A2 October 2020 Paper 1 Q14
14 Solve the simultaneous differential equations
\[\frac{\mathrm{d}x}{\mathrm{d}t} + 2x = 4y, \qquad \frac{\mathrm{d}y}{\mathrm{d}t} + 3x = 5y,\]given that when \(t = 0\), \(x = 0\) and \(y = 1\). [11]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2\dfrac{\mathrm{d}x}{\mathrm{d}t} = 4\dfrac{\mathrm{d}y}{\mathrm{d}t} = 20y - 12x\) | M1 | 3.1a |
| \(\Rightarrow \dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2\dfrac{\mathrm{d}x}{\mathrm{d}t} = 5\dfrac{\mathrm{d}x}{\mathrm{d}t} + 10x - 12x\) | M1 | 3.1a |
| \(\Rightarrow \dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} - 3\dfrac{\mathrm{d}x}{\mathrm{d}t} + 2x = 0\) | A1 | 1.1 |
| AE \(\lambda^2 - 3\lambda + 2 = 0\) \(\Rightarrow \lambda = 1\) or \(2\) GS \(x = A\mathrm{e}^t + B\mathrm{e}^{2t}\) | B1ft | 1.1 |
| \(y = \dfrac{1}{4}\left(\dfrac{\mathrm{d}x}{\mathrm{d}t} + 2x\right) = \dfrac{1}{4}\left(A\mathrm{e}^t + 2B\mathrm{e}^{2t} + 2A\mathrm{e}^t + 2B\mathrm{e}^{2t}\right)\) | M1 | 2.1 |
| \(= \dfrac{3}{4}A\mathrm{e}^t + B\mathrm{e}^{2t}\) | A1 | 2.2a |
| when \(t = 0\), \(x = A + B = 0\) | M1 | 1.1 |
| \(1 = \dfrac{3}{4}A + B\) | A1 | 1.1 |
| M1 | 1.1 | |
| \(\Rightarrow A = -4,\ B = 4\) so \(x = 4\mathrm{e}^{2t} - 4\mathrm{e}^t\) \(y = 4\mathrm{e}^{2t} - 3\mathrm{e}^t\) | A1 A1 | 3.2a 3.2a |
| [11] |
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
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}t^2} + 3\dfrac{\mathrm{d}x}{\mathrm{d}t} = 5\dfrac{\mathrm{d}y}{\mathrm{d}t}\) | |
| \(\frac{1}{3}\left(5\frac{\mathrm{d}y}{\mathrm{d}t} - \frac{\mathrm{d}^2y}{\mathrm{d}t^2}\right) + \frac{2}{3}\left(5y - \frac{\mathrm{d}y}{\mathrm{d}t}\right) = 4y\) | |
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}t^2} - 3\dfrac{\mathrm{d}y}{\mathrm{d}t} + 2y = 0\) AE \(\lambda^2 - 3\lambda + 2 = 0\) \(\Rightarrow \lambda = 1\) or \(2\) GS \(y = C\mathrm{e}^t + D\mathrm{e}^{2t}\) | |
| \(x = \frac{4}{3}C\mathrm{e}^t + D\mathrm{e}^{2t}\) | |
| when \(t = 0\), \(4C + 3D = 0\) \(C + D = 1\) | |
| \(C = -3,\ D = 4\) |