Which of the following is the correct solution to y′=x2yy' = x^2 yy′=x2y with y(0)=1y(0) = 1y(0)=1?
y=ex3/3y = e^{x^3/3}y=ex3/3
y=x3/3+1y = x^3/3 + 1y=x3/3+1
y=ex2y = e^{x^2}y=ex2
y=ex3y = e^{x^3}y=ex3