Standard +0.8 This is a separable differential equation requiring partial fraction decomposition of y²+4/y(y+4), integration involving both logarithmic and arctangent forms, and application of initial conditions to find a specific constant. The algebraic manipulation and integration techniques go beyond routine exercises, though it follows a standard method for this topic.
8 The variables \(x\) and \(y\) satisfy the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y ^ { 2 } + 4 } { x ( y + 4 ) }$$
for \(x > 0\). It is given that \(x = 4\) when \(y = 2 \sqrt { 3 }\).
Solve the differential equation to obtain the value of \(x\) when \(y = 2\).
Split the fraction and integrate to obtain \(p\ln(y^2+4)\) or \(q\tan^{-1}\frac{y}{2}\) correctly
\*M1
Only following subdivision into \(\frac{y}{y^2+4} + \frac{4}{y^2+4}\). If no subdivision seen then both terms \(p\ln(y^2+4)\) and \(q\tan^{-1}\frac{y}{2}\) must be present
Obtain \(\frac{1}{2}\ln(y^2+4)\)
A1
Obtain \(2\tan^{-1}\frac{y}{2}\)
A1
Use \((4, 2\sqrt{3})\) in expression containing at least 2 of \(a\ln x,\ b\ln(y^2+4)\) and \(c\tan^{-1}\frac{y}{2}\) to obtain constant of integration
DM1
Allow one sign or arithmetic error e.g. \(\frac{2\pi}{3}\). May use \((4, 2\sqrt{3})\) and \((x, 2)\) as limits to find \(x\) for the final 3 marks
Correct solution (any form) e.g. \(\frac{1}{2}\ln(y^2+4) + 2\tan^{-1}\frac{y}{2} = \ln x + \frac{2\pi}{3}\) or \(\frac{1}{2}\ln(y^2+4) + 2\tan^{-1}\frac{y}{2} = \ln x + 2\tan^{-1}\sqrt{3} + \frac{1}{2}\ln 16 - \ln 4\)
A1
However solution not asked for so allow \(\frac{1}{2}\ln 8 + 2\tan^{-1}1 = \ln x + 2\tan^{-1}\sqrt{3} + \frac{1}{2}\ln 16 - \ln 4\)
Obtain \(\sqrt{8}\,e^{-\frac{1}{6}\pi}\) or \(1.68\) or more accurate or \(2\sqrt{2}\,e^{-\frac{1}{6}\pi}\) or \(\frac{\sqrt{8}}{e^{\frac{1}{6}\pi}}\) or \(e^{0.516}\)
A1
ISW. Must remove \(\ln\) so \(x = e^{(\ln 2\sqrt{2} - \pi/6)}\) A0
## Question 8:
| Answer | Mark | Guidance |
|--------|------|----------|
| Separate the variables correctly | **B1** | $\frac{y+4}{y^2+4}\,dy = \frac{1}{x}\,dx$ |
| Obtain $\ln x$ | **B1** | |
| Split the fraction and integrate to obtain $p\ln(y^2+4)$ or $q\tan^{-1}\frac{y}{2}$ correctly | **\*M1** | Only following subdivision into $\frac{y}{y^2+4} + \frac{4}{y^2+4}$. If no subdivision seen then both terms $p\ln(y^2+4)$ and $q\tan^{-1}\frac{y}{2}$ must be present |
| Obtain $\frac{1}{2}\ln(y^2+4)$ | **A1** | |
| Obtain $2\tan^{-1}\frac{y}{2}$ | **A1** | |
| Use $(4, 2\sqrt{3})$ in expression containing at least 2 of $a\ln x,\ b\ln(y^2+4)$ and $c\tan^{-1}\frac{y}{2}$ to obtain constant of integration | **DM1** | Allow one sign or arithmetic error e.g. $\frac{2\pi}{3}$. May use $(4, 2\sqrt{3})$ and $(x, 2)$ as limits to find $x$ for the final 3 marks |
| Correct solution (any form) e.g. $\frac{1}{2}\ln(y^2+4) + 2\tan^{-1}\frac{y}{2} = \ln x + \frac{2\pi}{3}$ or $\frac{1}{2}\ln(y^2+4) + 2\tan^{-1}\frac{y}{2} = \ln x + 2\tan^{-1}\sqrt{3} + \frac{1}{2}\ln 16 - \ln 4$ | **A1** | However solution not asked for so allow $\frac{1}{2}\ln 8 + 2\tan^{-1}1 = \ln x + 2\tan^{-1}\sqrt{3} + \frac{1}{2}\ln 16 - \ln 4$ |
| Obtain $\sqrt{8}\,e^{-\frac{1}{6}\pi}$ or $1.68$ or more accurate or $2\sqrt{2}\,e^{-\frac{1}{6}\pi}$ or $\frac{\sqrt{8}}{e^{\frac{1}{6}\pi}}$ or $e^{0.516}$ | **A1** | ISW. Must remove $\ln$ so $x = e^{(\ln 2\sqrt{2} - \pi/6)}$ A0 |
**Alternative method for first \*M1 A1 A1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $p\left((y+4)\tan^{-1}\frac{y}{2} - \int\tan^{-1}\frac{y}{2}\,dy\right)$ | **\*M1** | Allow sign error |
| $(y+4)\frac{1}{2}\tan^{-1}\frac{y}{2} - \frac{1}{2}\tan^{-1}\frac{y}{2} + \int\frac{y}{y^2+4}\,dy$ | **A1** | |
| Obtain $2\tan^{-1}\frac{y}{2} + \frac{1}{2}\ln(y^2+4)$ | **A1** | |
8 The variables $x$ and $y$ satisfy the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y ^ { 2 } + 4 } { x ( y + 4 ) }$$
for $x > 0$. It is given that $x = 4$ when $y = 2 \sqrt { 3 }$.\\
Solve the differential equation to obtain the value of $x$ when $y = 2$.\\
\hfill \mbox{\textit{CAIE P3 2023 Q8 [8]}}