1 It is given that \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \frac { \ln ( x + y ) } { \ln y }$$ and $$y ( 6 ) = 3$$ 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 \(h = 0.4\), to obtain an approximation to \(y ( 6.4 )\), giving your answer to three decimal places.
[0pt] [5 marks]

## Question 1:

**Step 1:** $k_1 = 0.4 \times f(6, 3) = 0.4 \times \frac{\ln(9)}{\ln(3)}$ | M1 | Correct substitution into $k_1$

**Step 2:** Evaluate $k_2 = 0.4 \times f(6.4, 3 + k_1)$ | M1 | Correct substitution into $k_2$

**Step 3:** $y_1 = 3 + \frac{1}{2}(k_1 + k_2)$ | M1 | Correct application of formula

**Final answer:** $y(6.4) \approx 3.252$ | A1 A1 | 3 dp

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1 It is given that $y ( x )$ satisfies the differential equation

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

where

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

and

$$y ( 6 ) = 3$$

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 $h = 0.4$, to obtain an approximation to $y ( 6.4 )$, giving your answer to three decimal places.\\[0pt]
[5 marks]

\hfill \mbox{\textit{AQA FP3 2014 Q1 [5]}}