A2 June 2019 Paper 1 Q11
11 Find the general solution of the differential equation
\[x\frac{\mathrm{d}y}{\mathrm{d}x} - 2y = \frac{x^3}{\sqrt{4 - 2x - x^2}}\]where \(0 \lt x \lt \sqrt{5} - 1\) [7 marks]
| Scheme | Marks | AO |
|---|---|---|
| Divides through by \(x\) | M1 | 1.1a |
| Recognises that the Integrating Factor Method can be applied and finds correct integrating factor, accept \(\mathrm{e}^{-2\ln x}\) | M1 | 3.1a |
| Multiplies equation by their integrating factor | M1 | 1.1a |
| Integrates LHS to obtain \(\dfrac{y}{x^2}\) | A1 | 1.1b |
| Recognises the need to complete the square inside the square root. | M1 | 3.1a |
| Correctly uses the appropriate inverse sine, inverse cosh or inverse sinh function to integrate all (or part of) their RHS. | M1 | 3.1a |
| Finds correct solution including constant of integration. ACF Accept \(\dfrac{y}{x^2} = \ldots\) | A1 | 1.1b |
| (7 marks) |
Typical solution
\[\frac{\mathrm{d}y}{\mathrm{d}x} - \frac{2y}{x} = \frac{x^2}{\sqrt{4 - 2x - x^2}}\]\[\int P\,\mathrm{d}x = -\int \frac{2}{x}\,\mathrm{d}x = -2\ln x\]Integrating factor
\[= \mathrm{e}^{\int P\,\mathrm{d}x} = \mathrm{e}^{-2\ln x} = x^{-2}\]\[\frac{1}{x^2}\frac{\mathrm{d}y}{\mathrm{d}x} - \frac{2y}{x^3} = \frac{1}{\sqrt{4 - 2x - x^2}}\]\[\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{y}{x^2}\right) = \frac{1}{\sqrt{4 - 2x - x^2}}\]To find
\[\int \frac{1}{\sqrt{4 - 2x - x^2}}\,\mathrm{d}x\]\[4 - 2x - x^2 = 5 - (x + 1)^2\]\[\int \frac{1}{\sqrt{4 - 2x - x^2}}\,\mathrm{d}x = \int \frac{1}{\sqrt{5 - (x + 1)^2}}\,\mathrm{d}x\]\[\therefore y = x^2\left\{\sin^{-1}\left(\frac{x + 1}{\sqrt{5}}\right) + c\right\}\]