How do I apply integrating factor to solve this differential equation? $$x \frac{dy}{dx}  +  y  =  -2x^6y^4$$
I tried to find the general solution by dividing both sides by $x$  or $x^6$ but no solution I could get.. Do I even solve it with integrating factor?
 A: Instead of looking for an integrating factor, why don't you let $z(x):=xy(x)$? 
Then the ODE can be written as
$$z'  =  -2x^2z^4.$$
Now separate the variables and integrate. Can you take it from here?
A: $$x \frac{dy}{dx}  +  y  =  -2x^6y^4$$
Divide by $y^4$ and let $u=y^{-3}$ to get  a linear equation in $u$ as a function of $x$.
A: Contrary to other answers, you CAN find an integrating factor and manipulate your ODE via substitutions.
We have the ODE :
$$xy' + y = -2x^6y^4$$
Dividing both sides by $-\frac{1}{3}xy^4$, we yield :
$$-\frac{3y'}{y^4} - \frac{3}{xy^3} = 6x^5$$
Let $v(x) = \frac{1}{y^3(x)}$ and then this gives $v'(x) = -\frac{3y'(x)}{y^4(x)} $ and then the differential equation becomes :
$$v'(x) - \frac{3v(x)}{x} = 6x^5$$
Let $μ(x) = e^{\int -\frac{3}{x}\rm d x} = \frac{1}{x^3}$ and then multiply both sides by $μ(x)$ :
$$\frac{v'(x)}{x^3}-\frac{3v(x)}{x^4}=6x^2$$
By substituting $-\frac{3}{x^4}=\big(\frac{1}{x^3}\big)'$ we have :
$$\frac{v'(x)}{x^3}-\bigg(\frac{1}{x^3}\bigg)'v(x)=6x^2$$
Now, we shall apply the reverse product rule : $f\frac{\rm d g}{\rm d x}+g\frac{\rm d f}{\rm dx} = \frac{\rm{d}}{\rm d x}(f\;g)$ :
$$\int \frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac{v(x)}{x^3}\bigg)\mathrm{d}x=\int6x^2\mathrm{d}x \implies v(x) = x^3(2x^3+c_1)$$
Now, you can substitute $v(x) = \frac{1}{y^3(x)}$ and solve for $y(x)$ to yield the final result.
A: It's already a derivative so you don't need an integrating factor.$(\mu =1)$
$$x \frac{dy}{dx}  +  y  =  -2x^6y^4$$
$$ \implies (xy)'=  -2x^2(xy)^4$$
Then integrate
$$ \implies \int \frac  {d(xy)}{(xy)^4}=  -2\int x^2 dx$$
A: By inspection the LHS is
$$xy'+y=(xy)'$$ hence the change of variable $z=xy$,
$$z'=-2x^2z^4$$ which is a separable equation.

 $$z^{-3}=2x^3+C,\\y=\frac1{x\sqrt[4]{2x^3+C}}.$$

A: HINT: Recall the Bernoulli form for First Order ODEs
$$y'+p(x)y=q(x)y^n$$
Aim your ODE $$y'+x^{-1}y=-2x^5y^4$$
Divide everything by $y^4$
$$y^{-4}y'+x^{-1}y^{-3}=-2x^5$$
Your Integrating Factor for Bernoulli form is by a substitution $z=y^{1-n}\rightarrow z=y^{-3}$ this is for reduce to a Linear Form $y'+p(x)y=q(x)$ and then you'll find it.
Isolate $y'\quad $ from $\quad \frac{dz}{dx}=-3y^{-4}\frac{dy}{dx}$
and replace in $\quad (y^{-4}y'+x^{-1}y^{-3}=-2x^5)\quad $ from here you can do solve by yourself
