properly, \(\gt 4 \Rightarrow\) not operating properly). Condone ‘it is operating properly’.
Accept e.g.: ‘So it can be inferred that the cell is probably working well although we can’t be sure because it is close to 4 and we don’t know how well the model works’

Notes

B1: Properly reasoned argument. Could be related to \(C - P \lt 0\). ie
• some conditional
• consequence in terms of movement of children
• consequence in terms of sign of derivative
Ignore reference to \(\mathrm{d}P/\mathrm{d}t\).

Notes

M1: Correctly differentiating the \(\mathrm{d}P/\mathrm{d}t\) equation and then substituting from the \(\mathrm{d}C/\mathrm{d}t\) equation. Condone sign/coefficient errors when substituting.

A1: AG. Substitution for \(C\) (or eg elimination of \(C - P\)) must be seen.
Must reach AG.
Could see eg \(P'\) or \(\dot{P}\) etc.
Allow in terms of \(\alpha\) and \(\gamma\) if values of \(\alpha\) and \(\gamma\) quoted at end:

\[\begin{aligned} \dfrac{\mathrm{d}P}{\mathrm{d}t} &= \alpha(P - C) + \gamma t \\ \therefore \dfrac{\mathrm{d}^2P}{\mathrm{d}t^2} &= \alpha\left(\dfrac{\mathrm{d}P}{\mathrm{d}t} - \dfrac{\mathrm{d}C}{\mathrm{d}t}\right) + \gamma \\ \therefore \dfrac{\mathrm{d}^2P}{\mathrm{d}t^2} &= \alpha\dfrac{\mathrm{d}P}{\mathrm{d}t} - \alpha\dfrac{\mathrm{d}C}{\mathrm{d}t} + \gamma \\ &= \alpha\dfrac{\mathrm{d}P}{\mathrm{d}t} - \alpha(\alpha C - \alpha P) + \gamma \\ \dfrac{\mathrm{d}P}{\mathrm{d}t} &= \alpha(P - C) + \gamma t \\ \therefore \alpha C &= \alpha P + \gamma t - \dfrac{\mathrm{d}P}{\mathrm{d}t} \\ \therefore \dfrac{\mathrm{d}^2P}{\mathrm{d}t^2} &= \alpha\dfrac{\mathrm{d}P}{\mathrm{d}t} - \alpha^2C + \alpha^2P + \gamma \\ &= \alpha\dfrac{\mathrm{d}P}{\mathrm{d}t} - \alpha\left(\alpha P + \gamma t - \dfrac{\mathrm{d}P}{\mathrm{d}t}\right) + \alpha^2P + \gamma \\ &= \alpha\dfrac{\mathrm{d}P}{\mathrm{d}t} - \alpha\gamma t + \alpha\dfrac{\mathrm{d}P}{\mathrm{d}t} + \gamma \\ \therefore \dfrac{\mathrm{d}^2P}{\mathrm{d}t^2} - 2\alpha\dfrac{\mathrm{d}P}{\mathrm{d}t} &= -\alpha\gamma t + \gamma \end{aligned}\]

Notes

(c)(i)

B1: Condone missing “\(P =\)” and/or \(x\) instead of \(t\) / \(y\) instead of \(P\) here. Condone \(A\mathrm{e}^{0t}\)
\(m^2 - 2\alpha m = 0\) so \(m = 0\) or \(m = 2\alpha\) so \((P =)\ A + B\mathrm{e}^{2\alpha t}\)

(c)(ii)

B1: or other correct reason eg this trial function doesn’t produce a term in \(t\) to compare with \(64t\).
If considering the CF, must address the fact that there is a constant term in the CF and so there cannot be a constant term in the trial function.
“The complementary function already has a constant in it”.
“Part of the trial function is included in the complementary function”.
Not “There is no \(P\) term in the differential equation”.
Not “It matches the form of the complementary function”.

(c)(iii)

M1: Correctly differentiating to find the first and second derivatives and substituting into the LHS of the DE in (b).

