| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2005 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Finding constants from data |
| Difficulty | Standard +0.3 This is a structured differential equations question with clear scaffolding through parts (i)-(iv). The partial fractions decomposition is straightforward, the integration follows standard techniques, and finding K from initial conditions is routine. While it requires multiple steps, each is a standard C4 technique with no novel insight needed, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08h Integration by substitution1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks |
|---|---|
| The length of the next interval = l, where \(\frac{0.0952...}{l} = 4.669...\) | M1 |
| \(\Rightarrow l = 0.0203\) | A1 |
| So next bifurcation at \(3.5437... + 0.0203... ≈ 3.564\) | DM1 A1 [4] |
The length of the next interval = l, where $\frac{0.0952...}{l} = 4.669...$ | M1 |
$\Rightarrow l = 0.0203$ | A1 |
So next bifurcation at $3.5437... + 0.0203... ≈ 3.564$ | DM1 A1 [4] |7 In a chemical process, the mass $M$ grams of a chemical at time $t$ minutes is modelled by the differential equation
$$\frac { \mathrm { d } M } { \mathrm {~d} t } = \frac { M } { t \left( 1 + t ^ { 2 } \right) }$$
(i) Find $\int \frac { t } { 1 + t ^ { 2 } } \mathrm {~d} t$.\\
(ii) Find constants $A , B$ and $C$ such that
$$\frac { 1 } { t \left( 1 + t ^ { 2 } \right) } = \frac { A } { t } + \frac { B t + C } { 1 + t ^ { 2 } } .$$
(iii) Use integration, together with your results in parts (i) and (ii), to show that
$$M = \frac { K t } { \sqrt { 1 + t ^ { 2 } } } ,$$
where $K$ is a constant.\\
(iv) When $t = 1 , M = 25$. Calculate $K$.
What is the mass of the chemical in the long term?
\hfill \mbox{\textit{OCR MEI C4 2005 Q7 [18]}}