= Solution
Integrating the <ordinary differential equation> four times gives a quartic <polynomial>. It is useful to use the <linearity> of the equation and split the solution into a gravity part and a force part:
$$
y=y_0+y_F,
$$
where
$$
y_0(x)
=-\frac{\rho g}{24A}x^2(6L^2-4Lx+x^2)
$$
satisfies the clamped and torque-free <boundary conditions> with $F=0$, while
$$
y_F(x)=\frac{F}{6A}x^2(3L-x)
$$
satisfies the homogeneous equation and contributes the endpoint <force>. Direct <differentiation> verifies
$$
y(0)=y'(0)=0,\qquad y''(L)=0,\qquad -Ay'''(L)=F.
$$
Thus
$$
\boxed{
y(x)=-\frac{\rho g}{24A}x^2(6L^2-4Lx+x^2)
+\frac{F}{6A}x^2(3L-x)
},
$$
the <clamped-free beam under uniform load and endpoint force>.
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