What is the general solution of y′+yx=1x2y' + \frac{y}{x} = \frac{1}{x^2}y′+xy=x21?
y=lnx+Cxy = \frac{\ln x + C}{x}y=xlnx+C
y=lnx+Cx2y = \frac{\ln x + C}{x^2}y=x2lnx+C
y=x+Cxy = \frac{x + C}{x}y=xx+C
y=lnx+Cy = \ln x + Cy=lnx+C