A2 June 2024 Q10
10. The motion of a particle \(P\) along the \(x\)-axis is modelled by the differential equation
\[t^2\frac{\mathrm{d}^2 x}{\mathrm{d}t^2} - 2t(t + 1)\frac{\mathrm{d}x}{\mathrm{d}t} + 2(t + 1)x = 8t^3\mathrm{e}^t \qquad \text{(I)}\]where \(P\) has displacement \(x\) metres from the origin \(O\) at time \(t\) minutes, \(t \gt 0\)
Given that \(P\) is at \(O\) when \(t = \ln 3\) and when \(t = \ln 5\)
| 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 | 1.1b |
| \(t^2\left[\dfrac{\mathrm{d}^{2}u}{\mathrm{d}t^{2}}t + 2\dfrac{\mathrm{d}u}{\mathrm{d}t}\right] - 2t(t + 1)\left[\dfrac{\mathrm{d}u}{\mathrm{d}t}t + u\right] + 2(t + 1)tu = 8t^3\mathrm{e}^t\) \(t^3\dfrac{\mathrm{d}^{2}u}{\mathrm{d}t^{2}} + 2t^2\dfrac{\mathrm{d}u}{\mathrm{d}t} - 2t(t + 1)\dfrac{\mathrm{d}u}{\mathrm{d}t}t - 2t(t + 1)u + 2t(t + 1)u = 8t^3\mathrm{e}^t\) \(t^3\dfrac{\mathrm{d}^{2}u}{\mathrm{d}t^{2}} + 2t^2\dfrac{\mathrm{d}u}{\mathrm{d}t} - 2t^3\dfrac{\mathrm{d}u}{\mathrm{d}t} - 2t^2\dfrac{\mathrm{d}u}{\mathrm{d}t} - 2t(t + 1)u + 2t(t + 1)u = 8t^3\mathrm{e}^t\) | M1 | 1.1b |
| \(\dfrac{\mathrm{d}^{2}u}{\mathrm{d}t^{2}} - 2\dfrac{\mathrm{d}u}{\mathrm{d}t} = 8\mathrm{e}^t\ *\) | A1* | 2.1 |
| (4) |
Notes
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.
| Scheme | Marks | AO |
|---|---|---|
| \(m^2 - 2m = 0 \Rightarrow m = 0, 2\) | M1 | 1.1b |
| \(u = A + B\mathrm{e}^{2t}\) | A1 | 1.1b |
| \(\text{PS} \Rightarrow u = \lambda\mathrm{e}^t\) | B1 | 2.2a |
| \(\dfrac{\mathrm{d}u}{\mathrm{d}t} = \lambda\mathrm{e}^t, \dfrac{\mathrm{d}^{2}u}{\mathrm{d}t^{2}} = \lambda\mathrm{e}^t \Rightarrow \lambda\mathrm{e}^t - 2\lambda\mathrm{e}^t = 8\mathrm{e}^t \Rightarrow \lambda = \ldots\{-8\}\) \(\Rightarrow u = \text{PS} + \text{CF}\) | M1 | 3.4 |
| \(u = \text{“}A + B\mathrm{e}^{2t}\text{”} - 8\mathrm{e}^t\) or \(x = \left(\text{“}A + B\mathrm{e}^{2t}\text{”} - 8\mathrm{e}^t\right)t\) | A1ft | 2.2a |
| \(x = \left(A + B\mathrm{e}^{2t} - 8\mathrm{e}^t\right)t\) \(x = 0, t = \ln 3\) and \(t = \ln 5\) \(0 = A + B\mathrm{e}^{2\ln 3} - 8\mathrm{e}^{\ln 3} \Rightarrow 0 = A + 9B - 24\) \(0 = A + B\mathrm{e}^{2\ln 5} - 8\mathrm{e}^{\ln 5} \Rightarrow 0 = A + 25B - 40\) | M1 | 3.4 |
| \(\left.\begin{aligned}A + 9B &= 24\\ A + 25B &= 40\end{aligned}\right\} \Rightarrow A = \ldots\{15\},\ B = \ldots\{1\}\) | dM1 | 3.1a |
| \(x = \left(15 + \mathrm{e}^{2t} - 8\mathrm{e}^t\right)t\) (oe) | A1 | 1.1b |
| (8) | ||
| (12 marks) |
Notes
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.
A1: Correct particular solution \(x = \left(15 + \mathrm{e}^{2t} - 8\mathrm{e}^t\right)t\) (may be expanded).