FP2 June 2014 Q7
7.
(a) Show that the substitution \(v = y^{-3}\) transforms the differential equation \[x\frac{\mathrm{d}y}{\mathrm{d}x} + y = 2x^4y^4 \qquad \text{(I)}\] into the differential equation \[\frac{\mathrm{d}v}{\mathrm{d}x} - \frac{3v}{x} = -6x^3 \qquad \text{(II)}\] (5)
(b) By solving differential equation (II), find a general solution of differential equation (I) in the form \(y^3 = \mathrm{f}(x)\). (6)
Way 1
| Scheme | Marks |
|---|---|
| \(v = y^{-3} \Rightarrow \dfrac{\mathrm{d}v}{\mathrm{d}y} = -3y^{-4}\) Correct derivative | B1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}v}\dfrac{\mathrm{d}v}{\mathrm{d}x} = -\dfrac{y^4}{3}\dfrac{\mathrm{d}v}{\mathrm{d}x}\) Or \(-3y^{-4}\dfrac{\mathrm{d}y}{\mathrm{d}x}x - 3y^{-3} = -6x^4\) M1: Correct use of the chain rule A1: Correct equation | M1A1 |
| \(-\dfrac{y^4}{3}\dfrac{\mathrm{d}v}{\mathrm{d}x}x + y = 2x^4y^4\) | |
| \(-\dfrac{y^4}{3}\dfrac{\mathrm{d}v}{\mathrm{d}x}x + y = 2x^4y^4 \Rightarrow \dfrac{\mathrm{d}v}{\mathrm{d}x} - \dfrac{3v}{x} = -6x^3\) dM1: Substitutes to obtain an equation in \(v\) and \(x\). A1: Correct completion with no errors seen | dM1A1 |
| (5) |
Way 2
| Scheme | Marks |
|---|---|
| \(y = v^{-\frac{1}{3}} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}v} = -\tfrac{1}{3}v^{-\frac{4}{3}}\) Correct derivative | B1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}v}\dfrac{\mathrm{d}v}{\mathrm{d}x} = -\tfrac{1}{3}v^{-\frac{4}{3}}\dfrac{\mathrm{d}v}{\mathrm{d}x}\) M1: Correct use of the chain rule A1: Correct equation | M1A1 |
| \(-\dfrac{v^{-\frac{4}{3}}}{3}\dfrac{\mathrm{d}v}{\mathrm{d}x}x + v^{-\frac{1}{3}} = 2x^4v^{-\frac{4}{3}}\) dM1: Substitutes to obtain an equation in \(v\) and \(x\). | dM1 |
| \(-\dfrac{v^{-\frac{4}{3}}}{3}\dfrac{\mathrm{d}v}{\mathrm{d}x}x + v^{-\frac{1}{3}} = 2x^4v^{-\frac{4}{3}} \Rightarrow \dfrac{\mathrm{d}v}{\mathrm{d}x} - \dfrac{3v}{x} = -6x^3\) A1: Correct completion with no errors seen | A1 |
Way 3 (Working in reverse)
| Scheme | Marks |
|---|---|
| \(v = y^{-3} \Rightarrow \dfrac{\mathrm{d}v}{\mathrm{d}y} = -3y^{-4}\) B1: Correct derivative | B1 |
| \(\dfrac{\mathrm{d}v}{\mathrm{d}x} = \dfrac{\mathrm{d}v}{\mathrm{d}y}\dfrac{\mathrm{d}y}{\mathrm{d}x} = -3y^{-4}\dfrac{\mathrm{d}y}{\mathrm{d}x}\) M1: Correct use of chain rule A1: Correct expression for \(\mathrm{d}v/\mathrm{d}x\) | M1A1 |
| \(-3y^{-4}\dfrac{\mathrm{d}y}{\mathrm{d}x} - \dfrac{3y^{-3}}{x} = -6x^3\) M1: Substitutes correctly for \(\dfrac{\mathrm{d}v}{\mathrm{d}x}\) and \(v\) in equation (II) to obtain a D.E. in terms of \(x\) and \(y\) only. A1: Correct completion to obtain equation (I) with no errors seen | dM1A1 |
| Scheme | Marks |
|---|---|
| \(I = \mathrm{e}^{\int -\frac{3}{x}\mathrm{d}x} = \mathrm{e}^{-3\ln x} = \dfrac{1}{x^3}\) M1: \(\mathrm{e}^{\int\pm\frac{3}{x}\mathrm{d}x}\) and attempt integration. If not correct, \(\ln x\) must be seen. A1: \(\tfrac{1}{x^3}\) | M1A1 |
| \(\dfrac{v}{x^3} = \displaystyle\int -6\,\mathrm{d}x = -6x\ (+c)\) M1: \(v \times\) their \(I = \displaystyle\int -6x^3 \times\) their \(I\,\mathrm{d}x\) A1: Correct equation with or without \(+ c\) | dM1A1 |
| \(\dfrac{1}{y^3x^3} = -6x + c \Rightarrow y^3 = \ldots\) Include the constant, then substitute for \(y\) and attempt to rearrange to \(y^3 = \ldots\) or \(y = \ldots\) with the constant treated correctly | ddM1 dep on both M marks of (b) |
| \(y^3 = \dfrac{1}{cx^3 - 6x^4}\) Or equivalent | A1 |
| (6) | |
| (11 marks) |