| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 17 |
| 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 challenging Further Maths question requiring substitution to transform a second-order DE with variable coefficients into constant coefficient form, then solving and back-substituting. The transformation involves careful chain rule application for second derivatives, solving a non-homogeneous DE with auxiliary equation, particular integral by inspection/undetermined coefficients, and converting back to x. While systematic, it demands technical proficiency across multiple steps with significant algebraic manipulation—substantially harder than typical A-level but standard for FP3. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = e^{2t}\), so \(t = \frac{1}{2}\ln x\) | ||
| \(\frac{dx}{dt} = 2e^{2t}\) | B1 | Correct expression for \(\frac{dx}{dt}\) |
| \(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{1}{2e^{2t}}\frac{dy}{dt}\) | M1 | Chain rule applied correctly |
| \(\frac{d^2y}{dx^2} = \frac{1}{4e^{4t}}\left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right)\) | M1 A1 | Correct second derivative expression |
| Substituting \(\sqrt{x} = e^t\), \(\sqrt{x^5} = e^{5t}\), \(\ln x = 2t\) into original equation | M1 | Correct substitution of terms |
| \(4e^{5t} \cdot \frac{1}{4e^{4t}}\left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right) + 2e^t \cdot \frac{1}{2e^{2t}}\frac{dy}{dt} \cdot 2e^{2t} \cdot y = e^t(2t)^2 + 5\) | M1 | Correct assembly of terms |
| \(\frac{d^2y}{dt^2} - 2\frac{dy}{dt} + 2y = 4t^2 + 5e^{-t}\) | A1 | Correct final equation obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Auxiliary equation: \(m^2 - 2m + 2 = 0\) | M1 | Correct auxiliary equation |
| \(m = \frac{2 \pm \sqrt{4-8}}{2} = 1 \pm i\) | A1 | Correct complex roots |
| Complementary function: \(y = e^t(A\cos t + B\sin t)\) | A1 | Correct CF from complex roots |
| For particular integral, try \(y = at^2 + bt + c\) | M1 | Appropriate PI form for \(4t^2\) |
| Substituting and comparing: \(2a = 4\), \(-4a + 2b = 0\), \(2a - 2b + 2c = 0\) | M1 | Correct substitution and equating |
| \(a = 2\), \(b = 4\), \(c = 2\) | A1 | Correct values |
| For \(5e^{-t}\): try \(y = ke^{-t}\), giving \(k + 2k + 2k = 5\), so \(k = 1\) | M1 A1 | Correct PI for exponential term |
| General solution in \(t\): \(y = e^t(A\cos t + B\sin t) + 2t^2 + 4t + 2 + e^{-t}\) | A1 | Correct GS in terms of \(t\) |
| Back-substitute \(t = \frac{1}{2}\ln x\): \(y = \sqrt{x}\left(A\cos(\frac{1}{2}\ln x) + B\sin(\frac{1}{2}\ln x)\right) + 2(\ln x)^2/... + x^{-1/2}\) | M1 A1 | Correct back-substitution using \(e^t = \sqrt{x}\), \(t = \frac{1}{2}\ln x\) |
# Question 6:
## Part (a): [7 marks]
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = e^{2t}$, so $t = \frac{1}{2}\ln x$ | | |
| $\frac{dx}{dt} = 2e^{2t}$ | B1 | Correct expression for $\frac{dx}{dt}$ |
| $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{1}{2e^{2t}}\frac{dy}{dt}$ | M1 | Chain rule applied correctly |
| $\frac{d^2y}{dx^2} = \frac{1}{4e^{4t}}\left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right)$ | M1 A1 | Correct second derivative expression |
| Substituting $\sqrt{x} = e^t$, $\sqrt{x^5} = e^{5t}$, $\ln x = 2t$ into original equation | M1 | Correct substitution of terms |
| $4e^{5t} \cdot \frac{1}{4e^{4t}}\left(\frac{d^2y}{dt^2} - \frac{dy}{dt}\right) + 2e^t \cdot \frac{1}{2e^{2t}}\frac{dy}{dt} \cdot 2e^{2t} \cdot y = e^t(2t)^2 + 5$ | M1 | Correct assembly of terms |
| $\frac{d^2y}{dt^2} - 2\frac{dy}{dt} + 2y = 4t^2 + 5e^{-t}$ | A1 | Correct final equation obtained |
## Part (b): [10 marks]
| Answer/Working | Mark | Guidance |
|---|---|---|
| Auxiliary equation: $m^2 - 2m + 2 = 0$ | M1 | Correct auxiliary equation |
| $m = \frac{2 \pm \sqrt{4-8}}{2} = 1 \pm i$ | A1 | Correct complex roots |
| Complementary function: $y = e^t(A\cos t + B\sin t)$ | A1 | Correct CF from complex roots |
| For particular integral, try $y = at^2 + bt + c$ | M1 | Appropriate PI form for $4t^2$ |
| Substituting and comparing: $2a = 4$, $-4a + 2b = 0$, $2a - 2b + 2c = 0$ | M1 | Correct substitution and equating |
| $a = 2$, $b = 4$, $c = 2$ | A1 | Correct values |
| For $5e^{-t}$: try $y = ke^{-t}$, giving $k + 2k + 2k = 5$, so $k = 1$ | M1 A1 | Correct PI for exponential term |
| General solution in $t$: $y = e^t(A\cos t + B\sin t) + 2t^2 + 4t + 2 + e^{-t}$ | A1 | Correct GS in terms of $t$ |
| Back-substitute $t = \frac{1}{2}\ln x$: $y = \sqrt{x}\left(A\cos(\frac{1}{2}\ln x) + B\sin(\frac{1}{2}\ln x)\right) + 2(\ln x)^2/... + x^{-1/2}$ | M1 A1 | Correct back-substitution using $e^t = \sqrt{x}$, $t = \frac{1}{2}\ln x$ |6 A differential equation is given by
$$4 \sqrt { x ^ { 5 } } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 2 \sqrt { x } ) y = \sqrt { x } ( \ln x ) ^ { 2 } + 5 , \quad x > 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that the substitution $x = \mathrm { e } ^ { 2 t }$ transforms this differential equation into
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 2 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 2 y = 4 t ^ { 2 } + 5 \mathrm { e } ^ { - t }$$
\item Hence find the general solution of the differential equation
$$4 \sqrt { x ^ { 5 } } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 2 \sqrt { x } ) y = \sqrt { x } ( \ln x ) ^ { 2 } + 5 , \quad x > 0$$
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{7b4a1237-bb28-4cba-84b1-35fa91d87408-14_1634_1709_1071_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2015 Q6 [17]}}