(du/dx)+u+au^(5/4)+u^4 = 0
(du/dx) = -(u+au^(5/4)+u^4)
-du/(u+au^(5/4)+u^4) = dx
so we reach the final conclusion
-x = Int 1/(u+au^(5/4)+u^4) du
So you now have a functional form x(u). To get u(x), you must find an inversion, at least locally.
The next problem is exactly the same...with one change...
-x = Int 1/(u+au^(5/4)+bu^4) du
Invert x(u) to get the desired functional form u(x)
Unfortunately, I'm not certain if I can get any further on this...
because even though the inverse function theorem states that there exist neighborhoods where this function is bijective with non-zero derivative, and invertible in such neighborhoods, admittedly it is far more difficult to find such functional inverses...I apologize for not being of more assistance, but at least this gives you a start...
Where the response above mine goes wrong is it integrates with respect to u, and not x. If we just integrate the LHS with respect to x, Int (du/dx) du...that is not in fact u.
Also, note: His answer was all u's...u=0 is indeed a solution to this, but there are others, namely constants satisfying
u+au+u^4 = 0, in the first problem,
u+au+bu^4 = 0, in the second problem.
Further, x(u) gives functional forms of x, which when inverted, will give non-trivial solutions...