TThe method I’ve used is not the requested method but it does yield a general solution.
F''''+F=0 means we need to solve the characteristic equation r⁴+1=0, which has complex roots:
(√2/2)(1+i), (√2/2)(1-i), -(√2/2)(1+i), -(√2/2)(1-i).
For legibility let p=√2/2.
These roots give us the general solution F(x)=e^(px)[Ae^(ipx)+Be^-(ipx)]+e^-(px)[Ce^(ipx)+De^-(ipx)], where A, B, C, D are constants.
e^(ipx)=cos(px)+isin(px) and e^-(ipx)=cos(px)-isin(px) by de Moivre’s Theorem.
So F(x)=e^(px)[A(cos(px)+isin(px)+B(cos(px)-isin(px)]+
e^-(px)[C(cos(px)+isin(px)+D(cos(px)-isin(px)].
If we want to remove the imaginary parts, A=B and C=D.
F(x)=2Ae^(px)cos(px)+2Ce^-(px)cos(px). Let a=2A and b=2C, so:
F(x)=cos(px)(ae^(px)+be^-(px)).
Replace p=√2/2:
F(x)=cos(x√2/2)(ae^(x√2/2)+be^-(x√2/2)) is the general non-complex solution.