A2 June 2022 Q9
9. A particle \(P\) moves along a straight line.
At time \(t\) minutes, the displacement, \(x\) metres, of \(P\) from a fixed point \(O\) on the line is modelled by the differential equation
\[t^2\frac{\mathrm{d}^2 x}{\mathrm{d}t^2} - 2t\frac{\mathrm{d}x}{\mathrm{d}t} + 2x + 16t^2 x = 4t^3\sin 2t \qquad \text{(I)}\]| 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 | 1.1b |
| \(\dfrac{\mathrm{d}^2 x}{\mathrm{d}t^2} = \dfrac{\mathrm{d}y}{\mathrm{d}t} + \dfrac{\mathrm{d}y}{\mathrm{d}t} + t\dfrac{\mathrm{d}^2 y}{\mathrm{d}t^2}\) or \(\dfrac{\mathrm{d}^2 y}{\mathrm{d}t^2} = -\dfrac{1}{t^2}\dfrac{\mathrm{d}x}{\mathrm{d}t} + \dfrac{2x}{t^3} + \dfrac{1}{t}\dfrac{\mathrm{d}^2 x}{\mathrm{d}t^2} - \dfrac{1}{t^2}\dfrac{\mathrm{d}x}{\mathrm{d}t}\) oe | M1 A1 | 2.1 1.1b |
| \(t^2\left[\text{their } \dfrac{\mathrm{d}^2 x}{\mathrm{d}t^2}\right] - 2t\left[\text{their } \dfrac{\mathrm{d}x}{\mathrm{d}t}\right] + 2[ty] + 16t^2[ty] = 4t^3\sin 2t\) Or \(\left[\text{their } \dfrac{\mathrm{d}^2 y}{\mathrm{d}t^2}\right] + 16\dfrac{x}{t} = 4\sin 2t\) | dM1 | 2.1 |
| \(t^2\left[\dfrac{\mathrm{d}y}{\mathrm{d}t} + \dfrac{\mathrm{d}y}{\mathrm{d}t} + t\dfrac{\mathrm{d}^2 y}{\mathrm{d}t^2}\right] - 2t\left[y + t\dfrac{\mathrm{d}y}{\mathrm{d}t}\right] + 2[ty] + 16t^2[ty] = 4t^3\sin 2t\) \(t^3\dfrac{\mathrm{d}^2 y}{\mathrm{d}t^2} + 16t^3 y = 4t^3\sin 2t \Rightarrow \dfrac{\mathrm{d}^2 y}{\mathrm{d}t^2} + 16y = 4\sin 2t\ *\) (oe in reverse) | A1* | 1.1b |
| (5) |
Notes
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.
| 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 \((y =)\,\underline{\lambda\sin 2t} + \mu\cos 2t\) | B1 | 2.2a |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 2\lambda\cos 2t - 2\mu\sin 2t\) and \(\dfrac{\mathrm{d}^2 y}{\mathrm{d}t^2} = -4\lambda\sin 2t - 4\mu\cos 2t\) | M1 | 1.1b |
| Substitutes into the differential equation and finds values for \(\lambda\) and \(\mu\) \([-4\lambda\sin 2t - 4\mu\cos 2t] + 16[\lambda\sin 2t + \mu\cos 2t] = 4\sin 2t\) \(\Rightarrow \lambda = \ldots,\ \mu = \ldots\) | dM1 | 2.1 |
| \(y = \text{“}A\cos 4t + B\sin 4t\text{”} + \dfrac{1}{3}\sin 2t\) | A1ft | 1.1b |
| \(x = t[\text{their } y]\) | M1 | 3.4 |
| \(x = t\left[A\cos 4t + B\sin 4t + \dfrac{1}{3}\sin 2t\right]\) | A1 | 2.2a |
| (8) | ||
| (13 marks) |
Notes
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 \(x = \ldots\)