A1: Finding \(a\) and \(b\) correctly.

\[\begin{aligned} &\therefore 2a - 2\alpha(2at + b) = -\alpha\gamma t + \gamma \\ &\therefore -4a\alpha = -\alpha\gamma \Rightarrow a = \frac{\gamma}{4} \\ &2a - 2\alpha b = \gamma \Rightarrow 2\alpha b = \frac{\gamma}{2} - \gamma = -\frac{\gamma}{2} \\ &\therefore b = \frac{-\gamma}{4\alpha} \end{aligned}\]

A1FT: FT their CF + their numerical PI.
Must be “\(P =\)”. Must be \(P(t)\).
\(P = A + B\mathrm{e}^{2\alpha t} + \dfrac{\gamma}{4}t^2 - \dfrac{\gamma}{4\alpha}t\)

Notes

M1: Using first initial condition.

M1: Correctly differentiating their \(P\) to find \(P'\) (can be implied by next line)

M1: Substituting \(t = 0\) correctly in \(P'\) to obtain the second initial condition and using the latter to solve for \(A\) and \(B\).

A1: Must be from correct \(P\).

Notes

B1: or other reasonable limitation of the model in the long-term which could be environmental. eg the room has a finite size; the centre will close.
\(P \sim 8t^2\) as \(t\) gets large which is not realistic.
Not just “Children are being added constantly”.
Not “The number of children grows exponentially”.

Notes

(a)(i)

B1: Convincingly shown

(corrected from the printed mark scheme: the printed answer says “sign of \(\frac{\mathrm{d}y}{\mathrm{d}t}\)”; the term in the equation is \(\frac{\mathrm{d}x}{\mathrm{d}t}\))

(a)(ii)

M1: Substitutes \(t = 0\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \pm 5\)

A1: Or 6 towards \(O\)

Notes

(b)(i)

B1: First two terms of Maclaurin’s formula, possibly in generalised form, must be seen
AG Convincingly shown.

(b)(ii)

B1: AG Convincingly shown

(b)(iii)

M1: For \(\frac{\mathrm{d}}{\mathrm{d}t}(\cosh t) = \sinh t\) and product rule attempted

A1: Fully correct differentiation

M1: Substitute \(t = 0\)

A1: AG. Allow embedded answer.

Notes

(c)(i)

B1: Substitutes \(t = 0.25\) to obtain an approximation and correct conclusion

(c)(ii)

B1: Allow any correct explanation that explains/implies that for \(t \gt 1\) some of the higher power terms are large and so non-negligible.
\(\mathrm{f}(10) = 156\) is too large is not enough

Notes

B1: Correct roots of auxiliary equation.

B1: ft complex \(k\) only.
Or \(y = R\mathrm{e}^x\cos(2x + \varphi)\) or \(y = R\mathrm{e}^x\sin(2x + \varphi)\)
Or \(y = A\mathrm{e}^{(2\mathrm{i} + 1)x} + B\mathrm{e}^{(-2\mathrm{i} + 1)x}\)
Final equation must be \(y = \mathrm{f}(x)\)

Notes

B1: Allow any extraneous terms in the trial function (eg. \(dx^3\)) as long as \(d\) shown to be zero, \(a, b, c \neq 0\)

M1: Differentiates their trial function to find \(\frac{\mathrm{d}y}{\mathrm{d}x}\) and \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) and substitutes

A1ft: ft their particular integral, and their CF from (a) (dependent on CF containing exactly two arbitrary constants)

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

B1: Correct form with 2 arbitrary constants. Must be “\(x =\)”.
Do not ISW; consider final answer as GS unless explicitly labelled otherwise.
Candidates may derive GS from, eg, auxiliary equation but GS must be in real form.

Notes

M1: Using one boundary condition (may be seen in (a)).

A1: Using other boundary condition

Notes

B1: AG. Sufficient working must be shown.

Notes

M1: Use of mean formula, over \(x\), with correct limits and \(z\) substituted

Notes

B1: Indication that for most, but not all, of the possible combinations of values of \(d\) and \(\omega\) the value of \(z_m\) exceeds 8.
\(7.99998129\ldots \lt z_m \lt 8.782758327\ldots\)

Notes

M1: Putting \(t = 0\) in expression for \(v\)
Or \(x = 0\) in expression for \(v^2\).

