Question

(1 point) Use Euler's method with step size 0.25 to compute the approximate y-values y1, y2, y3, and y4 of the solution of the initial-value problem y′=2+2x+4y, y(1)=2.

Answer 1

Answer:

$y_1=4.75\\y_2=10.625\\y_3=22.5\\y_4=46.375$

Step-by-step explanation:

Formula to calculate yn using Euler method is as follows:

$x_{n+1}=x_n+h\\\\h=step\,\,size\\\\y_{n+1}=y_n+F(x_n,y_n)\\\\where\\\\y'=F(x,y)\,\, and\,\,y(x_o)=y_o$

Given that:

$y'=2+2x+4y\\\\x_o=1\,,\,y_o=2\,,\,h=0.25$

$F(x_o,y_o)=1+2(1)+4(2) =11\\\\y_1=y_o+hF(x_o,y_o)\\\\y_1=2+(0.25)(11)=4.75\\\\x_1=1+0.25=1.25\\\\F(x_1,y_1)=2+2(1.25)+4(4.75)= 23.5\\\\y_2=y_1+hF(x_1,y_1)\\\\y_2=4.75+(0.25)(23.5)=10.625\\\\x_2=1.25+.25=1.5\\\\f(x_2,y_2)=2+2(1.5)+4(10.625)=47.5\\\\y_3=y_2+h(F(x_2,y_2)\\\\y_3=10.625+(0.25)(47.5)=22.5\\\\x_3=1.5+0.25=1.75\\\\F(x_3,y_3)=2+2(1.75)+4(22.5)=95.5\\\\y_4=y_3+hF(x_3,y_3)\\\\y_4=22.5+0.25(95.5)=46.375$