| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Particular solution with initial conditions |
| Difficulty | Standard +0.8 This is a Further Maths FP3 question requiring solution of a second-order linear ODE with non-homogeneous terms (exponential and constant). Students must find the complementary function (auxiliary equation with real roots), determine particular integrals for two different forcing terms, then apply initial conditions including an asymptotic constraint. While systematic, it requires multiple techniques and careful algebra across 10 marks, placing it moderately above average difficulty. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Aux. eqn: \(m^2 - 2m - 3 = 0\) | M1 | |
| \(m = -1,\ 3\) | A1 | PI |
| CF: \(y_C = Ae^{3x} + Be^{-x}\) | M1 | |
| Try \(y_{PI} = ae^{-2x}\ (+b)\) | M1 | |
| \(\dfrac{dy}{dx} = -2ae^{-2x}\) | A1 | |
| \(\dfrac{d^2y}{dx^2} = 4ae^{-2x}\) | A1 | |
| Substitute into DE: \(4ae^{-2x} + 4ae^{-2x} - 3ae^{-2x} - 3b = 10e^{-2x} - 9\) | M1 | |
| \(\Rightarrow a = 2\) | A1 | |
| \(b = 3\) | B1 | |
| \(y_{GS} = Ae^{3x} + Be^{-x} + 2e^{-2x} + 3\) | B1F | 10 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x=0,\ y=7 \Rightarrow 7 = A + B + 2 + 3\) | B1F | |
| \(\dfrac{dy}{dx} = 3Ae^{3x} - Be^{-x} - 4e^{-2x}\) | ||
| As \(x\to\infty\), \(e^{-kx}\to 0\), \(\dfrac{dy}{dx}\to 0\) so \(A = 0\) | B1 | |
| When \(A=0\): \(5 = 0 + B + 3 \Rightarrow B = 2\) | B1F | |
| \(y = 2e^{-x} + 2e^{-2x} + 3\) | A1 | 4 |
| Total | 14 |
## Question 6(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Aux. eqn: $m^2 - 2m - 3 = 0$ | M1 | |
| $m = -1,\ 3$ | A1 | PI |
| CF: $y_C = Ae^{3x} + Be^{-x}$ | M1 | |
| Try $y_{PI} = ae^{-2x}\ (+b)$ | M1 | |
| $\dfrac{dy}{dx} = -2ae^{-2x}$ | A1 | |
| $\dfrac{d^2y}{dx^2} = 4ae^{-2x}$ | A1 | |
| Substitute into DE: $4ae^{-2x} + 4ae^{-2x} - 3ae^{-2x} - 3b = 10e^{-2x} - 9$ | M1 | |
| $\Rightarrow a = 2$ | A1 | |
| $b = 3$ | B1 | |
| $y_{GS} = Ae^{3x} + Be^{-x} + 2e^{-2x} + 3$ | B1F | 10 | (c's CF + c's PI) with 2 arbitrary constants |
## Question 6(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x=0,\ y=7 \Rightarrow 7 = A + B + 2 + 3$ | B1F | | Only ft if exponentials in GS and two arbitrary constants remain |
| $\dfrac{dy}{dx} = 3Ae^{3x} - Be^{-x} - 4e^{-2x}$ | | |
| As $x\to\infty$, $e^{-kx}\to 0$, $\dfrac{dy}{dx}\to 0$ so $A = 0$ | B1 | |
| When $A=0$: $5 = 0 + B + 3 \Rightarrow B = 2$ | B1F | | Must be using $A = 0$ |
| $y = 2e^{-x} + 2e^{-2x} + 3$ | A1 | 4 | CSO |
| **Total** | **14** | |6
\begin{enumerate}[label=(\alph*)]
\item Find the general solution of the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } - 3 y = 10 \mathrm { e } ^ { - 2 x } - 9$$
(10 marks)
\item Hence express $y$ in terms of $x$, given that $y = 7$ when $x = 0$ and that $\frac { \mathrm { d } y } { \mathrm {~d} x } \rightarrow 0$ as $x \rightarrow \infty$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2008 Q6 [14]}}