A1: Must include units.

Notes

M1: Consider discriminant with \(\geqslant\) or \(\gt\)

Ignore \(\lambda \leqslant -2\sqrt{3}\)

Notes

B1: Graph of under-damped system.
Start anywhere non-zero on \(\theta\)-axis with zero gradient.
Each peak must be lower than before
At least two peaks (not including start point)
The graph must look as though it is approaching the \(t\) axis

Notes

B1: Or \(\mathrm{d}Q/\mathrm{d}t\) tends to zero as \(t \to \infty\). www

M1: Using initial condition to find \(A\) (or \(A + B\))

A1: soi

Notes

B1: Only if from \(Q = k\mathrm{e}^{-at},\ a \gt 0\)
It must be clear that the limit is 0; “\(Q\) is approximately 0” would not be sufficient for B1.

Notes

M1: For obtaining the AE

A1: Solving AE and interpreting. Or equivalent forms (i.e. exponential form)

M1: Correct form of PI and differentiate twice

M1: Dep on previous M. Substitute their PI with two constants into DE

A1: Values of \(a\) and \(b\)

A1: Correct GS

B1: For substituting into 2 eqns

B1: For \(A\)

M1: Dep on previous M. For diffn their GS. Their value for A could be substituted

A1: For B

Notes

M1: Consider behaviour of \(\mathrm{e}^{-t}\)

A1: Accept =
ft their equation from (a)
Accept this final line for 2 marks

Notes

B1: At least one cycle of \(A\cos\omega t\) graph

B1: Amplitude 0.1

B1: Period 0.4

B1: Intersect with the \(t\)-axis at 0.1 and 0.3 (values must be indicated or implied unambiguously (eg by tick-marks and a single value))

Graph must instantaneously horizontal at top/bottom, continuous and not vertical at any point.

Ignore any graph outside [0, 0.4]. Non-inverted cos graph can still get 4/4

Notes

M1: or by argument from sketch. Condone amplitude of 0.2 for M1

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: Forming AE

M1: Attempt to differentiate their PI twice and substituting. Do not condone \(y = C\mathrm{e}^{-2x}\) or \(y = C\mathrm{e}^{x}\).

A1: Must see \(y =\)

Notes

B1FT: Equating coefficient(s) of their \(\mathrm{e}^{kx}\) term(s) to 0 where \(k \gt 0\)

M1*: FT their GS. Accept \(\frac{\mathrm{d}y}{\mathrm{d}x} = -2A\mathrm{e}^{-2x} + 6\mathrm{e}^{-x}\) if \(B = 0\) already found. Allow a slip. Not implied by \(-2A + B + 6 = 0\).

A1: Do not FT.

M1dep: A complete method to solve their \(y = 0\) which leads to a solution

A1: or exact equivalent

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: AG

Notes

B1: Accept anything wrt 4.4

Notes

B1: Must come from correct GS

M1: Must be differentiating a function in terms of cos and sin

Notes

A1: Accept anything wrt 2.1

Notes

B1: AG

Notes

B1: Stating complex roots of auxillary equation is acceptable \((-1 \pm i)\)

Notes

A1: soi

Notes

B1: Has to come from a correct CF

M1*: Must include use of product rule from x = CF + PI

M1dep*: For substitution to lead to an equation in A and B

Notes

M1: Dependent on a function containing \(e^{-kt}\) and \(\cos pt\) and \(\sin pt\)

Notes

M1: diff and subst for \(\mathrm{d}y/\mathrm{d}t\) or \(\mathrm{d}x/\mathrm{d}t\)

M1: subst for \(y\) (or \(x\))

A1: Must be simplified

B1ft: ft their values of \(\lambda\)

M1: subst for \(x\), \(\mathrm{d}x/\mathrm{d}t\)

M1: subst \(t = 0\) in \(x\), \(y\) to find eqns in \(A\), \(B\)

A1: both equations correct

M1: solving the equations to find A and B

A1: for both A and B

A1: for correct equations for both \(x\) and \(y\)

Alternative method