7. The concentration, \(P\text{ mg m}^{-3}\), of a pollutant in a reservoir, \(t\) days after the pollutant entered the reservoir, is modelled by the differential equation
(a) Show that the transformation \(x = \dfrac{1}{P^2}\) transforms equation (I) into the equation\[\frac{\mathrm{d}x}{\mathrm{d}t} - 2x = -8t \qquad \text{(II)}\] (3)
Given that \(P = 0.5\) when \(t = 0\)
(b) solve differential equation (II) to show that, according to the model,\[P^2 = \frac{1}{4t + 2 + k\mathrm{e}^{2t}}\]where \(k\) is a constant to be determined. (6)
Given that the concentration of the pollutant in the reservoir, 3 days after the pollutant entered the reservoir, was \(0.034\text{ mg m}^{-3}\)
(c) comment on the reliability of the model, giving a reason for your answer. (2)
M1: Identifies and applies a correct strategy for the differentiation. This may be seen when \(\dfrac{\mathrm{d}P}{\mathrm{d}t}\) is substituted into the equation.
M1: Substitutes into the given differential equation and proceeds to an equation in \(x\) and \(t\) only.
A1*: Correct proof with sufficient working shown and no errors, cso
M1: Applies their integrating factor \(I\) to obtain \(Ix = \displaystyle\int \pm 8It\,\{\mathrm{d}t\}\) condone missing \(\mathrm{d}t\)
M1: Recognises that integration by parts is required and applies this correctly to reach the form \(= At\mathrm{e}^{-2t} - \displaystyle\int B\mathrm{e}^{-2t}\,\{\mathrm{d}t\}\)
A1: Correct integration including an arbitrary constant
M1: Uses the conditions given in the model to find the constant of integration. This may be seen before rearranging to get \(P^2 = \ldots\), condone a slip
A1: Correct equation with no errors seen, cso, missing \(\mathrm{d}t\) during working would lose this mark
Note that the first 4 marks in (b) can also be obtained as follows:
B1: AE: \(m - 2 = 0 \Rightarrow m = 2 \Rightarrow x = A\mathrm{e}^{2t}\) (Correct CF)
M1: PI: \(x = at + b \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}t} = a\) (Selects the correct PI form and differentiates)
M1: \(a - 2at - 2b = -8t \Rightarrow a = \ldots(4), b = \ldots(2)\) (Substitutes and compares coefficients to find \(a\) and \(b\))
where \(P\) has displacement \(x\) metres from the origin \(O\) at time \(t\) minutes, \(t \gt 0\)
(a) Show that the transformation \(x = tu\) transforms the differential equation (I) into the differential equation\[\frac{\mathrm{d}^2 u}{\mathrm{d}t^2} - 2\frac{\mathrm{d}u}{\mathrm{d}t} = 8\mathrm{e}^t\] (4)
Given that \(P\) is at \(O\) when \(t = \ln 3\) and when \(t = \ln 5\)
(b) determine the particular solution of the differential equation (I) (8)
Mark scheme (a)
Scheme
Marks
AO
\(x = tu \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{\mathrm{d}u}{\mathrm{d}t}t + u\) or \(\dfrac{\mathrm{d}u}{\mathrm{d}t} = \dfrac{t\dfrac{\mathrm{d}x}{\mathrm{d}t} - x}{t^2}\) (oe)
B1
2.2a
\(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{\mathrm{d}u}{\mathrm{d}t}t + u \Rightarrow \dfrac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}} = \dfrac{\mathrm{d}^{2}u}{\mathrm{d}t^{2}}t + 2\dfrac{\mathrm{d}u}{\mathrm{d}t}\) or e.g. \(\dfrac{\mathrm{d}^{2}u}{\mathrm{d}t^{2}} = -\dfrac{1}{t^2}\dfrac{\mathrm{d}x}{\mathrm{d}t} + \dfrac{1}{t}\dfrac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}} - \dfrac{1}{t^4}\left(t^2\dfrac{\mathrm{d}x}{\mathrm{d}t} - 2tx\right)\)
B1: Deduces a correct first derivative of \(x = tu\) connecting \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) and \(\dfrac{\mathrm{d}u}{\mathrm{d}t}\). Mos likely the main version shown, but alternatives are possible.
