 Solution

For the Cauchy-Euler equation $x^2{y}^{\prime \prime }-x{y}^{\prime }+y=0$, the power trial $y=x^r$ on $x>0$ gives $(r-1{)}^2=0$. The repeated root supplies the logarithmic second solution. On either half-line,One may also obtain this by setting $y=xu$, which reduces the equation to $(x{u}^{\prime }{)}^{\prime }=0$. Constants on opposite sides of the singular point need not agree.