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Substitute a Frobenius series $y={\sum }_{n\ge 0}{a}_n{z}^{n+\rho }$ into the transformed equation. For $n\ge 1$, equating the coefficient of $z^{n+\rho -1}$ gives the recurrence relationAt the larger indicial root $\rho =2$, the coefficient of $a_n$ is $n(n+2)\ne 0$, so a Frobenius solution always exists. At the smaller indicial root $\rho =0$, the first two equations areSince $a_0\ne 0$, the exceptional equation is consistent exactly whenThese are both admissible values, including the zero value; the singular wording of the question does not remove either. This is the undetermined coefficient at Frobenius resonance: at either value $a_2$ is free, whereas otherwise the second independent solution requires a Logarithmic Frobenius solution.

Explicit pairs demonstrate sufficiency and linear independence. In the variable $z=x+1$ one may takeBoth displayed $u_2$ have leading term $z^2/2$, while $u_0(0)=1$. They therefore give independent Frobenius solutions with the two indicial exponents, and direct differentiation verifies the equation. The point is an apparent regular singular point in these two cases.

For the requested closed forms, particularly short choices areIn addition, $(x+2)e^{-x}$ solves the zero-parameter equation and $e^{-x}$ solves the equation with $c=-1$.

Substituting $a_n=2^n/n!$ into the $r=0$ recurrence, and $a_n=1/n!$ into the $r=1/2$ recurrence, satisfies every coefficient equation. They give $e^{2x}$ and $\sqrt{x}{e}^x$. Thus on a positive interval not crossing a singular point,Their Wronskian is $e^{3x}(2x-1)/(2\sqrt{x})$, nonzero for $x>0$, $x\ne 1/2$, proving independence on each ordinary interval. Both functions are analytic near $1/2$, so this is an apparent regular singular point: a pole in normalized coefficients need not force a singular solution. The Wronskian can vanish at that singular point without making the two functions linearly dependent on an ordinary interval.