B1: Correct second derivative of \(x = tu\) connecting \(\dfrac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}\) and \(\dfrac{\mathrm{d}^{2}u}{\mathrm{d}t^{2}}\). There may be alternatives.
M1: Substitutes their first and second derivatives into the equation and makes some attempt to expand the brackets (need not reach the final answer). (Note if you see answers substituting into the second equation that you feel are worth merit use review.)
A1*: Fully correct proof with no errors and omissions. There must be a clear line of working where the relevant terms cancel.
M1: Forms and solves the quadratic auxiliary equation \(m^2 - 2m = 0\) (may be implied by the correct CF).
A1: Correct CF.
B1: Deduces the correct form of the PI.
M1: Uses the model to find the general solution. Differentiates the correct form of the PI twice (coefficient slips only permitted) and substitutes into the differential equation to find the value of \(\lambda\), leading to \(u = \text{PI} + \text{CF}\)
A1ft: Deduces the general solution for \(u\) or for \(x\), follow through their CF.
M1: Uses the model and \(x = 0, t = \ln 3\) and \(t = \ln 5\) to form two simultaneous equations (need not be fully correct, but correct substitution must be seen at least once in each equation, and allow if slips are made rearranging). May undo the substitution first and use \(x\) and \(t\), or may use \(x = 0 \Rightarrow u = 0\) and substitute to find the constants in their equation for \(u\), but they must form suitable equations using correct conditions.
dM1: Dependent on the previous method mark. Proceeds to solve to find the values of the constants in their equation.
2. The vertical height, \(h\) m, above horizontal ground, of a passenger on a fairground ride, \(t\) seconds after the ride starts, where \(t \leqslant 5\), is modelled by the differential equation
(ii) \(t^2\dfrac{\mathrm{d}^{2}h}{\mathrm{d}t^{2}} = \dfrac{\mathrm{d}^{2}h}{\mathrm{d}x^{2}} - \dfrac{\mathrm{d}h}{\mathrm{d}x}\) (4)
(b) Hence show that the transformation \(t = \mathrm{e}^x\) transforms equation (I) into the equation\[\frac{\mathrm{d}^2 h}{\mathrm{d}x^2} - 3\frac{\mathrm{d}h}{\mathrm{d}x} + 2h = \mathrm{e}^{3x}\] (1)
(c) Hence show that\[h = At + Bt^2 + \frac{1}{2}t^3\]where \(A\) and \(B\) are constants. (6)
Given that when \(t = 1\), \(h = 2.5\) and when \(t = 2\), \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = -1\)
(d) determine the height of the passenger above the ground 5 seconds after the start of the ride. (5)
M1: Uses the chain rule (must be a clear statement of this before substitution or clear implication of it use) with an attempt to differentiate \(t = \mathrm{e}^x\) to form an equation linking \(\dfrac{\mathrm{d}h}{\mathrm{d}t}\) and \(\dfrac{\mathrm{d}h}{\mathrm{d}x}\) (or their reciprocals) in terms of either \(x\) or \(t\). Note they may take \(\ln t\) first, which is fine. Note that \(\dfrac{\mathrm{d}h}{\mathrm{d}x} = \mathrm{e}^x\dfrac{\mathrm{d}h}{\mathrm{d}t}\) with no supporting working is M0.
A1*: Correct proof with no errors.
(ii)
M1: Differentiates again with the product and chain rule, fully and correctly on at least one product, in order to establish any second derivative equation linking \(\dfrac{\mathrm{d}^{2}h}{\mathrm{d}t^{2}}\) and \(\dfrac{\mathrm{d}^{2}h}{\mathrm{d}x^{2}}\)
\(h = A\mathrm{e}^x + B\mathrm{e}^{2x} + \tfrac{1}{2}\mathrm{e}^{3x} \Rightarrow h = At + Bt^2 + \tfrac{1}{2}t^3\ *\)
A1*
2.2a
(6)
Notes
M1: Forms and solves a quadratic auxiliary equation.
