Solve the differential equation: $6 \cos^2(x) \dfrac{dy}{dx} -y \sin(x)+2y^4 \sin^3(x)=0$ I have the following differential equation before me:
$6 \cos^2(x) \dfrac{dy}{dx} -y \sin(x)+2y^4 \sin^3(x)=0$
I tried solving it by reducing to Bernoulli form of first order differential equation.
I divided both sides of the equation by $6 \cos^2(x)y^4$ to get:
$\dfrac{1}{y^4} \dfrac{dy}{dx}-\dfrac{1}{y^3} \dfrac{\sin(x)}{6\cos^2(x)}=\dfrac{-\sin^3(x)}{3\cos^2(x)}$
Then I took $\dfrac{1}{y^3}=z$
so that $\dfrac{-3}{y^4}\dfrac{dy}{dx}=\dfrac{dz
}{dx}$
This gave me:
$\dfrac{dz}{dx}+\dfrac{\sin(x)}{2\cos^2(x)}z=\dfrac{\sin^3(x)}{\cos^2(x)}$
This is first order linear differential equation of the type
$ \dfrac{dz}{dx}+P(x)z=Q(x)$
Integrating Factor (IF) is given by:
$IF=e^{\int P(x) \,dx}= e^{\int \dfrac{\sin(x)}{2\cos^2(x)}\,dx}= e^{\dfrac{1}{2\cos(x)}}$
Next, solution is given by:
$z.IF= \int IF.Q(x)$
or
$z.e^{\dfrac{1}{2\cos(x)}}=\int e^{\dfrac{1}{2\cos(x)}}.\dfrac{\sin^3(x)}{\cos^2(x)} \,dx$
This is the step where I falter as I find myself unable to tackle the integral on the RHS of the above equation.
You are requested to help me with the evaluation of this integral or come up with another way of tackling this differential equation. Any help will be highly appreciated.
 A: I am writing $\frac{dy}{dx}$ as $y'$
Our equation is $$6\operatorname{cos}^2(x)y'=\operatorname{sin}(x)y-2\operatorname{sin}^3(x)y^4$$ $=$
$$y'-\frac{\operatorname{sin}(x)y}{6\operatorname{cos}^2(x)}=-\frac{\operatorname{sin}^3(x)y^4}{3\operatorname{cos}^2(x)}$$ Dividing by $y^4$,
$$\frac{y'}{y^4}-\frac{\operatorname{sin}(x)}{6\operatorname{cos}^2(x)y^3}=-\frac{\operatorname{sin}^3(x)}{3\operatorname{cos}^2(x)}$$
Let $u=\frac{1}{y^3}$ therefore $y=\frac{1}{u^{\frac13}}$ and $u'=\frac{-3y'}{y^4}$ and $y'=\frac{-u'y^4}{3}$
Substituting all the values, our equation becomes $$-\frac{u\operatorname{sin}(x)}{6\operatorname{cos}^2(x)}-\frac{u'}{3}=-\frac{\operatorname{sin}^3(x)}{3\operatorname{cos}^2(x)}$$ which is equivalent to $$\frac{u\operatorname{sin}(x)}{2\operatorname{cos}^2(x)}+u'=\frac{\operatorname{sin}^3(x)}{\operatorname{cos}^2(x)}$$ Let $u=tv$ where $t,v$ are two functions $\implies$ $u'=tv'+t'v$ Rewriting the equation in terms of functions and then grouping we get, $$tv'+v\left(\frac{t\operatorname{sin}(x)}{2\operatorname{cos}^2(x)}+t'\right)=\frac{\operatorname{sin}^3(x)}{\operatorname{cos}^2(x)}$$ Solving the first equation $$\left(\frac{t\operatorname{sin}(x)}{2\operatorname{cos}^2(x)}+t'\right)=0$$ By seeing the work that you showed in your question, I am pretty sure you can calculate this equation easily. So after solving we get $$t=\frac{1}{\sqrt[2\operatorname{cos}(x)]e}$$ Now solving the second equation $$tv'+v\left(\frac{t\operatorname{sin}(x)}{2\operatorname{cos}^2(x)}+t'\right)=\frac{\operatorname{sin}^3(x)}{\operatorname{cos}^2(x)}$$ at $$t=\frac{1}{\sqrt[2\operatorname{cos}(x)]e}$$ and$$\left(\frac{t\operatorname{sin}(x)}{2\operatorname{cos}^2(x)}+t'\right)=0$$ Therefore the second equation becomes $$\frac{v'}{\sqrt[2\operatorname{cos}(x)]e}=\frac{\operatorname{sin}^3(x)}{\operatorname{cos}^2(x)}$$ After transforming things $$dv=\frac{\sqrt[2\operatorname{cos}(x)]e\operatorname{sin}^3(x)\:\:dx}{\operatorname{cos}^2(x)}$$ Integrating both sides,
$$\int1\cdot dv=\int\frac{\sqrt[2\operatorname{cos}(x)]e\operatorname{sin}^3(x)\:\:dx}{\operatorname{cos}^2(x)}$$ $\implies$
$$v=-\frac{\operatorname{Ei}\left(\frac{1}{2\operatorname{cos}(x)}\right)}{2}+\sqrt[2\operatorname{cos}(x)]e\operatorname{cos}(x)+2\cdot\sqrt[2\operatorname{cos}(x)]e+C$$ or
$$u=-\frac{\operatorname{Ei}\left(\frac{1}{2\operatorname{cos}(x)}\right)-2\cdot\sqrt[2\operatorname{cos}(x)]e\operatorname{cos}(x)-4\cdot\sqrt[2\operatorname{cos}(x)]e-2C}{2\cdot\sqrt[2\operatorname{cos}(x)]e}$$ $\implies$
$$\frac{1}{y^3}=\frac{C-\operatorname{Ei}\left(\frac{1}{2\operatorname{cos}(x)}\right)}{2\cdot\sqrt[2\operatorname{cos}(x)]e}+\operatorname{cos}(x)+2$$ or
$$y=-\frac{\sqrt[3]2\cdot\sqrt[6\operatorname{cos}(x)]e}{\sqrt[3]{\operatorname{Ei}\left(\frac{1}{2\operatorname{cos}(x)}\right)-2\cdot\sqrt[2\operatorname{cos}(x)]e\operatorname{cos}(x)-4\cdot\sqrt[2\operatorname{cos}(x)]e-C}}$$ This is the required answer and the solution $y=0$ is achieved when $C=\infty$
A: As @abcdefu already did, after $y=\frac{1}{\sqrt[3]{u}}$, we have
$$2 \cos ^2(x)\, u'+ \sin (x)\, u=2 \sin ^3(x)$$
The solution of the homegeneous is simple
$$u=C\,e^{-\frac{\sec (x)}{2}}$$ Variation of parameters leads to
$$\cos ^2(x) e^{-\frac{\sec (x)}{2}} C'=\sin ^3(x) \implies C'=\sin (x) \tan ^2(x) \,e^{\frac{\sec (x)}{2}}$$
Now, using $\sec(x)=2t$
$$C'=2 e^t \left(1-\frac{1}{4 t^2}\right) \implies C=\frac{1}{2} \left(e^t \left(\frac{1}{t}+4\right)-\text{Ei}(t)\right)+\text{Constant}$$
