Solve the initial value problems in Exercises 47–52. dy 47. = e' sin (e' – 2), y(ln 2) = 0 dt dy 48. = esec²(Te¯™), y(ln 4) = 2/T dt d'y = 2e*, y(0) = 1 and y'(0) = 0 49. dx? d²y 50. dt? 1 - e2", y(1) = -1 and y'(1) = 0 %3D dy 51. dx x y(1) = 3 d²y = sec? x, y(0) = 0 and y'(0) = 1 52. dx²

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