Question

Solve the Initial Value Problem, using Euler’s Method for the differential Equation:

y'= 1+5x²y, given that y(0) = 1. Find y(1.0) taking (i) h = 0.20 and then (ii) h = 0.5

Answer 1

$(i)y'=1+5x^2y$

$y(0)=1$

Step length =0.2

We have to find y(1)=? By Euler Method

$y_1=y_0+hf(x_0,y_0)=1+(0.2)f(0.1)$

$=1+(0.2)\cdot(0.1)$

$=1.2$

$y_2=y_1+hf(x_1,y_1)=1+0.2f(0.2,1.2)$

$=1.2+(0.2)(1.24)=1.448$

$y_3=y_2+hf(x_2,y_2)=1.448+(0.2)f(0.4,1.448)$

$=1.448+(0.2)(2.1584)=1.8797$

$y_4=y_3+hf(x_3,y_3)=1.8797+(0.2)f(0.6,1.8797)$

$=1.8797+(0.2)(4.3834)=2.7564$

$y_5=y_4+hf(x_4,y_4)=2.7564+(0.2)f(0.8,2.7564)$

$=2.7564+(0.2)(2.7564)=4.7204$

$\therefore y_1=4.7204$

(ii)$y'=1+5x^2y$

y(0)=1

Step length=0.5

We have to find y(1)=? By Euler Method

$y'=y_0+hf(x_0,y_0)=1+(0.5)f(0.1)$

$=1+(0.5)\cdot1=1.5$

$y_2=y_1+hf(x_1,y_1)=1.5+(0.5)f(0.5,1.5)$

$=1.5+(0.5)(2.875)$

$=2.9375$

$\therefore y(1)=2.9375$