| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Solve via substitution then back-substitute |
| Difficulty | Challenging +1.8 This is a Further Maths FP1 question requiring a substitution to transform a variable-coefficient second-order ODE into constant-coefficient form, then solving and back-substituting. While the substitution is given, students must correctly apply the product rule twice, manage algebraic manipulation with multiple terms, solve the resulting constant-coefficient equation (requiring particular integral for resonance case), and back-substitute. The multi-step nature, algebraic complexity, and resonance handling make this significantly harder than average A-level questions, though the substitution being provided reduces the difficulty somewhat. |
| Spec | 4.10e Second order non-homogeneous: complementary + particular integral4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use of \(x = ty\) to give \(\frac{dx}{dt} = y + t\frac{dy}{dt}\) or \(y = \frac{x}{t} \rightarrow \frac{dy}{dt} = -\frac{x}{t^2} + \frac{1}{t}\frac{dx}{dt}\) | B1 | For a correct suitable first derivative expression linking \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) |
| \(\frac{d^2x}{dt^2} = \frac{dy}{dt} + \frac{dy}{dt} + t\frac{d^2y}{dt^2}\) or \(\frac{d^2y}{dt^2} = -\frac{1}{t^2}\frac{dx}{dt} + \frac{2x}{t^3} + \frac{1}{t}\frac{d^2x}{dt^2} - \frac{1}{t^2}\frac{dx}{dt}\) | M1 | Uses product rule to find equation linking second derivatives from their first derivative expression |
| Correct second derivative expression | A1 | A correct second derivative expression |
| \(t^2\left[\text{their } \frac{d^2x}{dt^2}\right] - 2t\left[\text{their } \frac{dx}{dt}\right] + 2[ty] + 16t^2[ty] = 4t^3\sin 2t\) or \(\left[\text{their } \frac{d^2y}{dt^2}\right] + 16\frac{x}{t} = 4\sin 2t\) | dM1 | Substitutes first and second derivatives and replaces \(x\) with \(ty\); dependent on previous M1 |
| \(t^3\frac{d^2y}{dt^2} + 16t^3y = 4t^3\sin 2t \Rightarrow \frac{d^2y}{dt^2} + 16y = 4\sin 2t\) | A1 | Simplifies with correct intermediate stage to reach printed answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Solves \(m^2 + 16 = 0\) to give \(m = \ldots\) | M1 | Forms correct auxiliary equation and attempts to solve |
| \(y = A\cos 4t + B\sin 4t\) | A1 | Correct complementary function; accept in terms of \(x\) |
| Particular integral \((y =)\ \lambda\sin 2t + \mu\cos 2t\) | B1 | Deduces correct form of PI (must include at least \(\lambda\sin 2t\)) |
| \(\frac{dy}{dt} = 2\lambda\cos 2t - 2\mu\sin 2t\) and \(\frac{d^2y}{dt^2} = -4\lambda\sin 2t - 4\mu\cos 2t\) | M1 | Differentiates the PI twice |
| \([-4\lambda\sin 2t - 4\mu\cos 2t] + 16[\lambda\sin 2t + \mu\cos 2t] = 4\sin 2t \Rightarrow \lambda = \ldots,\ \mu = \ldots\) | dM1 | Substitutes \(y\) and \(\frac{d^2y}{dt^2}\) into DE leading to values for constants; dependent on previous M1 |
| \(y = A\cos 4t + B\sin 4t + \frac{1}{3}\sin 2t\) | A1ft | Correct general equation for \(y\); must be in terms of \(t\), start \(y = \ldots\), CF must be non-constant function of \(t\) |
| \(x = t[\text{their } y]\) | M1 | Links solution to model equation to find general solution for displacement |
