Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 6

Year : 2015

By eulers modified method

$$y_1=y_0+hf(x_0,y_0)$$

$=y_0+h[2+\sqrt{x_0y_0}]\\ =1.6403+0.2[2+\sqrt{1.2\times 1.6403}]\\ =2.32093\\ \therefore x_1=x_0+h=1.2+0.2=1.4 \\ \text{ Now first approximation gives } \\ y_1^1=y_0+\dfrac h2[f(x_0,y_0)+f(x_1,y_1)]\\ y_1^1=1.6403+\dfrac {0.2}2[(2+\sqrt{1.2\times1.6403)}+(2+\sqrt{1.4\times 2.3209)}]\\ \therefore y_1^1=2.36086 $

Third approximation gives

$$y_1^{(3)}=y_0+\dfrac h2[f(x_0,y_0)+f(x_1,y_1^{(2)}]$$ $ =1.6403+\dfrac {0.2}2[2+\sqrt{1.2\times 1.6403}+2+\sqrt{1.4\times 2.624}]\\ =2.36426$

Fourth approximation gives

$$y_1^{(4)}=y_0+\dfrac h2[f(x_0,y_0)+f(x_1,y_1^{(3)}]$$ $ y_1^{(4)}=1.6403+\dfrac {0.2}2[2+\sqrt{1.2\times 1.6403}+2+\sqrt{1.4\times 2.36246}]\\ y_1^{(4)}=2.36426$

The correct value of y when $x=1.4$ is $2.36246$