A1ft: Correct form for the CF for their AE solutions which must be \(\pm 1, \pm 2\) (ie accept sign errors solving the AE). The \(h =\) is not needed and condone e.g. \(A\mathrm{e}^x + B\mathrm{e}^{2x} = 0\)
B1: Deduces the correct form for the PI. Must be in terms of \(x\) for the equation II.
M1: Differentiates their PI twice and substitutes their derivatives into the DE to find “\(k\)”.
A1: Correct PI (for equation II) seen or implied by working. May or may not be combined with the CF for this mark.
A1*: Forms the correct GS for \(h\) in terms of \(x\) and deduces the correct GS for the height in terms of \(t\) with no errors. This is a given answer, so must have seen the GS in terms of \(h\) before proceeding to this answer. Accept with the constants \(A\) and \(B\) either way round (or with other constants).
Mark scheme (d)
Scheme
Marks
AO
\(t = 1, h = 2.5 \Rightarrow 2.5 = A + B + \tfrac{1}{2}\)
M1
3.4
\(\dfrac{\mathrm{d}h}{\mathrm{d}t} = A + 2Bt + \tfrac{3}{2}t^2 \Rightarrow -1 = A + 4B + 6\)
M1
3.4
\(A = 5, B = -3 \Rightarrow h = 5t - 3t^2 + \tfrac{1}{2}t^3\)
M1: Uses the conditions of the model (\(t = 1\), \(h = 2.5\)) to form an equation in \(A\) and \(B\),
M1: Differentiates and shows of evidence of using the conditions of the model \(\left(t = 2,\ \dfrac{\mathrm{d}h}{\mathrm{d}t} = -1\right)\) to form another equation in \(A\) and \(B\).
A1: Solves simultaneously to obtain correct constants and hence a correct equation connecting \(h\) with \(t\).
M1: Substitutes \(t = 5\). If substitution is not seen you will need to check their answer matches their expression.
A1: Obtains 12.5 m using the model. Must include the units.
(a) Show that the transformation \(x = ty\) transforms equation (I) into the equation\[\frac{\mathrm{d}^2 y}{\mathrm{d}t^2} + 16y = 4\sin 2t\] (5)
(b) Hence find a general solution for the displacement of \(P\) from \(O\) at time \(t\) minutes. (8)
Mark scheme (a)
Scheme
Marks
AO
Use of \(x = ty\) to give \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = y + t\dfrac{\mathrm{d}y}{\mathrm{d}t}\) or \(y = \dfrac{x}{t} \to \dfrac{\mathrm{d}y}{\mathrm{d}t} = -\dfrac{x}{t^2} + \dfrac{1}{t}\dfrac{\mathrm{d}x}{\mathrm{d}t}\) oe
B1: For a correct suitable first derivative expression linking \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\).
M1: Uses the product rule to find an equation linking second derivatives from their first derivative expression.
A1: A correct second derivative expression.
dM1: Substitutes the first and second derivatives and replaces \(x\) with \(ty\) to obtain a differential equation in \(y\) and \(t\) only. Alternatively, may go in reverse and replace \(y\) with \(\dfrac{x}{t}\) etc in the second equation to obtain a differential equation in \(x\) and \(t\) only.
A1*: Simplifies their expression with a correct intermediate stage/working to reach the printed answer. Alternatively, correct working in the other direction to achieve equation (I) from the final equation.
Mark scheme (b)
Scheme
Marks
AO
Solves \(m^2 + 16 = 0\) to give \(m = \ldots\)
M1
1.1b
\((y =)\,A\cos 4t + B\sin 4t\)
A1
1.1b
Particular integral \((y =)\,\underline{\lambda\sin 2t} + \mu\cos 2t\)
M1: Forms the correct auxiliary equation and attempts to solve (any values after the correct AE seen)
A1: Correct complementary function. Accept for this mark if they give it in terms of \(x\) - you are looking for the correct form for the CF.
