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Use the fourth-order Runge–Kutta subroutine with h = 0.25 to approximate the solution to the initial value problemy'=x+1-y,y(0)=1, at x = 1. Compare this approximation with the one obtained in Problem 5 using the Taylor method of order 4.

Short Answer

Expert verified

ϕ(1)=1.3679

Step by step solution

01

Find the values of ki, i = 1, 2, 3, 4

Since f(x,y)=x+1-yand x = 0, y = 1, and h = 0.25

role="math" localid="1664324055813" k1=hf(x,y)=0.25(0+1-1)=0k2=hfx+h2,y+k12=0.03125k3=hfx+h2,y+k22=0.0273438k4=hfx+h,y+k3=0.0556641

02

Find the values of x and y

x=0+0.25=0.25y=1+16k1+2k2+2k3+k4=1+16(0-2(0.031125)-2(0.0273438)-0.0556641)=1.02881

03

Use values of x and y for finding values of ki, i = 1, 2, 3, 4

k1=hf(x,y)=0.25(0.25+1-1.02881)=0.0552975k2=hfx+h2,y+k12=0.0796353k3=hfx+h2,y+k22=0.0765931k4=hfx+h,y+k3=0.0986392

Now

x=0.25+0.25=0.5y=1.02881+16(0.0552975-2(0.0796353)-2(0.0765931)-0.0986492)=1.10654

04

Repeat the procedure for two times

k1=hf(x,y)=0.25(0.5+1-1.10654)=0.098365k2=hfx+h2,y+k12=0.117319k3=hfx+h2,y+k22=0.11495k4=hfx+h,y+k3=0.132127x=0.5+0.25=0.75y=1.10654+16k1+2k2+2k3+k4=1.10654+16(0.098365-2(0.117319)-2(0.11495)-0.132127)=1.22238

And

k1=hf(x,y)=0.25(0.75+1-1.22238)=0.131905k2=hfx+h2,y+k12=0.146667k3=hfx+h2,y+k22=0.144822k4=hfx+h,y+k3=0.1582x=0.75+0.25=1y=1.22238+16k1+2k2+2k3+k4=0.52761+16(0.131905-2(0.146667)-2(0.144822)-0.1582)=1.36789

Thereforeϕ(1)=1.3679

Hence the solution isϕ(1)=1.3679

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