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? 
 A: 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 
$$\frac 1{z^3}  =-3\int  \frac {dx} {\cos^2 x}$$
$$\frac 1{z^3}  =-3\tan x +K$$
$$z^3  =\frac 1 {-3\tan x +K}$$
$$\boxed{y^3(x)  =\frac 1{\cos^2(x)(K\cos(x)-3\sin x)}}$$
$$......$$
A: $y' - y\tan(x) = y^4\cos(x)$
divide throughout by $y^4$
$\frac{y'}{y^4} - \frac1{y^3}\tan(x) = \cos(x)$
let $z= \frac{1}{y^3}\implies z' = \frac{-3}{y^4}y'$
$\frac{-z'}{3} -z\tan(x) = \cos(x)$
$z' + 3\tan(x)\,z = -3\cos(x)$
it is now a Linear differential equation
Integrating factor , $I=e^{\int3\tan(x)}=e^{\ln(\sec^3(x))} = \sec^3(x)$
the solution is given by ;
$z\cdot I =\int-3\cos(x)\cdot I\,dx$
integrate and sub back for $z$
Can you proceed further?Ask if you need help.
EDIT:  on how we got $z\cdot I =\int-3\cos(x)\cdot I\,dx$;
we have $z' + 3\tan(x)\,z = -3\cos(x)$
multiply throughout by $\sec^3(x)$
we get ; $z'\cdot\sec^3(x) +3\cdot\tan(x)\cdot\sec^3(x)\cdot z = -3\cos(x)\cdot\sec^3(x)$
recognize that the LHS is a product rule derivative of  $z\cdot\sec^3(x)$
ie $d(z\cdot\sec^3(x)) =(z'\cdot\sec^3(x) +3\cdot\tan(x)\cdot\sec^3(x)\cdot)\,dx z$ 
therefore  the equation becomes ;
$\big(z\cdot\sec^3(x)\big)' = 3\cos(x)\cdot\sec^3(x)$
integrating on both sides gives us;
$z\cdot I =\int-3\cos(x)\cdot I\,dx$; $\quad$ where $I= \sec^3(x)$
A: Hint:
 Divide both side by $y^4$ to obtain
$$\frac{y'}{y^4}-\frac{\tan x}{y^3}=\cos x,$$
and set $u=\dfrac 1{y^3}$. The equation rewrites as
$$-\frac13 u'-u\sin x=\cos x\iff u'=-3u\tan x-3\cos x,$$
which is a classical linear differential equation.
