It is given the following BVP \begin{align} u_{t} &= u_{xxxx},\quad\quad x \in (0, \pi), t >0 \tag1\\ u(0,t) &= u(\pi,t) = 0 ,\quad t > 0 \tag 2\\ u_{x}(0,t) &= e^{-4t},\quad t > 0 \tag3\\ u_{x}(\pi,t) &= -\cosh(\pi) \cdot e^{-4t}, t > 0 \ \tag4 \end{align}
and initial condition:
$u(x,0) = \sin(x)\cosh(x), x \in (0, \pi)$
Furthermore, $u$ is bounded as $t \to \infty$.
The Laplace transformed IBVP with initials condition is $$s \bar{u}-\bar{u}_{xxxx}-u(x,0)=0 $$ This is essentially an "ODE" (Fourth order, linear) in form $$\bar{u}^{(4)}-s \bar{u}+\sin(x)\cosh(x)=0$$ Now we are trying to solve the IBVP with the boundary conditions \begin{align} \bar{u}(0,s)&=0=\bar{u}(\pi ,s)\\ \bar{u}_{x}(0,s)&=\frac{1}{s+4}\\ \bar{u}_{x}(\pi ,s)&=-\frac{\cosh (\pi )}{s+4} \end{align}
However, I can not find the general solution and apply the inverse transform.