 Solution

On an interval with $x>0$, use the power-law ansatz $y=x^r$. Substitution gives $r(r-1)+2r+1=r^2+r+1=0$, whose roots are $r=-1/2\pm i\sqrt{3}/2$. Taking real and imaginary parts of $x^r$ givesThese two real solutions are linearly independent: their Wronskian is $(\sqrt{3}/2)x^{-2}\ne 0$. They therefore span the solution space of the second-order linear ordinary differential equation on that interval.

For $x<0$, replace $x^{-1/2}$ by $|x{|}^{-1/2}$ and $\log x$ by $\log |x|$, with independent real constants on the negative interval. This follows equally by the change of variable $t=\log |x|$, which turns the Cauchy-Euler equation into $d^{2}y/d{t}^2+dy/dt+y=0$. The coefficient of $y^{\prime \prime }$ vanishes at zero, so the general formula is stated separately on either side. Every nonzero mode has an unbounded oscillatory envelope as $x\to 0$; a continuous extension through zero requires the solution to vanish identically on both sides.