Standard +0.3 This is a straightforward separable variables question requiring standard manipulation (separating variables, integrating both sides using substitution for the trigonometric term, applying initial conditions, and solving for tan θ). While it involves multiple steps and some algebraic manipulation with exponentials and trigonometric identities, it follows a completely standard template with no novel insight required, making it slightly easier than average.
11 The variables \(y\) and \(\theta\) satisfy the differential equation
$$( 1 + y ) ( 1 + \cos 2 \theta ) \frac { d y } { d \theta } = e ^ { 3 y }$$
It is given that \(y = 0\) when \(\theta = \frac { 1 } { 4 } \pi\).
Solve the differential equation and find the exact value of \(\tan \theta\) when \(y = 1\).
If you use the following page to complete the answer to any question, the question number must be clearly shown.
Use correct double angle formula to obtain \(\int\frac{1}{2\cos^2\theta}\,d\theta\)
B1
Obtain \(k\tan\theta\ [+B]\)
B1
Condone no constant of integration.
Use \(y=0\), \(\theta=\frac{\pi}{4}\) to evaluate a constant of integration in an expression of the form \(\alpha ye^{-3y}\), \(\beta e^{-3y}\) and \(\gamma\tan\theta\) only.
M1\*
\(\frac{1}{2} = -\frac{1}{3} - \frac{1}{9} + C \quad \left(C = \frac{17}{18}\right)\). Allow \(\alpha ye^{3y}\) and \(\beta e^{3y}\). Must have integrated LHS twice.
Use \(y=1\)
DM1
\(\frac{-(1+1)}{3e^3} - 1(9e^3) = \frac{1}{2}\tan\theta - \frac{17}{18}\). Must have integrated LHS.
Or exact equivalent. Exact ISW. Allow \(\theta = \tan^{-1}\!\left(\frac{17}{9} - \frac{14}{9}e^{-3}\right)\). If \(x\) instead of \(\theta\) then withhold final A1.
## Question 11:
| Answer | Mark | Guidance |
|--------|------|----------|
| Separate variables correctly | **B1** | $\int(1+y)e^{-3y}\,dy = \int\frac{1}{1+\cos 2\theta}\,d\theta$. Allow $1/e^{3y}$ and missing integral signs. |
| Integrate to obtain $p(1+y)e^{-3y} + \int qe^{-3y}\,dy$ | **M1** | Allow unless clear evidence that formula used has $a+$ sign. |
| Obtain $\frac{-1}{3}(1+y)e^{-3y} + \int\frac{1}{3}e^{-3y}\,dy$ | **A1** | Allow unsimplified. |
| Obtain $\frac{-1}{3}(1+y)e^{-3y} - \frac{1}{9}e^{-3y}\ (+A)$ | **A1** | Condone no constant of integration. |
| Use correct double angle formula to obtain $\int\frac{1}{2\cos^2\theta}\,d\theta$ | **B1** | |
| Obtain $k\tan\theta\ [+B]$ | **B1** | Condone no constant of integration. |
| Use $y=0$, $\theta=\frac{\pi}{4}$ to evaluate a constant of integration in an expression of the form $\alpha ye^{-3y}$, $\beta e^{-3y}$ and $\gamma\tan\theta$ only. | **M1\*** | $\frac{1}{2} = -\frac{1}{3} - \frac{1}{9} + C \quad \left(C = \frac{17}{18}\right)$. Allow $\alpha ye^{3y}$ and $\beta e^{3y}$. Must have integrated LHS twice. |
| Use $y=1$ | **DM1** | $\frac{-(1+1)}{3e^3} - 1(9e^3) = \frac{1}{2}\tan\theta - \frac{17}{18}$. Must have integrated LHS. |
| Obtain $\tan\theta = \frac{17}{9} - \frac{14}{9}e^{-3}$ | **A1** | Or exact equivalent. Exact ISW. Allow $\theta = \tan^{-1}\!\left(\frac{17}{9} - \frac{14}{9}e^{-3}\right)$. If $x$ instead of $\theta$ then withhold final **A1**. |
11 The variables $y$ and $\theta$ satisfy the differential equation
$$( 1 + y ) ( 1 + \cos 2 \theta ) \frac { d y } { d \theta } = e ^ { 3 y }$$
It is given that $y = 0$ when $\theta = \frac { 1 } { 4 } \pi$.\\
Solve the differential equation and find the exact value of $\tan \theta$ when $y = 1$.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.\\
\hfill \mbox{\textit{CAIE P3 2024 Q11 [9]}}