Question

Given 𝑑𝑦

𝑑𝑥

= 𝑥 𝑥

2 + 𝑦

2

𝑒

−𝑥

and 𝑦(0) = 1, 𝑦(0.1) = 1.005, find

i) 𝑦 0.2 by Euler’s method

ii) 𝑦 0.3 by Modified Euler’s method

iii) 𝑦(0.4) by Milne’s predictor - corrector method.

Answer 1

$\frac{dy}{dx}=x^2+y^2e^{-x}=f(x,y)\\$

i)

By Euler's method,

$y_{n+1}=y_n+hf(x_n,y_n)\\ y_2=y_1+0.1×[x_1^2+y_1^2e^{-x_1}]\\ =1.005+0.1×[0.1^2+1.005^2e^{-0.1}]\\ =1.005+0.1×0.9239 =1.0973$

ii)

By Euler's method,

$y_{n+1}=y_n+hf(x_n,y_n)\\ y_3=y_2+0.1×[x_2^2+y_2^2e^{-x_2}]\\ =1.0973+0.1×[0.2^2+1.0973^2e^{-0.2}]\\ =1.0973+0.1×0.9858 =1.1958$

iii)

By Milne’s predictor method,

$y_{n+1,p}=y_{n-3}+\frac{4h}{3}(2f_{n-2}-f_{n-1}+2f_n)\\ y_{4,p}=y_{0}+\frac{4×0.1}{3}(2f_{1}-f_{2}+2f_3)\\ =1+\frac{0.4}{3}(2×1.005-1.0973+2×1.1958)\\ =1+0.4405\\ =1.4405$

By Milne’s corrector method,

$y_{n+1,c}=y_{n-1}+\frac{h}{3}(f_{n+1}+f_{n-1}+4f_n)\\ y_{4,p}=y_{2}+\frac{0.1}{3}(f_{4}+f_{2}+4f_3)\\ =1.0973+\frac{0.1}{3}(1.4405+1.0973+4×1.1958)\\ =1.0973+2.1963\\ =3.2936$