Solve the initial value problem. = 2 - e-t, y(1) = , y'(0) = -5

A. y = t2 - e-t - 5t
B. y = 2t2 + e-t - 6t + 4 - 
C. y = t2 - e-t
D. y = t2 - e-t - 6t + 5

Answer: D

Mathematics

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