| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Solve via substitution then back-substitute |
| Difficulty | Challenging +1.2 This is a standard FP2 substitution question with clearly signposted steps. Part (a) requires careful but routine differentiation using the chain rule, part (b) is a standard second-order linear DE with constant coefficients, and parts (c)-(d) involve straightforward back-substitution and boundary conditions. While it requires multiple techniques and careful algebra, the structure is entirely predictable for FP2 students and follows a well-practiced template. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(y = x^{-2} \Rightarrow \frac{dy}{dt} = -2x^{-3} \frac{dx}{dt} = -2x^{-3}t\) [Use of chain rule; need \(\frac{dx}{dt}\)] | M1 | |
| \(\Rightarrow \frac{d^2y}{dt^2} = -2x^{-3} \frac{d^2x}{dt^2} + 6x^{-4}\left(\frac{dx}{dt}\right)^2\) | A1ft, M1A1 | |
| \((\div \text{ given d.e. by } x^4) \quad \frac{2}{x^3}\frac{d^2x}{dt^2} - \frac{6}{x^4}\left(\frac{dx}{dt}\right)^2 = \frac{1}{x^2} - 3\) | ||
| \(\left(\div \text{ given d.e. by } x^4: \frac{d^2y}{dt^2} = y - 3\right) \quad \frac{d^2y}{dt^2} + y = 3\) | AG | A1cso5 |
| Second M1 is for attempt at product rule (be generous). Final A1 requires all working correct and sufficient "substitution" work | ||
| (b) Auxiliary equation: \(m^2 + 1 = 0\) and produce Complementary Function \(y = \ldots\) | M1 | |
| \((y) = A\cos t + B\sin t\) | A1cao | |
| Particular integral: \(y = 3\) | B1 | |
| \(\therefore\) General solution: \((y) = A\cos t + B\sin t + 3\) | A1ft4 | |
| Answer can be stated; M1 is implied by correct C.F. stated (allow \(\theta\) for \(t\)). A1 f.t. for candidates CF + PI. Allow \(m^2 + m = 0\) and \(m^2 - 1 = 0\) for M1. Marks for (b) can be gained in (c) | ||
| (c) \(\frac{1}{x^2} = A\cos t + B\sin t + 3\) | ||
| \(x = \frac{1}{2}, t = 0 \Rightarrow (4 = A + 3)\) A = 1 | B1 | |
| Differentiating (to include \(\frac{dx}{dt}\)): \(-2x^{-3}\frac{dx}{dt} = -A\sin t + B\cos t\) | M1 | |
| \(\frac{dx}{dt} = 0, t = 0 \Rightarrow (0 = 0 + B)\) | M1 | |
| \(B = 0\) | M1 | |
| \(\therefore \frac{1}{x^2} = 3 + \cos t\) so \(x = \frac{1}{\sqrt{3+\cos t}}\) | A1cao4 | |
| Second M: complete method to find other constant (This may involve solving two equations in A and B) | ||
| (d) (Max. value of \(x\) when \(\cos t = -1\)) so max \(x = \frac{1}{\sqrt{2}}\) or AWRT 0.707 | B11 |
**(a)** $y = x^{-2} \Rightarrow \frac{dy}{dt} = -2x^{-3} \frac{dx}{dt} = -2x^{-3}t$ [Use of chain rule; need $\frac{dx}{dt}$] | M1 |
$\Rightarrow \frac{d^2y}{dt^2} = -2x^{-3} \frac{d^2x}{dt^2} + 6x^{-4}\left(\frac{dx}{dt}\right)^2$ | A1ft, M1A1 |
$(\div \text{ given d.e. by } x^4) \quad \frac{2}{x^3}\frac{d^2x}{dt^2} - \frac{6}{x^4}\left(\frac{dx}{dt}\right)^2 = \frac{1}{x^2} - 3$ | |
$\left(\div \text{ given d.e. by } x^4: \frac{d^2y}{dt^2} = y - 3\right) \quad \frac{d^2y}{dt^2} + y = 3$ | AG | A1cso5 |
Second M1 is for attempt at product rule (be generous). Final A1 requires all working correct and sufficient "substitution" work | |
**(b)** Auxiliary equation: $m^2 + 1 = 0$ and produce Complementary Function $y = \ldots$ | M1 |
$(y) = A\cos t + B\sin t$ | A1cao |
Particular integral: $y = 3$ | B1 |
$\therefore$ General solution: $(y) = A\cos t + B\sin t + 3$ | A1ft4 |
Answer can be stated; M1 is implied by correct C.F. stated (allow $\theta$ for $t$). A1 f.t. for candidates CF + PI. Allow $m^2 + m = 0$ and $m^2 - 1 = 0$ for M1. Marks for (b) can be gained in (c) | |
**(c)** $\frac{1}{x^2} = A\cos t + B\sin t + 3$ | |
$x = \frac{1}{2}, t = 0 \Rightarrow (4 = A + 3)$ A = 1 | B1 |
Differentiating (to include $\frac{dx}{dt}$): $-2x^{-3}\frac{dx}{dt} = -A\sin t + B\cos t$ | M1 |
$\frac{dx}{dt} = 0, t = 0 \Rightarrow (0 = 0 + B)$ | M1 |
$B = 0$ | M1 |
$\therefore \frac{1}{x^2} = 3 + \cos t$ so $x = \frac{1}{\sqrt{3+\cos t}}$ | A1cao4 |
Second M: complete method to find other constant (This may involve solving two equations in A and B) | |
**(d)** (Max. value of $x$ when $\cos t = -1$) so max $x = \frac{1}{\sqrt{2}}$ or AWRT 0.707 | B11 |
**Total: [14]**
---3. A scientist is modelling the amount of a chemical in the human bloodstream. The amount $x$ of the chemical, measured in $\mathrm { mg } l ^ { - 1 }$, at time $t$ hours satisfies the differential equation
$$2 x \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 6 \left( \frac { \mathrm { dx } } { \mathrm { dt } } \right) ^ { 2 } = x ^ { 2 } - 3 x ^ { 4 } , \quad x > 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that the substitution $\mathrm { y } = \frac { 1 } { x ^ { 2 } }$ transforms this differential equation into
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + y = 3$$
\item Find the general solution of differential equation $I$.
Given that at time $t = 0 , x = \frac { 1 } { 2 }$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = 0$,
\item find an expression for $x$ in terms of $t$,
\item write down the maximum value of $x$ as $t$ varies.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2007 Q3 [14]}}