Find the general solution of y′=2xyy' = \frac{2x}{y}y′=y2x where y(0)=1y(0) = 1y(0)=1.
y=2x2+1y = \sqrt{2x^2 + 1}y=2x2+1
y=x2+1y = x^2 + 1y=x2+1
y=x2+1y = \sqrt{x^2 + 1}y=x2+1
y=2x+1y = 2x + 1y=2x+1