Challenging +1.2 This is a structured Further Maths differential equations question with clear guidance through substitution. While it requires careful differentiation using the product rule and algebraic manipulation to verify the substitution, then solving a standard second-order DE with constant coefficients, the path is explicitly signposted. The back-substitution and application of initial conditions are routine. Harder than average A-level due to being Further Maths content, but straightforward within that context.
11 It is given that \(x \neq 0\) and
$$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 x y = 8 x ^ { 2 } + 16$$
Show that if \(z = x y\) then
$$\frac { \mathrm { d } ^ { 2 } z } { \mathrm {~d} x ^ { 2 } } + 4 z = 8 x ^ { 2 } + 16$$
Find \(y\) in terms of \(x\), given that \(y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - 2\) when \(x = \frac { 1 } { 2 } \pi\).
11 It is given that $x \neq 0$ and
$$x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 x y = 8 x ^ { 2 } + 16$$
Show that if $z = x y$ then
$$\frac { \mathrm { d } ^ { 2 } z } { \mathrm {~d} x ^ { 2 } } + 4 z = 8 x ^ { 2 } + 16$$
Find $y$ in terms of $x$, given that $y = 0$ and $\frac { \mathrm { d } y } { \mathrm {~d} x } = - 2$ when $x = \frac { 1 } { 2 } \pi$.
\hfill \mbox{\textit{CAIE FP1 2010 Q11 [12]}}