B1: Deduces a correct form of the particular integral (must include at least \(\lambda\sin 2t\) but may be no more than this). SC if by error the CF includes \(\sin 2t\) allow B1 for a PI of form \(\lambda t\sin 2t + \mu t\cos 2t\)
M1: Differentiates the PI twice.
dM1: Dependent on the previous method mark. Substitutes \(y\) and \(\dfrac{\mathrm{d}^2 y}{\mathrm{d}t^2}\) into the differential equation leading to values for the constant(s).
A1ft: Correct general equation for \(y\) following through their CF, which must be a (non-constant) function of \(t\). Must be in terms of \(t\) and start \(y = \ldots\)
M1: Links the solution to the solution of the model equation to find the general solution for the displacement.
A1: Deduces the correct general solution for the displacement. Must be \(x = \ldots\)
8. A community is concerned about the rising level of pollutant in its local pond and applies a chemical treatment to stop the increase of pollutant.
The concentration, \(x\) parts per million (ppm), of the pollutant in the pond water \(t\) days after the chemical treatment was applied, is modelled by the differential equation
When the chemical treatment was applied the concentration of pollutant was 3 ppm.
(a) Use the iteration formula\[\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)_n \approx \frac{(y_{n+1} - y_n)}{h}\]once to estimate the concentration of the pollutant in the pond water 6 hours after the chemical treatment was applied. (4)
(b) Show that the transformation \(u = x^3\) transforms the differential equation (I) into the differential equation\[\frac{\mathrm{d}u}{\mathrm{d}t} + u\tanh t = 1 + \frac{3}{\cosh t} \qquad \text{(II)}\] (3)
(c) Determine the general solution of equation (II) (4)
(d) Hence find an equation for the concentration of pollutant in the pond water \(t\) days after the chemical treatment was applied. (3)
(e) Find the percentage error of the estimate found in part (a) compared to the value predicted by the model, stating if it is an overestimate or an underestimate. (3)
So \(x_1 \approx 3 + \text{“}0.25\text{”} \times \text{“}\dfrac{4}{27}\text{”} = \ldots\)
M1
1.1b
After 6 hours concentration of the pollutant is approximately awrt 3.04 ppm (3 s.f.) or \(\dfrac{82}{27}\) ppm
A1
3.2a
(4)
Notes
B1: Identifies a correct step length for the situation – 6 hours is a quarter of a day, so \(h = 0.25\)
M1: Uses “\(y_0\)” \(= x(0) = 3\) and \(t = 0\) to find “\(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_0\)” \(= \left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)_0\). Accept with whichever notation used, as long as it is clear they are attempting the correct things.
M1: Applies the approximation formula with their “\(h\)” and their “\(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_0\)”
A1: For awrt 3.04 ppm. Accept \(\dfrac{82}{27}\) ppm
\(\dfrac{\mathrm{d}u}{\mathrm{d}t} + u\tanh t = 1 + \dfrac{3}{\cosh t}\ *\)
A1*
1.1b
(3)
Notes
B1: A correct equation relating \(\dfrac{\mathrm{d}u}{\mathrm{d}t}\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) from the chain rule.
M1: Makes a complete substitution for \(x\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) in equation (I) or a complete substitution for \(u\) and \(\dfrac{\mathrm{d}u}{\mathrm{d}t}\) in equation (II)
A1*: Simplifies correctly to achieve the given result.
