| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Solve via substitution then back-substitute |
| Difficulty | Challenging +1.2 This is a structured Further Maths question with clear guidance through substitution. Part (a) is algebraic verification (routine differentiation and substitution), part (b) requires solving a standard second-order ODE with constant coefficients plus particular integral, and part (c) involves interpretation. While it's Further Maths content, the heavy scaffolding and standard techniques make it moderately above average difficulty rather than highly challenging. |
| Spec | 4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use \(x = tv\) to give \(\frac{dx}{dt} = v + t\frac{dv}{dt}\) | M1 | Uses product rule to obtain first derivative |
| \(\frac{d^2x}{dt^2} = \frac{dv}{dt} + \frac{dv}{dt} + t\frac{d^2v}{dt^2}\) | M1, A1 | Continue differentiating with product and chain rule; correct second derivative |
| Uses \(t^2\left(\text{their } 2^\text{nd}\text{ derivative}\right) - 2t\left(\text{their } 1^\text{st}\text{ derivative}\right) + (2+t^2)x = t^4\) and simplifies LHS | M1 | Shows substitution clearly to form new equation, gathers like terms |
| \(t^3\frac{d^2v}{dt^2} + t^3 v = t^4\) leading to \(\frac{d^2v}{dt^2} + v = t\) | A1* | Fully correct solution leading to given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Solve \(\lambda^2 + 1 = 0\) giving \(\lambda^2 = -1\) | M1 | Form and solve quadratic Auxiliary Equation |
| \(v = A\cos t + B\sin t\) | A1ft | Correct form of Complementary Function |
| Particular Integral is \(v = kt + l\) | B1 | Deduces correct form for PI (note \(v = mt^2 + kt + l\) is fine) |
| \(\frac{dv}{dt} = k\), \(\frac{d^2v}{dt^2} = 0\); solve \(0 + kt + l = t\) giving \(k=1\), \(l=0\) | M1 | Differentiates PI and substitutes to find constants |
| Solution: \(v = A\cos t + B\sin t + t\) | A1 | Correct general solution |
| Displacement: \(x = tv = \ldots\) | M1 | Links solution to \(x = tv\) |
| \(x = t(A\cos t + B\sin t + t)\) | A1 | Correct general solution for displacement |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For large \(t\), displacement gets very large (and positive) | B1 | States as \(t \to \infty\), \(x \to \infty\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Model suggests midpoint of spring has large displacement when \(t\) is large, which is unrealistic; spring may reach elastic limit / will break | B1 | Reflect on context; accept 'model unrealistic' / 'spring will break' |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $x = tv$ to give $\frac{dx}{dt} = v + t\frac{dv}{dt}$ | M1 | Uses product rule to obtain first derivative |
| $\frac{d^2x}{dt^2} = \frac{dv}{dt} + \frac{dv}{dt} + t\frac{d^2v}{dt^2}$ | M1, A1 | Continue differentiating with product and chain rule; correct second derivative |
| Uses $t^2\left(\text{their } 2^\text{nd}\text{ derivative}\right) - 2t\left(\text{their } 1^\text{st}\text{ derivative}\right) + (2+t^2)x = t^4$ and simplifies LHS | M1 | Shows substitution clearly to form new equation, gathers like terms |
| $t^3\frac{d^2v}{dt^2} + t^3 v = t^4$ leading to $\frac{d^2v}{dt^2} + v = t$ | A1* | Fully correct solution leading to given answer |
---
## Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Solve $\lambda^2 + 1 = 0$ giving $\lambda^2 = -1$ | M1 | Form and solve quadratic Auxiliary Equation |
| $v = A\cos t + B\sin t$ | A1ft | Correct form of Complementary Function |
| Particular Integral is $v = kt + l$ | B1 | Deduces correct form for PI (note $v = mt^2 + kt + l$ is fine) |
| $\frac{dv}{dt} = k$, $\frac{d^2v}{dt^2} = 0$; solve $0 + kt + l = t$ giving $k=1$, $l=0$ | M1 | Differentiates PI and substitutes to find constants |
| Solution: $v = A\cos t + B\sin t + t$ | A1 | Correct general solution |
| Displacement: $x = tv = \ldots$ | M1 | Links solution to $x = tv$ |
| $x = t(A\cos t + B\sin t + t)$ | A1 | Correct general solution for displacement |
---
## Question 3(c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For large $t$, displacement gets very large (and positive) | B1 | States as $t \to \infty$, $x \to \infty$ |
## Question 3(c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Model suggests midpoint of spring has large displacement when $t$ is large, which is unrealistic; spring may reach elastic limit / will break | B1 | Reflect on context; accept 'model unrealistic' / 'spring will break' |
---\begin{enumerate}
\item A vibrating spring, fixed at one end, has an external force acting on it such that the centre of the spring moves in a straight line. At time $t$ seconds, $t \geqslant 0$, the displacement of the centre $C$ of the spring from a fixed point $O$ is $x$ micrometres.
\end{enumerate}
The displacement of $C$ from $O$ is modelled by the differential equation
$$t ^ { 2 } \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 2 t \frac { \mathrm {~d} x } { \mathrm {~d} t } + \left( 2 + t ^ { 2 } \right) x = t ^ { 4 }$$
(a) Show that the transformation $x = t v$ transforms equation (I) into the equation
$$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} t ^ { 2 } } + v = t$$
(b) Hence find the general equation for the displacement of $C$ from $O$ at time $t$ seconds.\\
(c) (i) State what happens to the displacement of $C$ from $O$ as $t$ becomes large.\\
(ii) Comment on the model with reference to this long term behaviour.
\hfill \mbox{\textit{Edexcel FP1 Q3 [14]}}