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After division by $z$, the coefficient functions areBoth $zP(z)=1-z$ and $z^{2}Q(z)=\lambda z$ are analytic at zero, so the regular singular point criterion for a second-order equation shows that $z=0$ is regular singular. Substitution of $y=z^r$ into the leading terms gives the indicial equationThe exponent is repeated. Directly substituting $y={\sum }_{n\ge 0}{a}_n{z}^n$ givesThus $a_0$ determines the entire power series, so there is only one such solution $y_1$ up to scale. The logarithmic solution from a repeated Frobenius exponent predictsso its leading nonanalytic term is proportional to $\log z$.

For the contour ansatz, differentiation under the integral and one integration by parts giveThe integral vanishes whenwhose solution is the Laguerre contour-integral amplitudeThe contour and branches must make $f$ single-valued along the traversed path and must kill the endpoint term

Now suppose $\lambda <0$ is nonintegral. Near $t=0$, $f(t)=O(t^{-\lambda -1})$ is integrable and the endpoint factor is $O(t^{-\lambda })\to 0$. We may therefore choosea finite loop based at the branch point $0$ and avoiding $1$ except by encirclement. A second choice isalong one bank of the negative real axis, with a consistent branch. At $-\infty$, $e^{zt}$ kills the algebraic endpoint factor because $\mathrm{Re}z>0$.

The ${\gamma }_1$ integral is analytic in $z$ because its contour is finite, and it is a nonzero solution. By uniqueness of the analytic local solution, it is a constant multiple of $y_1$. This is the Finite Laguerre contour solution construction.

Seek a Frobenius solution $y=z^r{\sum }_{n\ge 0}{a}_n{z}^n$ with $a_0\ne 0$. Since $(z+z^3{)}^2=z^2+2z^4+z^6$, its lowest coefficient gives the indicial equationThus the repeated indicial root is $\boxed{r=-1}$. Equating the coefficient of $z^{n+r}$ before setting $r=-1$ givesFor $n\ge 1$ this becomes the linear recurrence relationEvery coefficient is fixed by $a_0$, since the denominator never vanishes for $n\ge 1$. In particular $a_1=0$ and then every odd coefficient is zero. Therefore there is $\boxed{\text{exactly one independent pure Frobenius series}}$, not two merely because the indicial root repeats. If "power series" means an ordinary Taylor series with nonnegative integer powers, the only such solution is zero: its coefficient equation starts with $a_0=0$ and inductively forces all coefficients to vanish.

A second independent local solution is a logarithmic solution from a repeated Frobenius exponent. Indeed, Abel's identity gives $W=C{z}^{-3}[1+O(z^2)]$. For the normalized first solution $y_1=z^{-1}[1+O(z^2)]$, reduction of order givesAfter choosing the harmless multiplicative constant in the numerator, its integrand is $1/s+O(s)$, so $y_2=y_1\log z+z^{-1}\times \text{an analytic series}$. The nonzero logarithmic coefficient proves that this second solution is not another pure Frobenius solution.