Question

X Yeuler Y true Example: Euler's Method Solve numerically: dy - 2x +12x’ - 20x+8. 5 Error Global Error Local % From x=0 to x=4 with step size h=0.5 initial condition: (x=0; y=1) 0 0.5 1.0 5.250 5.875 3.218 3.000 63.1 95.8 63.1 28 5.125 2.218 131.0 1.41 Exact Solution: y = -0.5x4 + 4x - 10x2 + 8.5x +1 Numerical Solution: Vi+ 1 Yi+1 = y; + f(x, y)h Vit 1.5 2.0 4.500 2.000 125.0 20.5 y(0.5) = y(0)+ (0, 1)0.5 = 1+8.5*0.5 = 5.25 (true solution at x=0.5 is y(0.5) = 3.22 and €, = 63%) y(1)-y(0.5)+(0.5, 5.25)0.5 = 5.25+(-2(0.5)3+12(0.5)2-20(0.5)+8.5]*0.5 = 5.25+0.625 = 5.875 (true solution at x-1 is y(1) - 3 and &, -96%) y(1.5)=y(1)+1(1, 5.875)0.5 = 5.125 True solution Numerical Methods

explain how to find error local in this example

and difference between local and global

Answer 1

geven that, Euler's method to numerically; de -2x3+1242_ 20x+8.5 x=0 to x=4 with step size hoort Initial condition : (x=0; y = 1) - 2x3 + 12x²_20x +8.5 dx Say - S (-2x3 + 1222 - 20x +8.5) dx Y=0.5x4 + 4 13- 10x2 + 8.5x+c Apply initial B. Ya1 at no Y=0.5x4+4x3 - 10x2+8.5x+1 Euler's method :- Jitr = gi th. f (xi, Y;) we have dy = - 2x² + 12x² _ 20x + 8.5 f (mig)= - 2x3 + 12x² -20x48.5 Intial condition at fx=0 Y=1 Ji - Yo the creo, yo) no ao, Yo = 1 f(xo, y) = f(0, 1) = -2X0 + 12x0-20x0+8.5 f(0, 1) = 8.8

with a step size is h=0.5 y(o)=1 Kati = kith Xi = koth At x 20.5 with inital condition Y{0.5) = Y(0) + f(0, 1) *0.5 y (0.5) = 1+ 8.5*0.5 (0.5)=5.25 the soluction at r=0.5 y = -0.5 (0.5" +4 (0.583_10(0.532+ 85(0.5)+1=3-2187 Error: = y tuery Retatire = 3.21875-5.25 =-2.03125 snor T.R. (True -Y) x 100 Yorue - 2003125 XI 3.21875 2-95-8y. golanswer local and global: S(O)= Y(x)=5:875 y (0.5)=5.25 9(.5)= 8.125 Y (4)=7 Y (2)=4.5 Y (2-5)= 4.75 Y (3)= 5.875 9(35) = 7.125