Consider the following equation $$u_{tt}+a^2u_{xxxx}=f(t),$$ for $-L<x<L$ and $t>0$, with boundary conditions $u(\pm L,t)=0=u_{x}(\pm L,t)$ and $u(x,0)=0=u_{t}(x,0)$. Im asked to prove that the eigenfunctions corresponding to different eigenvalues λ are mutually orthogonal, that the eigenvalues λ of the boundary-value problem are all either strictly positive or strictly negative and to obtain the solution as a series of eigenfunctions.
Firts I tried to solve the homogeneus case with separation of variables, and I ended up with the equations $$T''+a^2\lambda T=0,$$ and $$X^{iv}-\lambda X=0, X(\pm L)=0=X'(\pm L).$$ About the sign of the eigenvalues i ended up with them being strictly positive, i don't know if i made a mistake but with energy methods we can prove that $$\lambda \int_{-L}^L |X|^2dx=\int_{-L}^L X^{iv}\overline{X}dx=-\int_{-L}^L X'''\overline{X'}dx=\int_{-L}^L X''\overline{X''}dx=\int_{-L}^L|X''|^2dx.$$ So $\lambda\geq 0$, and $\lambda=0$ leds up to the trivial solution. Now the solution to the problem is $$X(x)=A\cosh (\mu x)+B\sinh (\mu x)+C\cos (\mu x)+D\sin (\mu x),$$ where $\lambda=\mu^4$ for some $\mu>0$. Imposing the boundary conditions we end with the trascendetal equation for the eigenvalues $$\tanh (\mu L)=\tan (\mu L),$$ so there are a family of strictly increasing eigenvalues $\mu_1<\mu_2<\cdots$, and the family of eigenfunctions $$X_n(x)=A_n\cosh (\mu_n x)+B_n\sinh (\mu_n x)+C_n\cos (\mu_n x)+D_n\sin (\mu_n x).$$ Now im stuck trying to prove that these eigenfunctions are orthogonal, and also how to proceed in the non homogeneus case, in particular i want $f(t)=C \sin (\omega t).$ Any help will be very appreciated!
The proof that eigenfunctions corresponding to different eigenvalues are orthogonal is similar to your proof that the eigenvalues are strictly positive. Let $X_m$ and $X_n$ be the eigenfunctions corresponding to $\lambda_m$ and $\lambda_n$, respectively, with $\lambda_m\neq\lambda_n$. Then \begin{align} \require{cancel} \lambda_m\int_{-L}^LX_mX_n\,dx&=\int_{-L}^LX_m^{(4)}X_n\,dx \\ &=\cancelto{0}{\left.X_m'''X_n\right|_{-L}^L}-\int_{-L}^LX_m'''X_n'\,dx \\ &=-\cancelto{0}{\left.X_m''X_n'\right|_{-L}^L}+\int_{-L}^LX_m''X_n''\,dx \\ &=\cancelto{0}{\left.X_m'X_n''\right|_{-L}^L}-\int_{-L}^LX_m'X_n'''\,dx \\ &=-\cancelto{0}{\left.X_mX_n'''\right|_{-L}^L}+\int_{-L}^LX_mX_n^{(4)}\,dx \\ &=\lambda_n\int_{-L}^LX_mX_n\,dx. \tag{1} \end{align} Since $\lambda_m\neq\lambda_n$, it follows from $(1)$ that $\int_{-L}^LX_mX_n\,dx=0.$
Now let's find the solution to $u_{tt}+a^2u_{xxxx}=f(t)$ expressed as a series of eigenfunctions. For this, we write the solution as $$ u(x,t)=\sum_{n=1}^{\infty}T_n(t)X_n(x), \tag{2} $$ where we assume the eigenfunctions $X_n$ are normalized to $1$, i.e., $\int_{-L}^LX_n^2\,dx = 1.$ Plugging $(2)$ into the PDE, we obtain $$ \sum_{n=1}^{\infty}[T_n''(t)+a^2\lambda_nT_n(t)]X_n(x)=f(t). \tag{3} $$ Multiplying both sides of $(3)$ by $X_m$, and using the orthogonality relation $\int_{-L}^LX_mX_n\,dx=\delta_{mn}$, we obtain the family of ODEs $$ T_m''(t)+a^2\lambda_mT_m(t)=C_mf(t)\qquad(m\in\mathbb{N}^{*}), \tag{4} $$ where $C_m=\int_{-L}^L X_m\,dx.$ In order to satisfy the initial condition $u(x,0)=0=u_t(x,0)$, we must solve $(4)$ with the initial condition $T_m(0)=0=T_m'(0)$. The result is $$ T_m(t)=\frac{C_m}{\omega_m}\int_0^t\sin(\omega_m(t-s))f(s)\,ds \qquad\left(\omega_m=\sqrt{a^2\lambda_m}\right). \tag{5} $$ Eqs. $(2)$ and $(5)$ give the desired solution.
Finally, for completeness, I'll present the explicit expression of the eigenfunctions $X_n(x)$. Since the differential operator $\frac{\partial^4}{\partial x^4}$ and the boundary conditions $X(\pm L)=0=X'(\pm L)$ commute with the parity operator $\mathcal{P}:x\mapsto -x$, we can find a basis of eigenfunctions that have definite parity, i.e., are even or odd functions of $x$. The even eigenfunctions have the form $$ X_{\text{even}}(x)=A_{\mu}[\cosh(\mu L)\cos(\mu x)-\cos(\mu L)\cosh(\mu x)]. \tag{6} $$ It's trivial to verify that $X_{\text{even}}(\pm L)=0$; however, in order that $X_{\text{even}}'(\pm L)=0$, $\mu$ must satisfy the condition $$ -\cosh(\mu L)\sin(\pm\mu L)-\cos(\mu L)\sinh(\pm\mu L)=0 \implies \tan(\mu L)=-\tanh(\mu L). \tag{7} $$ Similarly, the odd eigenfunctions have the form $$ X_{\text{odd}}(x)=B_{\mu}[\sinh(\mu L)\sin(\mu x)-\sin(\mu L)\sinh(\mu x)], \tag{8} $$ with $\mu$ fixed by the boundary condition $X_{\text{odd}}'(\pm L)=0$: $$ \sinh(\mu L)\cos(\pm\mu x)-\sin(\mu L)\cosh(\pm\mu x)=0 \implies \tan(\mu L)=\tanh(\mu L). \tag{9} $$ Notice that the series $(2)$ contains only even eigenfunctions, as $C_n=\int_{-L}^LX_n(x)\,dx=0$ if $X_n$ is an odd function of $x$.