Why doesn't this method work? (Re: Simultaneous ODEs)

[Hoplite]

I have been working on a derivation in which the following simultateous ordinary differential equations have appeared:
$$f^{(4)}(x)-2 a^2 f''(x)+a^4 f(x)+b(g''(x)-a^2 g(x))=0,$$
$$g^{(4)}(x)-2 a^2 g''(x)+a^4 g(x)-b(f''(x)-a^2 f(x))=0,$$
where $$a$$ and $$b$$ are constants. I figured that I could solve this using Fourier transforms. First, I transform the above, using $$\nu$$ as the transform space analogue of $$x$$, into the following:
$$(\nu^2 +a^2 )\hat f(\nu ) -b \hat g(\nu )=0,$$
$$(\nu^2 +a^2 )\hat g(\nu ) +b \hat f(\nu )=0.$$
I then rearrange these equations via substitution into
$$[(\nu^2 +a^2 )^2 +b^2 ] \hat f(\nu ) =0,$$
$$[(\nu^2 +a^2 )^2 +b^2 ] \hat g(\nu )=0.$$
Taking the inverse Fourier transform then results in
$$f^{(4)}(x)-2a^2 f''(x)+(a^4 +b^2 )f(x)=0,$$
$$g^{(4)}(x)-2a^2 g''(x)+(a^4 +b^2 )g(x)=0.$$

However, after I solve for these two ODEs using the boundary conditions, I find that the resulting answer is erroneous. I can only conclude that the above methodoly doesn't work, although I can't see why.

Could anyone point out the mathematical error in the above steps?

 

[lanedance]

can you show your transform... shouldn't you get factors of $nu^4$?

 

[vela]

The OP is dividing out a common factor of $\nu^2+a^2$.

 

[Hoplite]

The OP is dividing out a common factor of $\nu^2+a^2$.

Yes, that's what I've done.