Consider the ODE y′+2xy=0y' + 2xy = 0y′+2xy=0. Which function is the solution passing through (0,1)(0, 1)(0,1)?
y=ex2y = e^{x^2}y=ex2
y=e−x2y = e^{-x^2}y=e−x2
y=e2x2y = e^{2x^2}y=e2x2
y=2e−x2y = 2e^{-x^2}y=2e−x2