| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Euler-Cauchy equations via exponential substitution |
| Difficulty | Challenging +1.2 This is a structured, multi-part FP3 question on Euler-Cauchy equations that guides students through the standard substitution method. Parts (a) and (b) are routine verification using chain rule, part (c) is a standard constant-coefficient second-order DE with repeated roots, and part (d) applies boundary conditions. While it requires multiple techniques (chain rule, auxiliary equations, particular integrals, boundary conditions), each step follows a well-established procedure taught in FP3 with no novel problem-solving required. The 5-mark allocation for part (d) confirms it's straightforward application rather than challenging synthesis. |
| 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 |
8
\begin{enumerate}[label=(\alph*)]
\item Given that $x = \mathrm { e } ^ { t }$ and that $y$ is a function of $x$, show that
$$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { d } y } { \mathrm {~d} t }$$
\item Hence show that the substitution $x = \mathrm { e } ^ { t }$ transforms the differential equation
$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 3 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 2 \ln x$$
into
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 4 y = 2 t$$
\item Find the general solution of the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 4 y = 2 t$$
\item Hence solve the differential equation $x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 3 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = 2 \ln x$, given that $y = \frac { 3 } { 2 }$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 }$ when $x = 1$.\\
(5 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2011 Q8 [18]}}