9.
$$\left( x ^ { 2 } + 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 2 y ^ { 2 } + ( 1 - 2 x ) \frac { \mathrm { d } y } { \mathrm {~d} x }$$
- By differentiating equation (I) with respect to \(x\), show that
$$\left( x ^ { 2 } + 1 \right) \frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = ( 1 - 4 x ) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + ( 4 y - 2 ) \frac { \mathrm { d } y } { \mathrm {~d} x }$$
Given that \(y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1\) at \(x = 0\),
- find the series solution for \(y\), in ascending powers of \(x\), up to and including the term in \(x _ { 3 }\).(4)
- Use your series to estimate the value of \(y\) at \(x = - 0.5\), giving your answer to two decimal places.(1)