\(u\cosh t = \sinh t + 3t + c\) or \(u = \tanh t + \dfrac{3t}{\cosh t} + \dfrac{c}{\cosh t}\) oe
A1
1.1b
(4)
Notes
B1: Correct integrating factor found or spotted. Allow for \(\mathrm{e}^{\ln\cosh t}\)
M1: Applies IF to achieve \(u\,\text{“}\cosh t\text{”} = \displaystyle\int \text{“}\cosh t\text{”}\left(1 + \frac{3}{\cosh t}\right)\mathrm{d}t\)
M1: A reasonable attempt to integrate the RHS. Need not include constant of integration. If I.F. correct allow for \(\pm\sinh t + 3t\,(+c)\)
A1: Correct general solution, either implicit or explicit form including the context of integration (award when first seen and isw)
Mark scheme (d)
Scheme
Marks
AO
\(t = 0 \Rightarrow x = 3, u = 27 \Rightarrow c = 27\cosh 0 - \sinh 0 - 3(0) = 27\)
M1
3.4
\(\Rightarrow x = \left(\tanh t + \dfrac{3t + \text{“}27\text{”}}{\cosh t}\right)^{\frac{1}{3}}\)
M1
3.4
\(x = \left(\tanh t + \dfrac{3t + 27}{\cosh t}\right)^{\frac{1}{3}}\) (oe)
A1
3.2a
(3)
Notes
M1: Uses the initial conditions in an appropriate equation to find the constant of integration. Either \(t = 0\) and \(u = 27\) in the answer to (c), or \(t = 0\) and \(x = 3\) if substitution for \(x\) occurs first.
M1: Reverses the substitution and rearranges to find equation for \(x\), with evaluated constant included.
A1: Correct equation, any equivalent form, but must be \(x = \ldots\)
6. The concentration of a drug in the bloodstream of a patient, \(t\) hours after the drug has been administered, where \(t \leqslant 6\), is modelled by the differential equation
(a) Show that the transformation \(t = \mathrm{e}^x\) transforms equation (I) into the equation \[\frac{\mathrm{d}^2C}{\mathrm{d}x^2} - 6\frac{\mathrm{d}C}{\mathrm{d}x} + 8C = \mathrm{e}^{3x} \qquad \text{(II)}\] (5)
(b) Hence find the general solution for the concentration \(C\) at time \(t\) hours. (7)
Given that when \(t = 6\), \(C = 0\) and \(\dfrac{\mathrm{d}C}{\mathrm{d}t} = -36\)
(c) find the maximum concentration of the drug in the bloodstream of the patient. (5)
Mark scheme (a)
Scheme
Marks
AO
Examples: \(t = \mathrm{e}^x \Rightarrow \dfrac{\mathrm{d}t}{\mathrm{d}C} = \mathrm{e}^x\dfrac{\mathrm{d}x}{\mathrm{d}C}\) or \(\dfrac{\mathrm{d}C}{\mathrm{d}x} = t\dfrac{\mathrm{d}C}{\mathrm{d}t}\) or \(\dfrac{\mathrm{d}C}{\mathrm{d}t} = \mathrm{e}^{-x}\dfrac{\mathrm{d}C}{\mathrm{d}x}\) or \(\dfrac{\mathrm{d}C}{\mathrm{d}t} = \dfrac{1}{t}\dfrac{\mathrm{d}C}{\mathrm{d}x}\)
M1: Uses \(t = \mathrm{e}^x\) to obtain a correct equation in terms of \(\dfrac{\mathrm{d}C}{\mathrm{d}x}\), \(\dfrac{\mathrm{d}C}{\mathrm{d}t}\) and \(t\) (or \(\mathrm{e}^x\)) or their reciprocals
dM1: Differentiates again correctly with the product rule and chain rule in order to obtain an equation involving \(\dfrac{\mathrm{d}^2C}{\mathrm{d}t^2}\) and \(\dfrac{\mathrm{d}^2C}{\mathrm{d}x^2}\). This needs to be fully correct calculus work allowing sign errors only.
A1: Correct equation.
