5.
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 0$$
- Show that
$$\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } } = \left( a x ^ { 2 } + b \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$$
where \(a\) and \(b\) are constants to be found.
Given that \(y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3\) at \(x = 0\)
- find a series solution for \(y\) in ascending powers of \(x\) up to and including the term in \(x ^ { 4 }\)
- use your series to estimate the value of \(y\) at \(x = - 0.2\), giving your answer to four decimal places.