4.
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = 0$$
- Show that
$$\frac { \mathrm { d } ^ { 5 } y } { \mathrm {~d} x ^ { 5 } } = a x \frac { \mathrm {~d} ^ { 4 } y } { \mathrm {~d} x ^ { 4 } } + b \frac { \mathrm {~d} ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }$$
where \(a\) and \(b\) are integers to be found.
- Hence find a series solution, in ascending powers of \(x\), as far as the term in \(x ^ { 5 }\), of the differential equation (I) where \(y = 0\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1\) at \(x = 0\)