AQA FP3 2012 June — Question 7 19 marks

Exam BoardAQA
ModuleFP3 (Further Pure Mathematics 3)
Year2012
SessionJune
Marks19
PaperDownload PDF ↗
TopicSecond order differential equations
TypeEuler-Cauchy equations via exponential substitution
DifficultyChallenging +1.3 This is a structured Further Maths question on Euler-Cauchy equations with clear scaffolding across three parts. Part (a) is a guided 'show that' requiring chain rule application (standard FP3 technique), part (b) involves solving a constant-coefficient second-order ODE with particular integral (routine but multi-step), and part (c) is immediate from parts (a) and (b). While the topic is advanced and requires careful algebraic manipulation, the heavy scaffolding and standard methods make it moderately above average difficulty rather than highly challenging.
Spec1.05l Double angle formulae: and compound angle formulae1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral

7
  1. Show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation $$\text { into } \quad \begin{aligned} x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y & = 3 + 20 \sin ( \ln x ) \\ \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 6 y & = 3 + 20 \sin t \end{aligned}$$ (7 marks)
  2. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 6 y = 3 + 20 \sin t$$ (11 marks)
  3. Write down the general solution of the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y = 3 + 20 \sin ( \ln x )$$

7
\begin{enumerate}[label=(\alph*)]
\item Show that the substitution $x = \mathrm { e } ^ { t }$ transforms the differential equation

$$\text { into } \quad \begin{aligned}
x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y & = 3 + 20 \sin ( \ln x ) \\
\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 6 y & = 3 + 20 \sin t
\end{aligned}$$

(7 marks)
\item Find the general solution of the differential equation

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 6 y = 3 + 20 \sin t$$

(11 marks)
\item Write down the general solution of the differential equation

$$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 6 y = 3 + 20 \sin ( \ln x )$$
\end{enumerate}

\hfill \mbox{\textit{AQA FP3 2012 Q7 [19]}}