4. Given that
$$y \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 4 \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + 3 y = 0$$
- show that
$$\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = \frac { 28 } { y ^ { 2 } } \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 3 } - \frac { 24 } { y } \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right)$$
Given also that \(y = 8\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1\) at \(x = 0\)
- find a series solution for \(y\) in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\), simplifying the coefficients where possible.