| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | January |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Euler-Cauchy equations via exponential substitution |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on Euler-Cauchy equations with clear scaffolding through parts (a)-(d). Part (a) is a standard chain rule derivation shown in textbooks, parts (b)-(c) involve routine substitution and solving a constant-coefficient DE, and part (d) applies boundary conditions. While it requires multiple techniques and careful algebra, the heavy scaffolding and standard nature of each step place it moderately above average difficulty. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dx}{dt} = e^t \{= x\}\) | B1 | OE |
| \(\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx} = e^{-t}\frac{dy}{dt}\) | M1 A1 | Chain rule; OE eg \(x\frac{dy}{dx} = \frac{dy}{dt}\) |
| \(\frac{d^2y}{dx^2} = \frac{d}{dx}\left(e^{-t}\frac{dy}{dt}\right) = \frac{dt}{dx}\frac{d}{dt}\left(e^{-t}\frac{dy}{dt}\right)\) | M1 | \(\frac{d}{dx}(\ ) = \frac{dt}{dx}\frac{d}{dt}(\ )\) OE |
| \(= \frac{dt}{dx}\left(-e^{-t}\frac{dy}{dt} + e^{-t}\frac{d^2y}{dt^2}\right)\) | M1 | Product rule OE |
| \(= e^{-t}\left(-e^{-t}\frac{dy}{dt} + e^{-t}\frac{d^2y}{dt^2}\right)\) | A1 | OE |
| \(\Rightarrow x^2\frac{d^2y}{dx^2} = \left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right)\) | A1 | 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right) - 4\left(\frac{dy}{dt}\right) = 10\) | M1 | |
| \(\frac{d^2y}{dt^2} - 5\frac{dy}{dt} = 10\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Auxiliary equation \(m^2 - 5m = 0\) | M1 | PI |
| \(m = 0\) and \(5\); CF: \(y_C = A + Be^{5t}\) | A1 M1 | ft wrong values of \(m\) provided 2 arbitrary constants in CF; condone \(x\) for \(t\) |
| PI: \(y_P = -2t\) | B1 | |
| GS of (*): \(\{y\} = A + Be^{5t} - 2t\) | B1ft | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\Rightarrow y = A + Bx^5 - 2\ln x\) | M1 | |
| \(y'(x) = 5Bx^4 - 2x^{-1}\) | A1ft | Must involve differentiating \(a\ln x\); ft slip |
| Using boundary conditions to find \(A\) & \(B\) | M1 | |
| \(B = 2;\ A = -2\); \(\{y = -2 + 2x^5 - 2\ln x\}\) | A1;A1ft | 5 |
# Question 7:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dx}{dt} = e^t \{= x\}$ | B1 | OE |
| $\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx} = e^{-t}\frac{dy}{dt}$ | M1 A1 | Chain rule; OE eg $x\frac{dy}{dx} = \frac{dy}{dt}$ |
| $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(e^{-t}\frac{dy}{dt}\right) = \frac{dt}{dx}\frac{d}{dt}\left(e^{-t}\frac{dy}{dt}\right)$ | M1 | $\frac{d}{dx}(\ ) = \frac{dt}{dx}\frac{d}{dt}(\ )$ OE |
| $= \frac{dt}{dx}\left(-e^{-t}\frac{dy}{dt} + e^{-t}\frac{d^2y}{dt^2}\right)$ | M1 | Product rule OE |
| $= e^{-t}\left(-e^{-t}\frac{dy}{dt} + e^{-t}\frac{d^2y}{dt^2}\right)$ | A1 | OE |
| $\Rightarrow x^2\frac{d^2y}{dx^2} = \left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right)$ | A1 | **7** | CSO AG Completion. Be convinced |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right) - 4\left(\frac{dy}{dt}\right) = 10$ | M1 | |
| $\frac{d^2y}{dt^2} - 5\frac{dy}{dt} = 10$ | A1 | **2** | CSO AG Completion. Be convinced |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| Auxiliary equation $m^2 - 5m = 0$ | M1 | PI |
| $m = 0$ and $5$; CF: $y_C = A + Be^{5t}$ | A1 M1 | ft wrong values of $m$ provided 2 arbitrary constants in CF; condone $x$ for $t$ |
| PI: $y_P = -2t$ | B1 | |
| GS of (*): $\{y\} = A + Be^{5t} - 2t$ | B1ft | **5** | ft on c's CF + PI, provided PI is non-zero and CF has two arbitrary constants |
## Part (d):
| Working | Mark | Guidance |
|---------|------|----------|
| $\Rightarrow y = A + Bx^5 - 2\ln x$ | M1 | |
| $y'(x) = 5Bx^4 - 2x^{-1}$ | A1ft | Must involve differentiating $a\ln x$; ft slip |
| Using boundary conditions to find $A$ & $B$ | M1 | |
| $B = 2;\ A = -2$; $\{y = -2 + 2x^5 - 2\ln x\}$ | A1;A1ft | **5** | ft a slip |7
\begin{enumerate}[label=(\alph*)]
\item Given that $x = \mathrm { e } ^ { t }$ and that $y$ is a function of $x$, show that
$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t }$$
\item Hence show that the substitution $x = \mathrm { e } ^ { t }$ transforms the differential equation
$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 10$$
into
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } = 10$$
\item Find the general solution of the differential equation $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } = 10$.
\item Hence solve the differential equation $x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 10$, given that $y = 0$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = 8$ when $x = 1$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2009 Q7 [19]}}