It is given that \(y ( x )\) satisfies the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$
where
$$\mathrm { f } ( x , y ) = ( 2 x + 1 ) \ln ( x + y )$$
and
$$y ( 0 ) = 2$$
Use the improved Euler formula
$$y _ { r + 1 } = y _ { r } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right)$$
where \(k _ { 1 } = h \mathrm { f } \left( x _ { r } , y _ { r } \right)\) and \(k _ { 2 } = h \mathrm { f } \left( x _ { r } + h , y _ { r } + k _ { 1 } \right)\) and \(h = 0.1\), to obtain an approximation to \(y ( 0.1 )\), giving your answer to three decimal places.
It is given that \(y ( x )\) satisfies the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = ( 2 x + 1 ) \ln ( x + y )$$
and \(y = 2\) when \(x = 0\).
Use implicit differentiation to find \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\), giving your answer in terms of \(x\) and \(y\).
Hence find the first three non-zero terms in the expansion, in ascending powers of \(x\), of \(y ( x )\). Give your answer in an exact form.
Use your answer to part (b)(ii) to obtain an approximation to \(y ( 0.1 )\), giving your answer to three decimal places. [0pt]
[1 mark]