Given the differential equation y' = 6 x y and y(0) = 10 and its next few implicit derivatives y" = 6y + 6x y' y"' = 12y' + 6x y" y4 = 18y" + 6 x y"" Use the initial value "y(0) = 10" to find the value of "y'(0)", "y"(0)", "y"(0)", and "y*(0)", then plug them into the Taylor Series Polynomial formula. 6 x2 + 10 12 18 A y(x) O! x! + 2! | ... 1! 3! 4! 60 x! + + 1! 3660 x2 22680 201960 10 y(x) = O! (B + +... 2! 3! 4! C) y(x) = 0! x2 + 2! %3! ... 1! 3! 4! 60 x2 10 1080 (D) D y(x): to + + 3! | O! 1! 2! 4! (E) y(x) = 0 10 x! + 60 x2 + 2! 720 x3 1080 x4 | | ... 0! 1! 3! 4! 10 x° + 60 x + 60 x2 + 720 x + 3! 1080 F) y(x) x4 + 4! ... O! 1! 2! + + +

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