This is a part of the vibration of an elastic beam solution
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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.

by Top Rated User (640k points)

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