Determine the solution of y′=3x2y' = 3x^2y′=3x2 with y(0)=2y(0) = 2y(0)=2.
y=x3+2y = x^3 + 2y=x3+2
y=3x3+2y = 3x^3 + 2y=3x3+2
y=x3−2y = x^3 - 2y=x3−2
y=x2+2y = x^2 + 2y=x2+2