Standard +0.3 This is a separable differential equation requiring straightforward separation of variables, integration of standard forms (1/(x²+3) gives arctan), and application of initial conditions. While it involves trigonometric substitution and algebraic manipulation, it follows a standard template with no novel insight required, making it slightly easier than average.
7 The variables \(x\) and \(\theta\) satisfy the differential equation
$$\frac { x } { \tan \theta } \frac { \mathrm {~d} x } { \mathrm {~d} \theta } = x ^ { 2 } + 3$$
It is given that \(x = 1\) when \(\theta = 0\).
Solve the differential equation, obtaining an expression for \(x ^ { 2 }\) in terms of \(\theta\).
\(\int \tan\theta \, d\theta = \int \frac{x}{x^2+3} \, dx\). Condone missing integral signs or missing \(dx\), \(d\theta\). Can be implied by later work.
Obtain term \(-\ln(\cos\theta)\)
B1
Or equivalent e.g. \(\ln(\sec\theta)\)
Obtain term of the form \(a\ln(x^2+3)\)
M1
Obtain term \(\frac{1}{2}\ln(x^2+3)\)
A1
Use \(x=1\), \(\theta=0\) to evaluate a constant or as limits in a solution containing terms of the form \(a\ln(x^2+3)\) and \(b\ln(\cos\theta)\)
M1
If they have rearranged then the constant must be of the correct form.
Obtain final answer \(x^2 = \frac{4}{\cos^2\theta} - 3\)
A1
Or equivalent e.g. \(x^2 = 4\sec^2 x - 3\). lns removed.
## Question 7:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Separate variables correctly | **B1** | $\int \tan\theta \, d\theta = \int \frac{x}{x^2+3} \, dx$. Condone missing integral signs or missing $dx$, $d\theta$. Can be implied by later work. |
| Obtain term $-\ln(\cos\theta)$ | **B1** | Or equivalent e.g. $\ln(\sec\theta)$ |
| Obtain term of the form $a\ln(x^2+3)$ | **M1** | |
| Obtain term $\frac{1}{2}\ln(x^2+3)$ | **A1** | |
| Use $x=1$, $\theta=0$ to evaluate a constant or as limits in a solution containing terms of the form $a\ln(x^2+3)$ and $b\ln(\cos\theta)$ | **M1** | If they have rearranged then the constant must be of the correct form. |
| Obtain correct answer in any form | **A1** | $\frac{1}{2}\ln(x^2+3) = -\ln\cos\theta + \ln 2$ |
| Obtain final answer $x^2 = \frac{4}{\cos^2\theta} - 3$ | **A1** | Or equivalent e.g. $x^2 = 4\sec^2 x - 3$. lns removed. |
---
7 The variables $x$ and $\theta$ satisfy the differential equation
$$\frac { x } { \tan \theta } \frac { \mathrm {~d} x } { \mathrm {~d} \theta } = x ^ { 2 } + 3$$
It is given that $x = 1$ when $\theta = 0$.\\
Solve the differential equation, obtaining an expression for $x ^ { 2 }$ in terms of $\theta$.\\
\hfill \mbox{\textit{CAIE P3 2023 Q7 [7]}}