| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Year | 2018 |
| Session | September |
| Marks | 8 |
| Topic | First order differential equations (integrating factor) |
| Type | Applied/modelling contexts |
| Difficulty | Standard +0.8 This is a Further Maths question requiring integrating factor method with a non-trivial RHS (e^(-0.1t)cos t), leading to integration by parts twice. The multi-part structure, applied context requiring interpretation, and the computational complexity of solving with the trigonometric forcing function place it moderately above average difficulty, though it follows a standard template for FM first-order linear ODEs. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dx}{dt} + 0.1x = \cos e^{-0.1t} \Rightarrow \text{I.F.} = e^{0.1t}\) | B1 [3] | |
| \(\Rightarrow e^{0.1t} \times \frac{dx}{dt} + 0.1e^{0.1t} \times x = \cos t\) | M1 [3] | Multiplying throughout by IF and writing RHS as an exact derivative of a product |
| \(\Rightarrow \frac{d}{dx}(e^{0.1t}x) = \cos t\) | M1 [3] | |
| \(\Rightarrow e^{0.1t}x = \sin t + c \Rightarrow x = (\sin t + c)e^{-0.1t}\) | A1 [3] | |
| (ii) \(x = (\sin t + c)e^{-0.1t}; x = 10, t = 0 \Rightarrow c = 10\) | M1 [2] | 3.4 |
| \(x = (\sin t + 10)e^{-0.1t}\) | A1 [2] | 3.4 |
| (iii) (a) 5.33477 | B1 [1] | 3.4 |
| (iii) (b) 5.33486 | B1 [1] | 3.4 |
| (iv) E.g. the amount at 6.25 hours is higher than at 6 hours, so the amount is not always decreasing, and so model is not appropriate (for predicting decay) | B1 [1] | 3.5a |
**(i)** $\frac{dx}{dt} + 0.1x = \cos e^{-0.1t} \Rightarrow \text{I.F.} = e^{0.1t}$ | B1 [3] |
$\Rightarrow e^{0.1t} \times \frac{dx}{dt} + 0.1e^{0.1t} \times x = \cos t$ | M1 [3] | Multiplying throughout by IF and writing RHS as an exact derivative of a product
$\Rightarrow \frac{d}{dx}(e^{0.1t}x) = \cos t$ | M1 [3] |
$\Rightarrow e^{0.1t}x = \sin t + c \Rightarrow x = (\sin t + c)e^{-0.1t}$ | A1 [3]
**(ii)** $x = (\sin t + c)e^{-0.1t}; x = 10, t = 0 \Rightarrow c = 10$ | M1 [2] | 3.4 | Substituting $x = 10$ and $t = 0$ to find a value for $c$
$x = (\sin t + 10)e^{-0.1t}$ | A1 [2] | 3.4
**(iii) (a)** 5.33477 | B1 [1] | 3.4 | to 6sf or better
**(iii) (b)** 5.33486 | B1 [1] | 3.4 | to 6sf or better
**(iv)** E.g. the amount at 6.25 hours is higher than at 6 hours, so the amount is not always decreasing, and so model is not appropriate (for predicting decay) | B1 [1] | 3.5a | Conclusion required11 A particular radioactive substance decays over time.\\
A scientist models the amount of substance, $x$ grams, at time $t$ hours by the differential equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } + \frac { 1 } { 10 } x = \mathrm { e } ^ { - 0.1 t } \cos t .$$
(i) Solve the differential equation to find the general solution for $x$ in terms of $t$.
Initially there was 10 g of the substance.\\
(ii) Find the particular solution of the differential equation.\\
(iii) Find to 6 significant figures the amount of substance that would be predicted by the model at
\begin{enumerate}[label=(\alph*)]
\item 6 hours,
\item 6.25 hours.\\
(iv) Comment on the appropriateness of the model for predicting the amount of substance over time.
\section*{END OF QUESTION PAPER}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 1 2018 Q11 [8]}}