Use the Nonlinear Finite-Difference Algorithm with TOL = 104 to approximate the solution to the following boundary-value
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a. y" = y3 -yy', 1≤ x ≤ 2, y(1) = 1 / 2, y(2) = 1 / 3; use h = 0.1; actual solution y(x) = (x+1)−1.
b. y" = 2y3 − 6y − 2x3, 1 ≤ x ≤ 2, y(1) = 2, y(2) = 5/2 ; use h = 0.1; actual solution y(x) = x + x−1.
c. y" = y' + 2(y − ln x)3 − x−1, 2 ≤ x ≤ 3, y(2) = 1/ 2 + ln 2, y(3) = 1/3 + ln 3; use h = 0.1; actual solution y(x) = x−1 + ln x.
d. y" = (y')2x−3 − 9y2x−5 + 4x, 1 ≤ x ≤ 2, y(1) = 0, y(2) = ln 256; use h = 0.05; actual solution y(x) = x3 ln x.
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