Exercises 3.7solution to the differential equation atx= 1. Since the actual solution isy=ex, this meansthatyn≈e. To find the equation we are looking for, we see thaty1=1 +1n+12n2y0=1 +1n+12n2,y2=1 +1n+12n2y1=1 +1n+12n22,y3=1 +1n+12n2y2=1 +1n+12n23,y4=1 +1n+12n2y3=1 +1n+12n24,...yn=1 +1n+12n2yn−1=1 +1n+12n2n.(This can be proved rigorously by mathematical induction.) As we observed above,yn≈e,and so we havee≈1 +1n+12n2n.19.In this initial value problem, the independent variable isu, the dependent variable isv,u0= 2,v0= 0.1, andf(u, v) =uu2+ 1v3+u+52v2.We will use the classical 4th order Runge-Kutta algorithm with tolerance given on page 139 ofthe text but, since the stopping criteria should be based on the relative error, we will replace
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