Homework Answers
solution:
The general Solution for the Euler method is
yn+1= yn+ h f( xn, yn) xn= xn-1+ h .
where h is the step size and f(xn , yn) = (dy/dx) at xn .
We have given initial value y0 = y(0) = 4 and h = 0.1 , We have to calculate for four step
f(x0 , y0) = y’(0) = -2x0+2y0
= -2(0)+2(4)
= 0+8 =8
Now we have to calculate y1
y1 = y0 + h f(x0 , y0)
= 4 + 0.1(8)
=4.8
x1 = x0+h
= 0+0.1=0.1
f(x1 , y1) = -2x1+2y1
= -2(0.1)+2(4.8)
= -0.2+9.6 = 9.4
y2 = y1 + h f(x1 , y1)
= 4 .8+ 0.1(9.4)
=4.8 +0.94 = 5.74
x2 = x1+h
= 0.1+0.1=0.2
f(x2 , y2) = -2x2+2y2
= -2(0.2)+2(5.74)
= -0.4+11.48 = 11.08
y3 = y2 + h f(x2 , y2)
= 5.74 + 0.1(11.08)
=5.74 + 1.108 = 6.848
x3 = x2+h
= 0.2+0.1=0.3
f(x3 , y3) = -2x3+2y3
= -2(0.3)+2(6.848)
= -0.4+13.696 = 13.296
y4 = y3 + h f(x3 , y3)
= 6.848 + 0.1(13.296)
=6.848 +1.3296 = 8.1776
Now we have to caluculate the error value at y4 , By Taylor Series expansion , We know that error term is always is order higher derivative involve in the approximation method.
So error value E = h2/2 ( y’’) where y’’ is second order derivative
differentiation of given equation y’’ = -2
so E = {(0.1)2 (-2)}/2 = {(0.01)(-1)} = -0.01 .
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