Well, you could use Variation of Parameters.
The fourth order ODE $$ y^{(4)} - 16 y = u' + u $$ has the homogeneous solution $$ y_h(x) = c_1 \cosh(2 x) + c_2 \sinh(2 x) + c_3 \cos(2 x) +c_4 \sin(2 x). $$ Now, for the particular solution, take $$ y_p(x) = a_1(x) \cosh(2 x) + a_2(x) \sinh(2 x) + a_3(x) \cos(2 x) + a_4(x) \sin(2 x). $$ where \begin{align} a_1'(x) \cosh(2 x) + a_2'(x) \sinh(2 x) + a_3'(x) \cos(2 x) + a_4'(x) \sin(2 x) &=0 \\ a_1'(x) \sinh(2 x) + a_2'(x) \cosh(2 x) - a_3'(x) \sin(2 x) + a_4'(x) \cos(2 x)&=0 \\ a_1'(x) \cosh(2 x) + a_2'(x) \sinh(2 x) - a_3'(x) \cos(2 x) - a_4'(x) \sin(2 x)&=0 \end{align}
Substituting in the ODE we have $$ a_1'(x) \sinh(2 x) + a_2'(x) \cosh(2 x) + a_3'(x) \sin(2 x) - a_4'(x) \cos(2 x) = \frac{1}{8}\big(u'(x) +u(x)\big). $$ This four equations can be solved for $a_1'(x)$, $a_2'(x)$,$a_3'(x)$ and $a_4'(x)$, leading to \begin{align} a_1'(x) &= -\frac{\sinh(2 x)}{16} \big(u'(x) + u(x)\big)\\ a_2'(x) &= \frac{\cosh(2 x)}{16} \big(u'(x) + u(x)\big)\\ a_3'(x) &= \frac{\sin(2 x)}{16} \big(u'(x) + u(x)\big)\\ a_4'(x) &= -\frac{\cos(2 x)}{16} \big(u'(x) + u(x)\big) \end{align}
and, assuming that it's an Initial Value Problem given at $x = 0$ (w.l.g), \begin{align} a_1(x) &= -\frac{\sinh(2 x) u(x)}{16} - \int_0^x \frac{\sinh(2 t) - 2\cosh(2 t)}{16} u(t)\, dt\\ a_2(x) &= \frac{\cosh(2 x) u(x) - u(0)}{16} + \int_0^x \frac{\cosh(2 t) - 2\sinh(2 t)}{16} u(t)\, dt\\ a_3(x) &= \frac{\sin(2 x) u(x)}{16} + \int_0^x \frac{\sin(2 t) - 2 \cos(2 t)}{16} u(t)\, dt\\ a_4(x) &= -\frac{\cos(2 x) u(x) - u(0)}{16} - \int_0^x \frac{\cos(2 t) + 2\sin(2 t)}{16} u(t)\, dt \end{align}
the solution is \begin{multline} y(x) = \big(c_1 + a_1(x)\big) \cosh (2 x) + \big(c_2 + a_2(x)\big) \sinh (2 x) \\ + \big(c_3 + a_3(x)\big) \cos (2 x) + \big(c_4 + a_4(x)\big) \sin (2 x). \end{multline}
For a Boundary Value Problem the construction is a bit different, but the same principle can be applied.