Challenging +1.8 This is a challenging Further Maths question requiring substitution to transform a non-linear second-order ODE into a linear form, then solving the resulting linear ODE and back-substituting. The algebraic manipulation to verify the substitution is intricate (involving product/chain rules with y², y³ terms), and students must then solve a second-order linear ODE with particular integral. While the techniques are standard for FP1, the multi-stage process and careful algebra place this well above average difficulty.
7 The variables \(x\) and \(y\) are related by the differential equation
$$y ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } - 5 y ^ { 3 } = 8 \mathrm { e } ^ { - x }$$
Given that \(v = y ^ { 3 }\), show that
$$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} v } { \mathrm {~d} x } - 15 v = 24 \mathrm { e } ^ { - x }$$
Hence find the general solution for \(y\) in terms of \(x\).
7 The variables $x$ and $y$ are related by the differential equation
$$y ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } - 5 y ^ { 3 } = 8 \mathrm { e } ^ { - x }$$
Given that $v = y ^ { 3 }$, show that
$$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} v } { \mathrm {~d} x } - 15 v = 24 \mathrm { e } ^ { - x }$$
Hence find the general solution for $y$ in terms of $x$.
\hfill \mbox{\textit{CAIE FP1 2011 Q7 [11]}}