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}}