1 The function \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \ln ( x + y )$$ and $$y ( 2 ) = 3$$ 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 \(y ( 2.1 )\), giving your answer to four decimal places.
(6 marks)

## Question 1:

| Working/Answer | Mark | Guidance |
|---|---|---|
| $k_1 = 0.1 \times \ln(2+3)$ | M1 | |
| $= 0.1609(4379...)$ | A1 | PI |
| $k_2 = 0.1 \times f(2.1, 3 + k_1...)= 0.1 \times \ln(2.1 + 3.16094...)$ | M1 | |
| $= 0.1660(31...)$ | A1 | PI |
| $y(2.1) = y(2) + \frac{1}{2}[k_1 + k_2] = 3 + 0.5 \times 0.3269748...$ | m1 | Dep on previous two M marks and numerical values for $k$'s |
| $= 3.163487... = 3.1635$ to 4dp | A1 | Must be 3.1635 |
| **Total** | **6** | |

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1 The function $y ( x )$ satisfies the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$

where

$$\mathrm { f } ( x , y ) = \ln ( x + y )$$

and

$$y ( 2 ) = 3$$

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 $y ( 2.1 )$, giving your answer to four decimal places.\\
(6 marks)

\hfill \mbox{\textit{AQA FP3 2008 Q1 [6]}}