Find the solution to y′=3x2y' = 3x^2y′=3x2 with y(1)=2y(1) = 2y(1)=2.
y=x3+1y = x^3 + 1y=x3+1
y=x3+2y = x^3 + 2y=x3+2
y=3x3y = 3x^3y=3x3
y=x3y = x^3y=x3