A2 June 2022 Paper 1 Q12
12 Solve the differential equation \(\left(4 - x^2\right)\dfrac{\mathrm{d}y}{\mathrm{d}x} - xy = 1\), given that \(y = 1\) when \(x = 0\), giving your answer in the form \(y = \mathrm{f}(x)\). [9]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} - \dfrac{x}{4 - x^2}y = \dfrac{1}{4 - x^2}\) | B1 | 2.1 |
| IF \(\mathrm{e}^{-\int\frac{x}{4 - x^2}\,\mathrm{d}x}\) | M1 | 2.1 |
| \(= \mathrm{e}^{\frac{1}{2}\ln\left(4 - x^2\right)}\) | M1 | 2.1 |
| \(= \sqrt{4 - x^2}\) | A1 | 2.2a |
| \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\sqrt{4 - x^2}\,y\right) = \dfrac{1}{\sqrt{4 - x^2}}\) \(\displaystyle\sqrt{4 - x^2}\,y = \int\frac{1}{\sqrt{4 - x^2}}\,\mathrm{d}x\) | M1 | 2.1 |
| \(= \arcsin\dfrac{x}{2} + c\) | A1 | 1.1 |
| \(y = \dfrac{\arcsin\left(\frac{1}{2}x\right) + c}{\sqrt{4 - x^2}}\) | M1 | 2.2a |
| when \(x = 0,\ y = 1 \Rightarrow c = 2\) | M1 | 2.1 |
| \(y = \dfrac{\arcsin\left(\frac{1}{2}x\right) + 2}{\sqrt{4 - x^2}}\) | A1 | 2.2a |
| [9] |
Notes
M1: Integral must come from an attempt to get \(\frac{\mathrm{d}y}{\mathrm{d}x}\) on its own
M1: For integrating
M1: Multiplying both sides by their IF
M1: Rearranging into the form \(y =\), equation must come from an attempt at integration having used IF and include \(c\)
M1: Substituting in \(x = 0\), \(y = 1\) to lead to a value of \(c\)