FP2 June 2008 Q7
7.
(a) Show that the substitution \(y = vx\) transforms the differential equation \[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x}{y} + \frac{3y}{x}, \quad x > 0,\ \ y > 0 \qquad \text{(I)}\] into the differential equation \[x\frac{\mathrm{d}v}{\mathrm{d}x} = 2v + \frac{1}{v}. \qquad \text{(II)}\] (3)
(b) By solving differential equation (II), find a general solution of differential equation (I) in the form \(y = \mathrm{f}(x)\). (7)
Given that \(y = 3\) at \(x = 1\),
(c) find the particular solution of differential equation (I). (2)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = v + x\dfrac{\mathrm{d}v}{\mathrm{d}x}\) | B1 |
| \(\left(v + x\dfrac{\mathrm{d}v}{\mathrm{d}x}\right) = \dfrac{x}{vx} + \dfrac{3vx}{x} \Rightarrow x\dfrac{\mathrm{d}v}{\mathrm{d}x} = 2v + \dfrac{1}{v}\) (*) | M1A1 |
| (3) |
Notes
B1 for statement printed or for \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \left(x + v\dfrac{\mathrm{d}x}{\mathrm{d}v}\right)\dfrac{\mathrm{d}v}{\mathrm{d}x}\)
First M1 is for RHS of equation only but for A1 need whole answer correct.
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \frac{v}{1 + 2v^2}\,\mathrm{d}v = \int \frac{1}{x}\,\mathrm{d}x\) | M1 |
| \(\dfrac{1}{4}\ln(1 + 2v^2),\ = \ln x\ (+C)\) | dM1A1, B1 |
| \(Ax^4 = 1 + 2v^2\) | dM1 |
| \(Ax^4 = 1 + 2\left(\dfrac{y}{x}\right)^2\) so \(y = \sqrt{\dfrac{Ax^6 - x^2}{2}}\) or \(y = x\sqrt{\dfrac{Ax^4 - 1}{2}}\) or \(y = x\sqrt{\left(\dfrac{1}{2}\mathrm{e}^{4\ln x + 4c} - \dfrac{1}{2}\right)}\) | M1A1 |
| (7) |
Notes
First M1 accept \(\displaystyle\int \frac{1}{2v + \frac{1}{v}}\,\mathrm{d}v = \int \frac{1}{x}\,\mathrm{d}x\)
Second M1 requires an integration of correct form ¼ may be missing
A1 for LHS correct with ¼ and B1 is independent and is for \(\ln x\)
Third M1 is dependent and needs correct application of log laws
Fourth M1 is independent and merely requires return to \(y/x\) for \(v\)
N.B. There is an IF method possible after suitable rearrangement – see note.
| Scheme | Marks |
|---|---|
| \(x = 1\) at \(y = 3\): \(3 = \sqrt{\dfrac{A - 1}{2}}\) \(A = \ldots\) | M1 |
| \(y = \sqrt{\dfrac{19x^6 - x^2}{2}}\) or \(y = x\sqrt{\dfrac{19x^4 - 1}{2}}\) | A1 |
| (2) | |
| (12 marks) |