June 2024 Paper 2 Q16
16 In this question you must show detailed reasoning.
Find the particular solution of the differential equation
\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{9y}{(x-1)(x+2)},\]given that \(x = 2\) when \(y = 16\). [12]
| Scheme | Marks | AO |
|---|---|---|
| \(\int \frac{\mathrm{d}y}{y} = \int \frac{9\,\mathrm{d}x}{(x-1)(x+2)}\) oe | M1 | 3.1a |
| \(\frac{A}{x-1} + \frac{B}{x+2}\) | M1 | 3.1a |
| \(\frac{9}{(x-1)(x+2)} = \frac{3}{(x-1)} - \frac{3}{(x+2)}\) oe | A1 A1 | 1.1 1.1 |
| \(\ln y = A\ln(x-1) + B\ln(x+2) + c\) | M1* | 2.1 |
| \(\ln y = 3\ln(x-1) - 3\ln(x+2) + c\) oe | A1 | 1.1 |
| \(\ln 16 = 3\ln(2-1) - 3\ln(2+2) + c\) | M1dep* | 2.1 |
| \(c = 5\ln 4\) or \(\ln 4^5\) or \(\ln 1024\) | A1 | 1.1 |
| \(\ln\frac{\text{their}\,1024(x-1)^A}{(x+2)^B}\) oe | M1 | 3.1a |
| \(\ln y = \ln\frac{1024(x-1)^3}{(x+2)^3}\) | A1 | 1.1 |
| \(\mathrm{e}^{\ln y} = \mathrm{e}^{\ln\frac{A(x-1)^3}{(x+2)^3}}\) oe | M1 | 2.1 |
| \(y = \frac{1024(x-1)^3}{(x+2)^3}\) | A1 | 3.2a |
| [12] |
Notes
M1: separation of variables; condone omission of integral signs or of \(\mathrm{d}x\) and/or \(\mathrm{d}y\); allow 1 slip such as omission of 9 or sign error in bracket
M1: allow 1 sign error in bracket
A1: one of two terms correct
A1: all correct
M1*: any ln integral correct; FT their \(A\) and \(B\); condone omission of \(+\,c\)
A1: all three terms correct including \(+\,c\);
may see \(\frac{1}{9}\ln y = \frac{1}{3}\ln(x-1) - \frac{1}{3}\ln(x+2) + c\)
NB \(\ln y + c\) is equivalent to \(\ln Ay\) where \(A\) is a constant
M1dep*: substitution of (2, 16) in their expression; may be implied by eg \(\ln 16 = 3\ln 1 - 3\ln 4 + c\); must see substitution for incorrect expressions
A1: may see \(c = \frac{5}{9}\ln 4\) oe
allow exact equivalents only
M1: correctly combines their RHS into a single logarithm; their \(+\,c\) must be correctly incorporated into their \(\ln\mathrm{f}(y)\) or their \(\ln\mathrm{f}(x)\)
A1: all correct
M1: correctly exponentiates their expressions, may be awarded before combination into single logarithm
A1: all correct; must see “\(y =\)” at some stage
Alternatively, for the last 6 marks
| Scheme | Marks | AO |
|---|---|---|
| \(\ln\left\{\frac{(x-1)^A}{(x+2)^B} \times \mathrm{e}^c\right\}\) oe | M1dep* | |
| \(\ln y = \ln\left[\frac{(x-1)^3}{(x+2)^3} \times \mathrm{e}^c\right]\) oe | A1 | |
| \(\mathrm{e}^{\ln y} = \mathrm{e}^{\ln\frac{(x-1)^A}{(x+2)^B} \times D}\) oe | M1 | |
| \(y = \frac{(x-1)^3}{(x+2)^3} \times D\) | A1 | |
| \(16 = \frac{(2-1)^A}{(2+2)^B} \times D\) oe | M1 | |
| \(y = \frac{1024(x-1)^3}{(x+2)^3}\) | A1 |
M1dep*: correctly combines their RHS into a single logarithm;
A1: all correct
M1: correctly exponentiates their expressions; may be awarded before combining into single logarithm
A1: all correct
M1: substitution of (2, 16) in their expression; may be implied by eg \(16 = \frac{1^3}{4^3} \times D\); must see substitution for incorrect expressions; may be awarded before exponentiating
A1: all correct; must see “\(y =\)” at some stage