| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2010 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Resonance cases requiring modified PI |
| Difficulty | Challenging +1.2 This is a standard Further Maths resonance case where the RHS matches a complementary function root, requiring the modified PI form y = pxe^(-2x). Part (a) is routine substitution and differentiation, while part (b) requires finding the complementary function and applying initial conditions. The question is methodical rather than insightful, making it moderately above average difficulty for A-level but standard for FP3. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| \(-pe^{-2x} = 2e^{-2x} \Rightarrow p = -2\) | M1, A1, M1, A1F | Product Rule used; Sub. into DE; ft one slip in differentiation; 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Aux. eqn. \(m^2 + 3m + 2 = 0 \Rightarrow m = -1, -2\) | B1 | |
| CF is \(Ae^{-x} + Be^{-2x}\) | M1 | ft on real values of m only |
| GS \(y = Ae^{-x} + Be^{-2x} - 2xe^{-2x}\) | B1F | Their CF + their PI must have 2 arb consts |
| When \(x = 0, y = 2 \Rightarrow A + B = 2\) | B1F | Must be using GS; ft on wrong non-zero values for p and m |
| \(\frac{dy}{dx} = -Ae^{-x} - 2Be^{-2x} - 2e^{-2x} + 4xe^{-2x}\) | B1F | Must be using GS; ft on wrong non-zero values for p and m |
| When \(x = 0, \frac{dy}{dx} = 0 \Rightarrow -A - 2B - 2 = 0\) | B1F | Must be using GS; ft on wrong non-zero values for p and m and slips in finding \(y'(x)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = 6, B = -4; \quad y = 6e^{-x} - 4e^{-2x} - 2xe^{-2x}\) | m1, A1 | CSO; 8 marks total |
**(a)**
$y\mathbf{r1} = pe^{-2x} \Rightarrow \frac{dy}{dx} = pe^{-2x} - 2pxe^{-2x}$
$\Rightarrow \frac{d^2y}{dx^2} = -2pe^{-2x} - 2pe^{-2x} + 4pxe^{-2x} - 3pe^{-2x} - 6pxe^{-2x} +$
$-4pe^{-2x} + 4pxe^{-2x} + 3pe^{-2x} - 6pxe^{-2x} +$
$2pxe^{-2x} = 2e^{-2x}.$
$-pe^{-2x} = 2e^{-2x} \Rightarrow p = -2$ | M1, A1, M1, A1F | Product Rule used; Sub. into DE; ft one slip in differentiation; 4 marks total
**(b)**
Aux. eqn. $m^2 + 3m + 2 = 0 \Rightarrow m = -1, -2$ | B1 |
CF is $Ae^{-x} + Be^{-2x}$ | M1 | ft on real values of m only
GS $y = Ae^{-x} + Be^{-2x} - 2xe^{-2x}$ | B1F | Their CF + their PI must have 2 arb consts
When $x = 0, y = 2 \Rightarrow A + B = 2$ | B1F | Must be using GS; ft on wrong non-zero values for p and m
$\frac{dy}{dx} = -Ae^{-x} - 2Be^{-2x} - 2e^{-2x} + 4xe^{-2x}$ | B1F | Must be using GS; ft on wrong non-zero values for p and m
When $x = 0, \frac{dy}{dx} = 0 \Rightarrow -A - 2B - 2 = 0$ | B1F | Must be using GS; ft on wrong non-zero values for p and m and slips in finding $y'(x)$
Solving simultaneously, 2 eqns each in two arbitrary constants
$A = 6, B = -4; \quad y = 6e^{-x} - 4e^{-2x} - 2xe^{-2x}$ | m1, A1 | CSO; 8 marks total
**Total: 12 marks**
---5 It is given that $y$ satisfies the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 3 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 2 \mathrm { e } ^ { - 2 x }$$
\begin{enumerate}[label=(\alph*)]
\item Find the value of the constant $p$ for which $y = p x \mathrm { e } ^ { - 2 x }$ is a particular integral of the given differential equation.
\item Solve the differential equation, expressing $y$ in terms of $x$, given that $y = 2$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$ when $x = 0$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2010 Q5 [12]}}