Standard +0.3 This is a straightforward separable variables question requiring separation, integration using partial fractions, and applying an initial condition. While it involves multiple steps (separation, partial fractions decomposition, integration, exponentiating, and substitution), these are all standard techniques with no novel insight required. The partial fractions are routine, making this slightly easier than average.
9 The variables \(x\) and \(y\) satisfy the differential equation
$$( x + 1 ) ( 3 x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } = y$$
and it is given that \(y = 1\) when \(x = 1\).
Solve the differential equation and find the exact value of \(y\) when \(x = 3\), giving your answer in a simplified form.
Correctly separate variables and integrate at least one side
M1
To obtain \(a\ln y\) or \(b\ln(x+1) + c\ln(3x+1)\)
Obtain term \(\ln y\) from integral of \(\frac{1}{y}\)
B1
State or imply the form \(\frac{A}{x+1} + \frac{B}{3x+1}\) and use correct method to find a constant
M1
Obtain \(A = -\frac{1}{2}\) and \(B = \frac{3}{2}\)
A1
Obtain terms \(-\frac{1}{2}\ln(x+1) + \frac{1}{2}\ln(3x+1)\) or \(-\frac{1}{2}\ln(2x+2) + \frac{1}{2}\ln(6x+2)\) or combination of these terms
A1 FT + A1 FT
The FT is on the values of \(A\) and \(B\)
Use \(x = 1\) and \(y = 1\) to evaluate a constant, or expression for a constant (decimal equivalent of ln terms allowed) or as limits, in solution containing terms \(a\ln y\), \(b\ln(x+1)\) and \(c\ln(3x+1)\), where \(abc \neq 0\)
*M1
e.g. \(\ln y = -\frac{1}{2}\ln(x+1) + \frac{1}{2}\ln(3x+1) - \frac{1}{2}\ln 2\)
Obtain an expression for \(y\) or \(y^2\) and substitute \(x = 3\)
DM1
Do not accept decimal equivalent of ln terms
Obtain answer \(y = \frac{1}{2}\sqrt{5}\) or \(\sqrt{\frac{5}{4}}\) or \(\sqrt{\frac{10}{8}}\)
A1
ISW; must be simplified and exact; do not allow \(1.118\) or \(e^{\frac{1}{2}\ln\frac{5}{4}}\)
9
## Question 9:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correctly separate variables and integrate at least one side | M1 | To obtain $a\ln y$ or $b\ln(x+1) + c\ln(3x+1)$ |
| Obtain term $\ln y$ from integral of $\frac{1}{y}$ | B1 | |
| State or imply the form $\frac{A}{x+1} + \frac{B}{3x+1}$ and use correct method to find a constant | M1 | |
| Obtain $A = -\frac{1}{2}$ and $B = \frac{3}{2}$ | A1 | |
| Obtain terms $-\frac{1}{2}\ln(x+1) + \frac{1}{2}\ln(3x+1)$ or $-\frac{1}{2}\ln(2x+2) + \frac{1}{2}\ln(6x+2)$ or combination of these terms | A1 FT + A1 FT | The FT is on the values of $A$ and $B$ |
| Use $x = 1$ and $y = 1$ to evaluate a constant, or expression for a constant (decimal equivalent of ln terms allowed) or as limits, in solution containing terms $a\ln y$, $b\ln(x+1)$ and $c\ln(3x+1)$, where $abc \neq 0$ | *M1 | e.g. $\ln y = -\frac{1}{2}\ln(x+1) + \frac{1}{2}\ln(3x+1) - \frac{1}{2}\ln 2$ |
| Obtain an expression for $y$ or $y^2$ and substitute $x = 3$ | DM1 | Do not accept decimal equivalent of ln terms |
| Obtain answer $y = \frac{1}{2}\sqrt{5}$ or $\sqrt{\frac{5}{4}}$ or $\sqrt{\frac{10}{8}}$ | A1 | ISW; must be simplified and exact; do not allow $1.118$ or $e^{\frac{1}{2}\ln\frac{5}{4}}$ |
| | **9** | |
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9 The variables $x$ and $y$ satisfy the differential equation
$$( x + 1 ) ( 3 x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } = y$$
and it is given that $y = 1$ when $x = 1$.\\
Solve the differential equation and find the exact value of $y$ when $x = 3$, giving your answer in a simplified form.\\
\hfill \mbox{\textit{CAIE P3 2022 Q9 [9]}}