1. **Problem statement:** We consider the odd-order linear differential equation $$y^{(2n+1)} + \sum_{k=0}^{2n} a_k y^{(k)} = 0, \quad x \in (0,L),$$ with constants $a_k \in \mathbb{C}$ and parameter $L>0$. Boundary conditions are $$y^{(k)}(0) = 0, \quad y^{(k)}(L) = 0, \quad k=0,1,\dots,n.$$ We want to determine all $L>0$ for which there exists a nontrivial solution. 2. **Key idea:** This is a linear boundary value problem (BVP) for an odd-order ODE with $2n+2$ boundary conditions. 3. **General solution:** The characteristic polynomial is $$p(\lambda) = \lambda^{2n+1} + \sum_{k=0}^{2n} a_k \lambda^k = 0.$$ Let its roots be $\lambda_1, \lambda_2, \dots, \lambda_{2n+1}$ (counted with multiplicity). 4. The general solution is $$y(x) = \sum_{j=1}^{2n+1} c_j e^{\lambda_j x}$$ for constants $c_j$. 5. **Applying boundary conditions:** For each $k=0,1,\dots,n$, $$y^{(k)}(0) = \sum_{j=1}^{2n+1} c_j \lambda_j^k = 0,$$ $$y^{(k)}(L) = \sum_{j=1}^{2n+1} c_j \lambda_j^k e^{\lambda_j L} = 0.$$ 6. This gives a homogeneous linear system of $2n+2$ equations in $2n+1$ unknowns $c_j$: $$\begin{cases} \sum_{j=1}^{2n+1} c_j \lambda_j^k = 0, & k=0,\dots,n \\ \sum_{j=1}^{2n+1} c_j \lambda_j^k e^{\lambda_j L} = 0, & k=0,\dots,n \end{cases}$$ 7. Since there are more equations than unknowns, the trivial solution $c_j=0$ always satisfies this. 8. For a nontrivial solution to exist, the coefficient matrix of this system must be singular. 9. The matrix is $$M(L) = \begin{pmatrix} 1 & 1 & \cdots & 1 \\ \lambda_1 & \lambda_2 & \cdots & \lambda_{2n+1} \\ \vdots & \vdots & & \vdots \\ \lambda_1^n & \lambda_2^n & \cdots & \lambda_{2n+1}^n \\ 1 & 1 & \cdots & 1 \\ \lambda_1 e^{\lambda_1 L} & \lambda_2 e^{\lambda_2 L} & \cdots & \lambda_{2n+1} e^{\lambda_{2n+1} L} \\ \vdots & \vdots & & \vdots \\ \lambda_1^n e^{\lambda_1 L} & \lambda_2^n e^{\lambda_2 L} & \cdots & \lambda_{2n+1}^n e^{\lambda_{2n+1} L} \end{pmatrix}^T$$ 10. The condition for nontrivial solutions is $$\det M(L) = 0.$$ 11. **Interpretation:** The values of $L>0$ for which $\det M(L) = 0$ are the eigenvalues of the boundary value problem. 12. **Summary:** - Find roots $\lambda_j$ of characteristic polynomial. - Form matrix $M(L)$ from boundary conditions. - Solve $\det M(L) = 0$ for $L>0$. These $L$ are exactly the values for which nontrivial solutions exist. **Final answer:** $$\boxed{\text{Nontrivial solutions exist if and only if } \det M(L) = 0, \text{ where } M(L) \text{ is defined above.}}$$