Challenging +1.2 This is a structured multi-part question where part (a) is verification (straightforward differentiation and substitution), part (b) is a standard first-order linear ODE with integrating factor (core A-level technique), and part (c) requires solving another first-order ODE after back-substitution. While it involves Further Maths content and multiple steps, the question provides the substitution and guides students through each stage methodically, making it more accessible than typical FP3 questions requiring independent problem-solving or insight.
Show that the substitution
$$u = \frac { \mathrm { d } y } { \mathrm {~d} x } + 2 y$$
transforms the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = \mathrm { e } ^ { - 2 x }$$
into
$$\frac { \mathrm { d } u } { \mathrm {~d} x } + 2 u = \mathrm { e } ^ { - 2 x }$$
(4 marks)
By using an integrating factor, or otherwise, find the general solution of
$$\frac { \mathrm { d } u } { \mathrm {~d} x } + 2 u = \mathrm { e } ^ { - 2 x }$$
giving your answer in the form \(u = \mathrm { f } ( x )\).
Hence find the general solution of the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = \mathrm { e } ^ { - 2 x }$$
giving your answer in the form \(y = \mathrm { g } ( x )\).
6
\begin{enumerate}[label=(\alph*)]
\item Show that the substitution
$$u = \frac { \mathrm { d } y } { \mathrm {~d} x } + 2 y$$
transforms the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = \mathrm { e } ^ { - 2 x }$$
into
$$\frac { \mathrm { d } u } { \mathrm {~d} x } + 2 u = \mathrm { e } ^ { - 2 x }$$
(4 marks)
\item By using an integrating factor, or otherwise, find the general solution of
$$\frac { \mathrm { d } u } { \mathrm {~d} x } + 2 u = \mathrm { e } ^ { - 2 x }$$
giving your answer in the form $u = \mathrm { f } ( x )$.
\item Hence find the general solution of the differential equation
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y = \mathrm { e } ^ { - 2 x }$$
giving your answer in the form $y = \mathrm { g } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2006 Q6 [14]}}