| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Solve via substitution then back-substitute |
| Difficulty | Challenging +1.2 This is a structured FP3 question where part (a) guides students through verifying a given substitution using the product rule and chain rule, then part (b) requires solving a standard constant-coefficient second-order DE and back-substituting. While it involves multiple differentiation steps and solving a non-homogeneous DE, the substitution is provided and the techniques are all standard for Further Maths students. The 8-mark allocation and scaffolded structure make it moderately above average difficulty but not exceptionally challenging. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
7 It is given that, for $x \neq 0 , y$ satisfies the differential equation
$$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 3 x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } + 3 y ( 3 x + 2 ) = 18 x$$
\begin{enumerate}[label=(\alph*)]
\item Show that the substitution $u = x y$ transforms this differential equation into
$$\frac { \mathrm { d } ^ { 2 } u } { \mathrm {~d} x ^ { 2 } } + 6 \frac { \mathrm {~d} u } { \mathrm {~d} x } + 9 u = 18 x$$
\item Hence find the general solution of the differential equation
$$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 ( 3 x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } + 3 y ( 3 x + 2 ) = 18 x$$
giving your answer in the form $y = \mathrm { f } ( x )$.\\
(8 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2012 Q7 [12]}}