Solve the initial value problem y′=2x(1+y2)y' = 2x(1+y^2)y′=2x(1+y2) with y(0)=0y(0)=0y(0)=0.
y=tan(x2)y = \tan(x^2)y=tan(x2)
y=sin(x2)y = \sin(x^2)y=sin(x2)
y=arctan(x2)y = \arctan(x^2)y=arctan(x2)
y=ex2y = e^{x^2}y=ex2