dM1: Shows clearly their substitution into the differential equation (or equivalent work) in order to form the new equation. Dependent on the first method mark and dependent on having obtained two terms for the second derivative. Allow substitution for \(\dfrac{\mathrm{d}C}{\mathrm{d}x}\) and \(\dfrac{\mathrm{d}^2C}{\mathrm{d}x^2}\) into equation (II) to achieve equation (I)
M1: Uses the conditions of the model (\(t = 6\), \(C = 0\)) to form an equation in \(A\) and \(B\). ***Note that is acceptable to use their \(C\) in terms of \(x\) for this mark as long as they use \(x = \ln 6\) when \(C = 0\)
M1: Uses the conditions of the model \(\left(t = 6,\ \dfrac{\mathrm{d}C}{\mathrm{d}t} = -36\right)\) to form another equation in \(A\) and \(B\). ***Note that it is not acceptable to use \(\dfrac{\mathrm{d}C}{\mathrm{d}x} = -36\) with \(x = \ln 6\), as it is necessary to use \(\dfrac{\mathrm{d}C}{\mathrm{d}t} = \dfrac{\mathrm{d}C}{\mathrm{d}x}\dfrac{\mathrm{d}x}{\mathrm{d}t}\) e.g. \(-36 = \left(4A\mathrm{e}^{4\ln 6} + 2B\mathrm{e}^{2\ln 6} - 3\mathrm{e}^{3\ln 6}\right) \times \mathrm{e}^{-\ln 6}\) or \(-216 = 4A\mathrm{e}^{4\ln 6} + 2B\mathrm{e}^{2\ln 6} - 3\mathrm{e}^{3\ln 6}\)
A1: Correct equation connecting \(C\) with \(t\)
ddM1: Uses a suitable method to find the maximum concentration. E.g. solves \(\dfrac{\mathrm{d}C}{\mathrm{d}t} = 0\) for \(t\) and substitutes to find \(C\). Allow a solution that solves \(\dfrac{\mathrm{d}C}{\mathrm{d}x} = 0\) for \(x\) and uses this correctly to find \(C\). Dependent on both previous method marks.
A1: Obtains \(32\ \mu\mathrm{g}\mathrm{L}^{-1}\) using the model. Units are required but allow e.g.
(a) Show that the transformation \(z = y^{\frac{1}{2}}\) transforms the differential equation \[\frac{\mathrm{d}y}{\mathrm{d}x} - 4y\tan x = 2y^{\frac{1}{2}} \qquad \text{(I)}\] into the differential equation \[\frac{\mathrm{d}z}{\mathrm{d}x} - 2z\tan x = 1 \qquad \text{(II)}\] (5)
(b) Solve the differential equation (II) to find \(z\) as a function of \(x\). (6)
(c) Hence obtain the general solution of the differential equation (I). (1)
Mark scheme (a)
Scheme
Marks
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}z} \cdot \dfrac{\mathrm{d}z}{\mathrm{d}x}\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}z} = 2z\) so \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2z \cdot \dfrac{\mathrm{d}z}{\mathrm{d}x}\)
M1 M1 A1
Substituting to get \(2z \cdot \dfrac{\mathrm{d}z}{\mathrm{d}x} - 4z^2\tan x = 2z\) and thus \(\dfrac{\mathrm{d}z}{\mathrm{d}x} - 2z\tan x = 1\) *
(a) Show that the substitution \(y = vx\) transforms the differential equation \[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x}{y} + \frac{3y}{x}, \quad x > 0,\ \ y > 0 \qquad \text{(I)}\] into the differential equation \[x\frac{\mathrm{d}v}{\mathrm{d}x} = 2v + \frac{1}{v}. \qquad \text{(II)}\] (3)
(b) By solving differential equation (II), find a general solution of differential equation (I) in the form \(y = \mathrm{f}(x)\). (7)
Given that \(y = 3\) at \(x = 1\),
(c) find the particular solution of differential equation (I). (2)
Mark scheme (a)
Scheme
Marks
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = v + x\dfrac{\mathrm{d}v}{\mathrm{d}x}\)
B1 for statement printed or for \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \left(x + v\dfrac{\mathrm{d}x}{\mathrm{d}v}\right)\dfrac{\mathrm{d}v}{\mathrm{d}x}\)
First M1 is for RHS of equation only but for A1 need whole answer correct.
