= Solution
Put $B=EI_2$, the <filament bending modulus> in the easier bending direction; $I_2$ is the relevant <second moment of area>. Take <filament tension> as positive in extension. A cut at height $z$ supports the weight of the material above it, so its axial force is compressive:
$$
\boxed{T(z)=-\lambda g(L-z)},\qquad T(L)=0.
$$
The small transverse displacement induces a transverse tension force density $(TX_z)_z$, while bending produces $-BX_{zzzz}$. Their balance gives
$$
\boxed{BX_{zzzz}-(TX_z)_z=0}.
$$
The same result follows from <elastic energy> and <gravitational energy>. To quadratic order, local <inextensibility> shortens the height of the material point labelled $z$ by $\frac12\int_0^zX_z(s)^2ds$. Integrating its weight and interchanging the integrals gives the second-order <potential energy>
$$
\mathcal E_2[X]=\frac12\int_0^L\left[B X_{zz}^2-\lambda g(L-z)X_z^2\right]dz
=\frac12\int_0^L\left[B X_{zz}^2+T X_z^2\right]dz.
$$
Two <integration by parts> steps in its first variation give
$$
\delta\mathcal E_2=\int_0^L[BX_{zzzz}-(TX_z)_z]\delta X\,dz
+\left[BX_{zz}\delta X_z+(TX_z-BX_{zzz})\delta X\right]_0^L.
$$
The <clamped boundary conditions> fix displacement and slope at the base, and the unconstrained tip must have zero bending moment and transverse end force. Thus the complete four conditions are
$$
\boxed{X(0)=X_z(0)=0,\qquad X_{zz}(L)=0,\qquad
BX_{zzz}(L)-T(L)X_z(L)=0}.
$$
Since the <filament tension> vanishes at the tip, the last condition is $X_{zzz}(L)=0$. Integrate the <ordinary differential equation> once: $BX_{zzz}-TX_z$ is constant, and its tip value makes that constant zero. For the slope $u=X_z$, it follows that
$$
B u_{zz}+\lambda g(L-z)u=0,\qquad
\boxed{u(0)=0,\quad u_z(L)=0}.
$$
The other tip condition $u_{zz}(L)=0$ already follows from this equation for a regular solution; it is not a third independent condition on the second-order slope problem. Finally reconstruct $X(z)=\int_0^zu(s)ds$ to impose the base position.
Set $x=L-z$, $\alpha=\lambda g/B$, and $\eta=(2/3)\sqrt\alpha\,x^{3/2}$. Then $d\eta/dx=3\eta/(2x)$ and $d^2\eta/dx^2=3\eta/(4x^2)$, so $u_{xx}+\alpha xu=0$ becomes
$$
\eta^2u_{\eta\eta}+\frac13\eta u_\eta+\eta^2u=0.
$$
Substituting $u=\eta^{1/3}F(\eta)$ gives
$$
\eta^2F''+\eta F'+\left(\eta^2-\frac19\right)F=0.
$$
This is the <Bessel differential equation> of order $1/3$, whose two independent <Bessel functions of the first kind> give
$$
\boxed{u=\eta^{1/3}\left[aJ_{-1/3}(\eta)+bJ_{1/3}(\eta)\right]}.
$$
The tip is $x=\eta=0$. To decide which branch is admissible, retain both leading behaviours, not just the fact that $u$ stays finite:
$$
\eta^{1/3}J_{-1/3}(\eta)
=\frac{2^{1/3}}{\Gamma(2/3)}\left[1-\frac38\eta^2+O(\eta^4)\right],\qquad
\eta^{1/3}J_{1/3}(\eta)
=\frac{\eta^{2/3}}{2^{1/3}\Gamma(4/3)}\left[1+O(\eta^2)\right].
$$
Since $\eta^2\propto x^3$ and $\eta^{2/3}\propto x$, the first branch has zero tip derivative, while the second has a nonzero tip derivative. The moment-free condition therefore sets $b=0$, and it does not require the tip slope itself to vanish. The base condition now requires
$$
J_{-1/3}\left(\frac23\sqrt{\frac{\lambda gL^3}{EI_2}}\right)=0.
$$
Let $j_{-1/3,1}$ be the first positive zero. The <Bessel threshold for self-buckling of a vertical rod> is
$$
\boxed{\frac{\lambda gL_c^3}{EI_2}=\frac94j_{-1/3,1}^2\simeq7.83735,
\qquad L_c\simeq1.98635\left(\frac{EI_2}{\lambda g}\right)^{1/3}}.
$$
To identify this as the first loss of stability, write the quadratic energy in terms of the slope as $\frac12[B\int u_z^2dz-\lambda g\int(L-z)u^2dz]$. On functions with $u(0)=0$, its critical coefficient is the minimum <Rayleigh quotient> $\int u_z^2dz/\int(L-z)u^2dz$; variation supplies the moment-free condition at the tip and the same <Sturm-Liouville problem>. The first zero gives the lowest <eigenvalue>, so \b[the straight state is stable below this load, neutral at it, and unstable above it], with the easier $I_2$ direction buckling first.
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