| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Verify particular integral form |
| Difficulty | Standard +0.8 This is a Further Maths F2 question requiring differentiation of products involving exponentials and trigonometric functions (twice), substitution into a second-order DE, coefficient matching to find A, then finding complementary function from auxiliary equation, and finally applying initial conditions. While systematic, it demands careful algebraic manipulation across multiple steps and is more demanding than standard A-level questions. |
| Spec | 4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(\frac{dy}{dx} = -3e^{-x}\cos 3x - 9e^{-x}\sin 3x - Ae^{-x}\sin 3x + 3Ae^{-x}\cos 3x\) | M1 | Attempts to differentiate given expression using product rule on \(3e^{-x}\cos 3x\) or \(Ae^{-x}\sin 3x\) |
| \(\frac{d^2y}{dx^2} = (-24-6A)e^{-x}\cos 3x + (18-8A)e^{-x}\sin 3x\) | dM1 | Uses product rule again on expression of form \(e^{-x}\sin 3x\) or \(e^{-x}\cos 3x\). Dependent on first M mark |
| \(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 10y = (12-12A)e^{-x}\cos 3x + (36+4A)e^{-x}\sin 3x\) | M1 | Substitutes results into the differential equation |
| \(12 - 12A = 0\) or \(36 + 4A = 40 \Rightarrow A = \ldots\) | dM1 | Compares coefficients of \(e^{-x}\sin 3x\) or \(e^{-x}\cos 3x\) and attempts to find \(A\). Dependent on previous M mark |
| \(A = 1\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(m^2 - 2m + 10 = 0 \Rightarrow m = 1 \pm 3i\) | M1A1 | M1: Forms and attempts to solve Auxiliary Equation; A1: Correct solution |
| \(y = e^x(C\cos 3x + D\sin 3x)\) | M1 | Correct form for CF using complex roots from AE |
| \(y = e^x(C\cos 3x + D\sin 3x) + 3e^{-x}\cos 3x + e^{-x}\sin 3x\) | A1ft | GS = CF + PI. Must start \(y=\ldots\) and depend on at least one M being scored, and must have PI of the form given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| \(x=0, y=3 \Rightarrow 3 = C + 3\ (\Rightarrow C=0)\) | M1 | Attempts to substitute \(x=0\) and \(y=3\) into answer to (b) |
| \(\frac{dy}{dx} = (C+3D)e^x\cos 3x + (-3C+D)e^x\sin 3x - 10e^{-x}\sin 3x\) | M1 | Attempt to differentiate their GS with or without their \(C\) |
| \(3 = C + 3D\) | M1 | Attempt to substitute \(x=0\) and \(\frac{dy}{dx}=3\) into their \(\frac{dy}{dx}\) |
| \(y = e^x\sin 3x + 3e^{-x}\cos 3x + e^{-x}\sin 3x\) | A1cao | Correct answer. Must start \(y=\ldots\) |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $\frac{dy}{dx} = -3e^{-x}\cos 3x - 9e^{-x}\sin 3x - Ae^{-x}\sin 3x + 3Ae^{-x}\cos 3x$ | M1 | Attempts to differentiate given expression using product rule on $3e^{-x}\cos 3x$ or $Ae^{-x}\sin 3x$ |
| $\frac{d^2y}{dx^2} = (-24-6A)e^{-x}\cos 3x + (18-8A)e^{-x}\sin 3x$ | dM1 | Uses product rule again on expression of form $e^{-x}\sin 3x$ or $e^{-x}\cos 3x$. **Dependent on first M mark** |
| $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 10y = (12-12A)e^{-x}\cos 3x + (36+4A)e^{-x}\sin 3x$ | M1 | Substitutes results into the differential equation |
| $12 - 12A = 0$ or $36 + 4A = 40 \Rightarrow A = \ldots$ | dM1 | Compares coefficients of $e^{-x}\sin 3x$ or $e^{-x}\cos 3x$ and attempts to find $A$. **Dependent on previous M mark** |
| $A = 1$ | A1 | cao |
## Part (b):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $m^2 - 2m + 10 = 0 \Rightarrow m = 1 \pm 3i$ | M1A1 | M1: Forms and attempts to solve Auxiliary Equation; A1: Correct solution |
| $y = e^x(C\cos 3x + D\sin 3x)$ | M1 | Correct form for CF using complex roots from AE |
| $y = e^x(C\cos 3x + D\sin 3x) + 3e^{-x}\cos 3x + e^{-x}\sin 3x$ | A1ft | GS = CF + PI. Must start $y=\ldots$ and depend on at least one M being scored, and must have PI of the form given |
## Part (c):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $x=0, y=3 \Rightarrow 3 = C + 3\ (\Rightarrow C=0)$ | M1 | Attempts to substitute $x=0$ and $y=3$ into answer to (b) |
| $\frac{dy}{dx} = (C+3D)e^x\cos 3x + (-3C+D)e^x\sin 3x - 10e^{-x}\sin 3x$ | M1 | Attempt to differentiate their GS with or without their $C$ |
| $3 = C + 3D$ | M1 | Attempt to substitute $x=0$ and $\frac{dy}{dx}=3$ into their $\frac{dy}{dx}$ |
| $y = e^x\sin 3x + 3e^{-x}\cos 3x + e^{-x}\sin 3x$ | A1cao | Correct answer. **Must start $y=\ldots$** |
**[Total: 13]**4.
$$y = 3 \mathrm { e } ^ { - x } \cos 3 x + A \mathrm { e } ^ { - x } \sin 3 x$$
is a particular integral of the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 10 y = 40 \mathrm { e } ^ { - x } \sin 3 x$$
where $A$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $A$.
\item Hence find the general solution of this differential equation.
\item Find the particular solution of this differential equation for which both $y = 3$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 3$ at $x = 0$
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2017 Q4 [13]}}