3. A scientist is modelling the amount of a chemical in the human bloodstream. The amount \(x\) of the chemical, measured in mg \(l^{-1}\), at time \(t\) hours satisfies the differential equation \[2x\frac{\mathrm{d}^2x}{\mathrm{d}t^2} - 6\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 = x^2 - 3x^4, \qquad x \gt 0.\]
(a) Show that the substitution \(y = \dfrac{1}{x^2}\) transforms this differential equation into \[\frac{\mathrm{d}^2y}{\mathrm{d}t^2} + y = 3. \qquad \boxed{\boldsymbol{I}}\] (5)
(b) Find the general solution of differential equation \(\boxed{\boldsymbol{I}}\). (4)
Given that at time \(t = 0\), \(x = \dfrac{1}{2}\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0\),
(c) find an expression for \(x\) in terms of \(t\), (4)
(d) write down the maximum value of \(x\) as \(t\) varies. (1)
Mark scheme (a)
Scheme
Marks
\(y = x^{-2} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}t} = -2x^{-3}\dfrac{\mathrm{d}x}{\mathrm{d}t} = -2x^{-3}\dot{x}\) [Use of chain rule; need \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\)]
(\(\div\) given d.e. by \(x^4\)) \(\ \dfrac{2}{x^3}\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} - \dfrac{6}{x^4}\left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)^2 = \dfrac{1}{x^2} - 3\) becomes \(\left(-\dfrac{\mathrm{d}^2y}{\mathrm{d}t^2} = y - 3\right) \qquad \dfrac{\mathrm{d}^2y}{\mathrm{d}t^2} + y = 3\) AG
A1 cso
(5)
Notes
Second M1 is for attempt at product rule. (be generous) Final A1 requires all working correct and sufficient “substitution” work
(corrected from the printed mark scheme: the end of the first line is printed as “\(= -2x - 3\,t\)”, which reads as \(-2x^{-3}\dot{x}\))
Mark scheme (b)
Scheme
Marks
Auxiliary equation: \(m^2 + 1 = 0\) and produce Complementary Function \(y = \ldots\)
M1
\((y) = A\cos t + B\sin t\)
A1cao
Particular integral: \(y = 3\)
B1
\(\therefore\) General solution: \((y) = A\cos t + B\sin t + 3\)
A1ft
(4)
Notes
Answer can be stated; M1 is implied by correct C.F. stated (allow \(\theta\) for \(t\)) A1 f.t. for candidates CF + PI Allow \(m^2 + m = 0\) and \(m^2 - 1 = 0\) for M1. Marks for (b) can be gained in (c)
Mark scheme (c)
Scheme
Marks
\(\dfrac{1}{x^2} = A\cos t + B\sin t + 3\)
\(x = \dfrac{1}{2},\ t = 0 \Rightarrow (4 = A + 3)\ A = 1\)
B1
Differentiating (to include \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\)): \(\ -2x^{-3}\dfrac{\mathrm{d}x}{\mathrm{d}t} = -A\sin t + B\cos t\)
M1
\(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0,\ t = 0 \Rightarrow (0 = 0 + B) \qquad B = 0\)
(a) Show that the substitution \(y = vx\) transforms the differential equation \[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3x - 4y}{4x + 3y} \qquad \text{(I)}\] into the differential equation \[x\frac{\mathrm{d}v}{\mathrm{d}x} = -\frac{3v^2 + 8v - 3}{3v + 4} \qquad \text{(II)}.\] (4)
(b) By solving differential equation (II), find a general solution of differential equation (I). (5)
(c) Given that \(y = 7\) at \(x = 1\), show that the particular solution of differential equation (I) can be written as \[(3y - x)(y + 3x) = 200.\] (5)
Mark scheme (a)
Scheme
Marks
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = v + x\dfrac{\mathrm{d}v}{\mathrm{d}x}\)
B1
\(v + x\dfrac{\mathrm{d}v}{\mathrm{d}x} = \dfrac{3x - 4vx}{4x + 3vx}\) (all in terms of \(v\) and \(x\))
M1
\(x\dfrac{\mathrm{d}v}{\mathrm{d}x} = \dfrac{3 - 4v - v(4 + 3v)}{4 + 3v}\) (Requires \(x\dfrac{\mathrm{d}v}{\mathrm{d}x} = \mathrm{f}(v)\), 2 terms over common denom.)
(corrected from the printed mark scheme: the last line is printed without the minus sign, as \(x\dfrac{\mathrm{d}v}{\mathrm{d}x} = \dfrac{3v^2 + 8v - 3}{3v + 4}\))
