To follow up on my remark about DSolve[] not solving for parameters, if one makes mu a dependent variable, then DSolve[] will solve for it. This can be done by replacing μ by μ[x] and adding the differential equation μ'[x] == 0. Unfortunately, the default solver uses inverse functions in this case. That yields only one solution, the trivial solution, instead of all solutions:

ode = yy''''[x] + \[Mu]^2 yy''[x] == 0;
bcs = {yy[0] == 0, yy[l] == 0, yy''[0] == 0, yy''[l] == 0};
DSolve[{ode /. \[Mu] -> \[Mu][x], \[Mu]'[x] == 0, bcs}, {yy[x], \[Mu][x]}, x,
 Assumptions -> \[Mu][x] > 0 && l > 0]

(*  `DSolve::ifun`: Inverse functions are being used by DSolve, so some solutions may not be found.  *)
(*  {{yy[x] -> 0, \[Mu][x] -> C[1]}}  *)

However, we can use what is often called the Villegas-Gayley trick to adjust how Solve[] works:

Internal`InheritedBlock[{Solve},
 Unprotect[Solve];
 Solve[eq_, v_, opts___] /; ! TrueQ[$in] := Block[{$in = True},
   Simplify@Solve[Flatten@{eq, $Assumptions}, v, Method -> Reduce, opts]];
 Protect[Solve];
 DSolve[{ode /. \[Mu] -> \[Mu][x], \[Mu]'[x] == 0, bcs}, {yy[x], \[Mu][x]}, x,
  Assumptions -> \[Mu][x] > 0 && l > 0 && C[1] > 0]
 ]
(*
{{yy[x] -> 0, \[Mu][x] -> C[1]}, 
  {yy[x] -> ConditionalExpression[
     -((l^3*C[3] Sin[(2*Pi*x*C[6])/l])/(8*Pi^3*C[6]^3)),
     Element[C[6], Integers] && C[6] >= 1], 
   \[Mu][x] -> ConditionalExpression[
     (2*Pi*C[6])/l, 
     Element[C[6], Integers] && C[6] >= 1]}, 
  {yy[x] -> ConditionalExpression[
     -((l^3*C[3] Sin[(x*(Pi + 2*Pi*C[6]))/l])/(Pi^3*(1 + 2*C[6])^3)), 
     Element[C[6], Integers] && C[6] >= 0], 
   \[Mu][x] -> ConditionalExpression[
     (Pi + 2*Pi*C[6])/l, 
     Element[C[6], Integers] && C[6] >= 0]}}
*)

The extra condition C[1] > 0 is not necessary but simplifies the output. It was added as an afterthought, after seeing \[Mu][x] was solved as \[Mu][x] -> C[1] in terms of C[1]. The assumption \[Mu][x] > 0 is not automatically extended to include C[1].

One could simplify the values in the solution to C[3] Sin[2 π x C[6] / l] and C[3] Sin[x (π + 2 π C[6]) / l], if one wanted to. A general code to do so takes some work.