What is the general solution of y′+1xy=0y' + \frac{1}{x}y = 0y′+x1y=0 for x>0x > 0x>0?
y=Cxy = \frac{C}{x}y=xC
y=Cxy = Cxy=Cx
y=C+ln(x)y = C + \ln(x)y=C+ln(x)
y=e−xy = e^{-x}y=e−x