attempt to solve a Bernoulli equation I tried solving the Bernoulli equation $y'-y\tan x = y^4 \cos x $ by equating the left hand side to 0 and finding the homogeneous solution, however the equation turned out to be too complex and without the ability to isolate x. Any hints on how to solve this? 

Another trick
$$y'-y\tan x = y^4 \cos x$$
$$\cos(x)y'-y\sin x = y^4 \cos^2 x$$
$$(\cos(x)y)' = y^4 \cos^2 x$$
Substitute $z=\cos(x)y$
$$z' = \frac {z^4} {\cos^2 x}$$
It's separable 
$$\frac 1{z^3}  =-3\int  \frac {dx} {\cos^2 x}$$
$$\frac 1{z^3}  =-3\tan x +K$$
$$z^3  =\frac 1 {-3\tan x +K}$$
$$\boxed{y^3(x)  =\frac 1{\cos^2(x)(K\cos(x)-3\sin x)}}$$
$$......$$
