<s>[INST] 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? 
 [/INST] $y' - y\tan(x) = y^4\cos(x)$
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 $I= \sec^3(x)$
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