Use the Extrapolation Algorithm with tolerance TOL = 10~6 , hmax = 0.5, and hmin = 0.05 to approximate the solutions to the following initial-value problems. Compare the results to the actual values. , 2 2ly (2/ + 1) a. y = t2 + \ ' 0 ^ ? 5 3- >'(0) = 1; actual solution y = ^ ^ . y 2 . 1 1 ^ ^ A ./1\ /!._ 1. b. y' , 1 < t < 4, y(l) = -(In2) actual solution y(r) = . 1 + r " ln(r + 1) 4t , c. y' ty -I , 0 < ? < 1, >'(D) = 1; actual solution y(t) \/4 3e~'2 . y d. >' = > + ryl/2 , 2 < r < 4, >(2) = 2; actual solution y{r) (t 2 + \/2 e e~'/2) 2 .

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