Standard +0.3 This is a straightforward application of the improved Euler method with all formulas provided. Students must substitute given values into the formula (f(x,y) = x + ln(1+y), h=0.2, initial condition y(2)=1) and perform one iteration of arithmetic. While it requires careful calculation with logarithms, it's essentially a procedural exercise with no problem-solving or conceptual insight required—slightly easier than average for Further Maths.
1 The function \(y ( x )\) satisfies the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$
where
$$\mathrm { f } ( x , y ) = x + \ln ( 1 + y )$$
and
$$y ( 2 ) = 1$$
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 \(h = 0.2\), to obtain an approximation to \(y ( 2.2 )\), giving your answer to four decimal places.
1 The function $y ( x )$ satisfies the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$
where
$$\mathrm { f } ( x , y ) = x + \ln ( 1 + y )$$
and
$$y ( 2 ) = 1$$
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 $h = 0.2$, to obtain an approximation to $y ( 2.2 )$, giving your answer to four decimal places.
\hfill \mbox{\textit{AQA FP3 2011 Q1 [5]}}