Q4Advanced Engineering Mathematics I

Question

Using Milne's Predictor-Corrector Method, obtain the value of $y$ for $x = 0.4$ for the following equation $\frac{d y}{d x} = 2 e^{x} - y$ , given that: | x | 0 | 0.1 | 0.2 | 0.3 | |---|---|-----|-----|-----| | y | 2 | 2.01 | 2.04 | 2.09 |

Answer 1

Given $y^{'} = 2 e^{x} - y$ . We calculate $f_{0}, f_{1}, f_{2}, f_{3}$ using the given table.

Predictor Formula: $y_{4}^{(P)} = y_{0} + \frac{4 h}{3} (2 f_{1} - f_{2} + 2 f_{3})$ .

Evaluate $f_{4}^{(P)} = 2 e^{x_{4}} - y_{4}^{(P)}$ .

Corrector Formula: $y_{4}^{(C)} = y_{2} + \frac{h}{3} (f_{2} + 4 f_{3} + f_{4}^{(P)})$ .

Repeat the corrector step until the value of $y_{4}$ converges.

Q4Advanced Engineering Mathematics I

Question

Answer