3.
$$2 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + \frac { \mathrm { d } y } { \mathrm {~d} x } - x y = 1$$
- Show that
$$\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } } = \frac { 1 } { 2 } \left( a \frac { \mathrm {~d} y } { \mathrm {~d} x } + b x \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + c \frac { \mathrm {~d} ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } \right)$$
where \(a , b\) and \(c\) are constants to be found.
Given that \(y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1\) when \(x = 2\)
- find a series solution for \(y\) in ascending powers of ( \(x - 2\) ), up to and including the term in \(( x - 2 ) ^ { 4 }\). Write each term in its simplest form.
- Use the solution to part (b) to find an approximate value for \(y\) when \(x = 2.1\), giving your answer to 3 decimal places.