How to use boundary condition for Euler-Bernoulli equation How to use enter boundary condition into a Euler-Bernoulli fourth order equation
 A: OK, first integrate step by step to obtain
$$\begin{array}{*{20}{l}}
  {\frac{{{d^4}y}}{{d{x^4}}} = F} \\ 
  {\frac{{{d^3}y}}{{d{x^3}}} = Fx + A} \\ 
  {\frac{{{d^2}y}}{{d{x^2}}} = \frac{F}{2}{x^2} + Ax + B} \\ 
  {\frac{{dy}}{{dx}} = \frac{F}{6}{x^3} + \frac{A}{2}{x^2} + Bx + C} \\ 
  {y = \frac{F}{{24}}{x^4} + \frac{A}{6}{x^3} + \frac{B}{2}{x^2} + Cx + D} 
\end{array}$$
Next, apply your boundary conditions to obtain a linear algebraic system for the unknown constants $A$, $B$, $C$, $D$
$$\begin{array}{*{20}{l}}
  {y(0) = 0}& \to &{D = 0} \\ 
  {y(l) = 0}& \to &{\frac{F}{{24}}{l^4} + \frac{A}{6}{l^3} + \frac{B}{2}{l^2} + Cl + D = 0} \\ 
  {\frac{{dy}}{{dx}}(0) = 0}& \to &{C = 0} \\ 
  {\frac{{dy}}{{dx}}(l) = 0}& \to &{\frac{F}{6}{l^3} + \frac{A}{2}{l^2} + Bl + C = 0} 
\end{array}$$
and hence
$$\left\{ \begin{gathered}
  \frac{A}{6}{l^3} + \frac{B}{2}{l^2} =  - \frac{F}{{24}}{l^4} \hfill \\
  \frac{A}{2}{l^2} + Bl =  - \frac{F}{6}{l^3} \hfill \\ 
\end{gathered}  \right.\,\,\,\, \to \,\,\,\,\left\{ \begin{gathered}
  \frac{l}{6}A + \frac{1}{2}B =  - \frac{{F{l^2}}}{{24}} \hfill \\
  \frac{l}{2}A + B =  - \frac{{F{l^2}}}{6} \hfill \\ 
\end{gathered}  \right.$$
Solve for $A$ and $B$. I will leave obtaining the constants to you. :)
