Question

Use Eulers Method to make a table of values for the approximate solution of y2x +2y with initial condition y(0) 4. Determine
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Answer #1

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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