The variables \(z\) and \(x\) are related by the differential equation $$3 z ^ { 2 } \frac { \mathrm {~d} ^ { 2 } z } { \mathrm {~d} x ^ { 2 } } + 6 z ^ { 2 } \frac { \mathrm {~d} z } { \mathrm {~d} x } + 6 z \left( \frac { \mathrm {~d} z } { \mathrm {~d} x } \right) ^ { 2 } + 5 z ^ { 3 } = 5 x + 2$$ Use the substitution \(y = z ^ { 3 }\) to show that \(y\) and \(x\) are related by the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = 5 x + 2$$ Given that \(z = 1\) and \(\frac { \mathrm { d } z } { \mathrm {~d} x } = - \frac { 2 } { 3 }\) when \(x = 0\), find \(z\) in terms of \(x\). Deduce that, for large positive values of \(x , z \approx x ^ { \frac { 1 } { 3 } }\).

The variables $z$ and $x$ are related by the differential equation

$$3 z ^ { 2 } \frac { \mathrm {~d} ^ { 2 } z } { \mathrm {~d} x ^ { 2 } } + 6 z ^ { 2 } \frac { \mathrm {~d} z } { \mathrm {~d} x } + 6 z \left( \frac { \mathrm {~d} z } { \mathrm {~d} x } \right) ^ { 2 } + 5 z ^ { 3 } = 5 x + 2$$

Use the substitution $y = z ^ { 3 }$ to show that $y$ and $x$ are related by the differential equation

$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 5 y = 5 x + 2$$

Given that $z = 1$ and $\frac { \mathrm { d } z } { \mathrm {~d} x } = - \frac { 2 } { 3 }$ when $x = 0$, find $z$ in terms of $x$.

Deduce that, for large positive values of $x , z \approx x ^ { \frac { 1 } { 3 } }$.

\hfill \mbox{\textit{CAIE FP1 2010 Q11 EITHER}}