5. Given that
$$\left( 2 - x ^ { 2 } \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 5 x \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } = 3 y$$
- show that
$$\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = \frac { 1 } { \left( 2 - x ^ { 2 } \right) } \left( 2 x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } \left( 1 - 5 \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) - 5 \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + 3 \frac { \mathrm {~d} y } { \mathrm {~d} x } \right)$$
Given also that \(y = 3\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 4 }\) at \(x = 0\)
- obtain a series solution for \(y\) in ascending powers of \(x\) with simplified coefficients, up to and including the term in \(x ^ { 3 }\)