Given the differential equation y' = 8y2 and y (0) = - 1 and its next few implicit derivatives y" = 16y y' y"' = 16 (y')2 + 16y y" y4 = 48 y'y" + 16 y y"" %3D Use the initial value "y(0) = -1" to find the value of "y'(0)", "y"(0)", "y"(0)", and "y (0)", then plug them into the Taylor Series Polynomial formula. 32 16 x2 + 2! 64 A y(x) 8 x! + 1! x° + x3 + x4 ... O! 3! 4! 8. 128 x2 128 384 By(x). O! x° + x! 1! 2! 3! 4! - 16 x2 + 112 x + 3! -4352 y(x). x! + 1! O! 2! 4! - 98304 8. x' + 1! -128 3072 Dy(x)= 0! x2 + x* +... 2! 3! 4! 2176 83968 Ey(x) 8. x' + 1! 128 x2 + x° + 0! 2! 3! 4! 480 960 F y(x) x° + O! x' + 1! x + 3! | ... 2! 4! + + + +

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