attempt to solve a Bernoulli equation

Question

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?

Answer 1

$y' - y\tan(x) = y^4\cos(x)$

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 $I= \sec^3(x)$

Answer 2

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.

Answer 3

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 $$\frac 1{z^3} =-3\int \frac {dx} {\cos^2 x}$$ $$\frac 1{z^3} =-3\tan x +K$$ $$z^3 =\frac 1 {-3\tan x +K}$$ $$\boxed{y^3(x) =\frac 1{\cos^2(x)(K\cos(x)-3\sin x)}}$$ $$......$$