| \(x = t\left[A\cos 4t + B\sin 4t + \frac{1}{3}\sin 2t\right]\) | A1 | Correct general solution for displacement; must be \(x = \ldots\) |
# Question 9:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of $x = ty$ to give $\frac{dx}{dt} = y + t\frac{dy}{dt}$ or $y = \frac{x}{t} \rightarrow \frac{dy}{dt} = -\frac{x}{t^2} + \frac{1}{t}\frac{dx}{dt}$ | B1 | For a correct suitable first derivative expression linking $\frac{dy}{dt}$ and $\frac{dx}{dt}$ |
| $\frac{d^2x}{dt^2} = \frac{dy}{dt} + \frac{dy}{dt} + t\frac{d^2y}{dt^2}$ or $\frac{d^2y}{dt^2} = -\frac{1}{t^2}\frac{dx}{dt} + \frac{2x}{t^3} + \frac{1}{t}\frac{d^2x}{dt^2} - \frac{1}{t^2}\frac{dx}{dt}$ | M1 | Uses product rule to find equation linking second derivatives from their first derivative expression |
| Correct second derivative expression | A1 | A correct second derivative expression |
| $t^2\left[\text{their } \frac{d^2x}{dt^2}\right] - 2t\left[\text{their } \frac{dx}{dt}\right] + 2[ty] + 16t^2[ty] = 4t^3\sin 2t$ or $\left[\text{their } \frac{d^2y}{dt^2}\right] + 16\frac{x}{t} = 4\sin 2t$ | dM1 | Substitutes first and second derivatives and replaces $x$ with $ty$; dependent on previous M1 |
| $t^3\frac{d^2y}{dt^2} + 16t^3y = 4t^3\sin 2t \Rightarrow \frac{d^2y}{dt^2} + 16y = 4\sin 2t$ | A1 | Simplifies with correct intermediate stage to reach printed answer |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Solves $m^2 + 16 = 0$ to give $m = \ldots$ | M1 | Forms correct auxiliary equation and attempts to solve |
| $y = A\cos 4t + B\sin 4t$ | A1 | Correct complementary function; accept in terms of $x$ |
| Particular integral $(y =)\ \lambda\sin 2t + \mu\cos 2t$ | B1 | Deduces correct form of PI (must include at least $\lambda\sin 2t$) |
| $\frac{dy}{dt} = 2\lambda\cos 2t - 2\mu\sin 2t$ and $\frac{d^2y}{dt^2} = -4\lambda\sin 2t - 4\mu\cos 2t$ | M1 | Differentiates the PI twice |
| $[-4\lambda\sin 2t - 4\mu\cos 2t] + 16[\lambda\sin 2t + \mu\cos 2t] = 4\sin 2t \Rightarrow \lambda = \ldots,\ \mu = \ldots$ | dM1 | Substitutes $y$ and $\frac{d^2y}{dt^2}$ into DE leading to values for constants; dependent on previous M1 |
| $y = A\cos 4t + B\sin 4t + \frac{1}{3}\sin 2t$ | A1ft | Correct general equation for $y$; must be in terms of $t$, start $y = \ldots$, CF must be non-constant function of $t$ |
| $x = t[\text{their } y]$ | M1 | Links solution to model equation to find general solution for displacement |
| $x = t\left[A\cos 4t + B\sin 4t + \frac{1}{3}\sin 2t\right]$ | A1 | Correct general solution for displacement; must be $x = \ldots$ |\begin{enumerate}
\item A particle $P$ moves along a straight line.
\end{enumerate}
At time $t$ minutes, the displacement, $x$ metres, of $P$ from a fixed point $O$ on the line is modelled by the differential equation
$$t ^ { 2 } \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 2 t \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 x + 16 t ^ { 2 } x = 4 t ^ { 3 } \sin 2 t$$
(a) Show that the transformation $x =$ ty transforms equation (I) into the equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 16 y = 4 \sin 2 t$$
(b) Hence find a general solution for the displacement of $P$ from $O$ at time $t$ minutes.
\hfill \mbox{\textit{Edexcel FP1 2022 Q9 [13]}}