Answer to Question #275683 in Quantitative Methods for khaled

Question

use Euler’s method to obtain a four-decimal approximation of the indicated value. First use h = 0.1 and then use h = 0.05. Find an explicit solution for each initial-value problem

Answer 1

$y'=2xy,y(1)=1,y(1.5)$

h = 0.1

$y_1=y_0+hf(x_0,y_0)=1+(0.1)f(1,1)=1.2$

$y_2=y_1+hf(x_1,y_1)=1.2+(0.1)f(1.1,1.2)=1.464$

$y_3=y_2+hf(x_2,y_2)=1.464+(0.1)f(1.2,1.464)=1.8154$

$y_4=y_3+hf(x_3,y_3)=1.8154+(0.1)f(1.3,1.8154)=2.2874$

 $y_5=y_4+hf(x_4,y_4)=2.2874+(0.1)f(1.4,2.2874)=2.9278$

$y(1.5)=2.9278$

h = 0.05

$y_1=y_0+hf(x_0,y_0)=1+(0.05)f(1,1)=1.1$

$y_2=y_1+hf(x_1,y_1)=1.1+(0.05)f(1.05,1.1)=1.2155$

$y_3=y_2+hf(x_2,y_2)=1.2155+(0.05)f(1.1,1.2155)=1.3492$

$y_4=y_3+hf(x_3,y_3)=1.3492+(0.05)f(1.15,1.3492)=1.5044$

$y_5=y_4+hf(x_4,y_4)=1.5044+(0.05)f(1.2,1.5044)=1.6849$

$y_6=y_5+hf(x_5,y_5)=1.6849+(0.05)f(1.25,1.6849)=1.8955$

$y_7=y_6+hf(x_6,y_6)=1.8955+(0.05)f(1.3,1.8955)=2.1419$

$y_8=y_7+hf(x_7,y_7)=2.1419+(0.05)f(1.35,2.1419)=2.4311$

$y_9=y_8+hf(x_8,y_8)=2.4311+(0.05)f(1.4,2.4311)=2.7714$

$y_{10}=y_9+hf(x_9,y_9)=2.7714+(0.05)f(1.45,2.7714)=3.1733$

$y(1.5)=3.1733$