Chapter 3y2=(1 + 4h+ 8h2)y1=(1 + 4h+ 8h2)13(1 + 4h+ 8h2)=13(1 + 4h+ 8h2)2,y3=(1 + 4h+ 8h2)y2=(1 + 4h+ 8h2)13(1 + 4h+ 8h2)2=13(1 + 4h+ 8h2)3.Continuing this way we see thatyn=13(1 + 4h+ 8h2)n.(3.16)(This can be proved by induction using equation (3.15) above.)We are looking for an ap-proximation to our solution at the pointx= 1/2. Therefore, we haveh=1/2−x0n=1/2−0n=12n⇒n=12h.Substituting this value forninto equation (3.16) yieldsyn=13(1 + 4h+ 8h2)1/(2h).7.For this problem,f(x, y) =x−y2. We need to approximate the solution on the interval [1,1.5]using a step size ofh= 0.1. Thus the number of steps needed isN= 5. The inputs to thesubroutine on page 129 are
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