FP2 June 2012 Q7
7.
(a) Show that the substitution \(y = vx\) transforms the differential equation \[3xy^2\frac{\mathrm{d}y}{\mathrm{d}x} = x^3 + y^3 \qquad \text{(I)}\] into the differential equation \[3v^2x\frac{\mathrm{d}v}{\mathrm{d}x} = 1 - 2v^3 \qquad \text{(II)}\] (3)
(b) By solving differential equation (II), find a general solution of differential equation (I) in the form \(y = \mathrm{f}(x)\). (6)
Given that \(y = 2\) at \(x = 1\),
(c) find the value of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) at \(x = 1\) (2)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = v + x\dfrac{\mathrm{d}v}{\mathrm{d}x}\) seen | B1 |
| \(3x^3v^2\left(v + x\dfrac{\mathrm{d}v}{\mathrm{d}x}\right) = x^3 + v^3x^3 \quad \Rightarrow \quad 3v^2x\dfrac{\mathrm{d}v}{\mathrm{d}x} = 1 - 2v^3\) (**ag**) | M1 A1 cso |
| (3) |
Notes
M1 for substituting \(y\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) obtaining an expression in \(v\) and \(x\) only
| Scheme | Marks |
|---|---|
| \(\displaystyle\int\dfrac{3v^2}{1 - 2v^3}\,\mathrm{d}v = \int\dfrac{1}{x}\,\mathrm{d}x\) | M1 |
| \(-\dfrac{1}{2}\ln(1 - 2v^3) = \ln x\ \ (+C)\) | M1 A1 |
| \(-\ln(1 - 2v^3) = \ln x^2 + \ln A\) | |
| \(Ax^2 = \dfrac{1}{1 - 2v^3}\) | M1 |
| \(1 - \dfrac{2y^3}{x^3} = \dfrac{1}{Ax^2}\) | |
| \(y = \sqrt[3]{\dfrac{x^3 - Bx}{2}}\) or equivalent | dM1 A1cso |
| (6) |
Notes
1st M1 for separating variables
2nd M1 for attempting to integrate both sides
1st A1 both sides required or equivalent expressions. (Modulus not required.)
3rd M1 Removing logs, dealing correctly with constant
4th M1 dep on 1st M. Substitute \(v = \dfrac{y}{x}\) and rearranging to \(y = \mathrm{f}(x)\)
| Scheme | Marks |
|---|---|
| Using \(y = 2\) at \(x = 1\): \(12\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1 + 8\) | M1 |
| At \(x = 1\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3}{4}\) | A1 |
| (2) | |
| (11 marks) |
Notes
M1 for finding a numerical value for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
A1 for correct numerical answer oe.