| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Series solution from differential equation |
| Difficulty | Challenging +1.2 This is a standard Further Pure 3 second-order differential equation question. Part (a) requires routine complementary function (repeated root) and particular integral (standard form for sin/cos). Part (b) involves straightforward differentiation and series expansion using initial conditions. While it requires multiple techniques, all are standard FP3 procedures with no novel insight needed, making it moderately above average difficulty. |
| Spec | 4.08a Maclaurin series: find series for function4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Auxiliary equation: \(m^2 + 6m + 9 = 0 \Rightarrow (m+3)^2 = 0 \Rightarrow m = -3\) (repeated) | M1 | |
| Complementary function: \((A + Bx)e^{-3x}\) | A1 | |
| Particular integral: try \(y = p\cos 3x + q\sin 3x\) | M1 | |
| \(\frac{dy}{dx} = -3p\sin 3x + 3q\cos 3x\), \(\frac{d^2y}{dx^2} = -9p\cos 3x - 9q\sin 3x\) | M1 | Differentiating PI twice |
| Substituting and comparing coefficients: | M1 | |
| \(\cos 3x\): \(-9p - 18q + 9p = 0 \Rightarrow q = 0\) | ||
| \(\sin 3x\): \(-9q + 18p + 9q = 36 \Rightarrow p = 2\) | A1 | Both correct |
| General solution: \(y = (A+Bx)e^{-3x} + 2\cos 3x\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| At \(x=0\): \(0 = A + 2 \Rightarrow A = -2\); \(f'(0)=0\): \(0 = B - 6A - 6 \Rightarrow B = -18+6 = ...\); substituting into ODE at \(x=0\): \(f''(0) + 0 + 0 = 0\) | B1 | Correct substitution showing \(f''(0)=0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Differentiating ODE to find \(f'''(0)\), \(f^{(4)}(0)\) etc. | M1 | |
| \(f'''(0) = -108\), giving term \(\frac{-108}{6}x^3 = -18x^3\) | A1 | |
| Next non-zero term found correctly | A1 | First two non-zero terms correct |
# Question 5:
## Part (a)
| Auxiliary equation: $m^2 + 6m + 9 = 0 \Rightarrow (m+3)^2 = 0 \Rightarrow m = -3$ (repeated) | M1 | |
| Complementary function: $(A + Bx)e^{-3x}$ | A1 | |
| Particular integral: try $y = p\cos 3x + q\sin 3x$ | M1 | |
| $\frac{dy}{dx} = -3p\sin 3x + 3q\cos 3x$, $\frac{d^2y}{dx^2} = -9p\cos 3x - 9q\sin 3x$ | M1 | Differentiating PI twice |
| Substituting and comparing coefficients: | M1 | |
| $\cos 3x$: $-9p - 18q + 9p = 0 \Rightarrow q = 0$ | | |
| $\sin 3x$: $-9q + 18p + 9q = 36 \Rightarrow p = 2$ | A1 | Both correct |
| General solution: $y = (A+Bx)e^{-3x} + 2\cos 3x$ | A1 | |
## Part (b)(i)
| At $x=0$: $0 = A + 2 \Rightarrow A = -2$; $f'(0)=0$: $0 = B - 6A - 6 \Rightarrow B = -18+6 = ...$; substituting into ODE at $x=0$: $f''(0) + 0 + 0 = 0$ | B1 | Correct substitution showing $f''(0)=0$ |
## Part (b)(ii)
| Differentiating ODE to find $f'''(0)$, $f^{(4)}(0)$ etc. | M1 | |
| $f'''(0) = -108$, giving term $\frac{-108}{6}x^3 = -18x^3$ | A1 | |
| Next non-zero term found correctly | A1 | First two non-zero terms correct |5
\begin{enumerate}[label=(\alph*)]
\item Find the general solution of the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 6 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 9 y = 36 \sin 3 x$$
\item It is given that $y = \mathrm { f } ( x )$ is the solution of the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 6 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 9 y = 36 \sin 3 x$$
such that $\mathrm { f } ( 0 ) = 0$ and $\mathrm { f } ^ { \prime } ( 0 ) = 0$.
\begin{enumerate}[label=(\roman*)]
\item Show that $f ^ { \prime \prime } ( 0 ) = 0$.
\item Find the first two non-zero terms in the expansion, in ascending powers of $x$, of $\mathrm { f } ( x )$.\\[0pt]
[3 marks]
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2015 Q5 [11]}}