FP2 June 2007 Q9
9. \[\frac{\mathrm{d}y}{\mathrm{d}x} = y\mathrm{e}^{x^2}.\]
It is given that \(y = 0.2\) at \(x = 0\).
(a) Use the approximation \(\dfrac{y_1 - y_0}{h} \approx \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_0\), with \(h = 0.1\), to obtain an estimate of the value of \(y\) at \(x = 0.1\). (2)
(b) Use your answer to part (a) and the approximation \(\dfrac{y_2 - y_0}{2h} \approx \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_1\), with \(h = 0.1\), to obtain an estimate of the value of \(y\) at \(x = 0.2\).
Gives your answer to 4 decimal places. (3)
Gives your answer to 4 decimal places. (3)
| Scheme | Marks |
|---|---|
| \(\dfrac{y_1 - 0.2}{0.1} \approx \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_0 = 0.2 \times \mathrm{e}^0\ (= 0.2)\) | M1 |
| \(y_1 \approx 0.22\) | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_1 \approx 0.22 \times \mathrm{e}^{0.01} \approx 0.2222\ldots\) | B1 |
| \(\dfrac{y_2 - 0.2}{0.2} \approx 0.2222\ldots\) | M1 |
| \(y_2 \approx 0.2444\) cao | A1 |
| (3) | |
| (5 marks) |