Solve $x(dy/dx) = x^4y^3 - y $ (Bernoulli's equations)

Question

Use method of Bernoulli's equations to solve the equation:

$$ x(dy/dx) = x^4y^3 - y $$

I don't really can understand how to use Bernoulli's equations as I had it only once and I wasn't able to understand it then. Can someone guide me through the way of solving?

Answer 1

$$xy'+y = x^4y^3 $$ $$y'+\frac yx = x^3y^3 $$ $$y'+\frac yx = x^3y^m \implies m=3$$ Divide by $y^3$ $$\frac {y'}{y^3}+\frac 1{xy^2} = x^3 $$ Then you can use Bernouilli's technics.

Substitute $u(x)=y^{1-m}=y^{-2} \implies u'(x)=-2y^{-3}y'$

The equation becomes linear of first order in u $$-\frac 12u'+\frac ux=x^3$$

Another approach, the original equation is separable : $$x(dy/dx) = x^4y^3 - y$$ $$ \implies xy'+y = x^4y^3 \implies (xy)' = x(xy)^3 $$ Simpli integrate : $$ \int \frac {d (xy)}{(xy)^3}=\int xdx $$

Answer 2

The idea is " separating the variables" :

So you write your original equation and set a set $A$ where $y$ doesn't take null values (and $x \neq 0 $) :

Let $y$ such a solution.

Let $u$ a function so that: $$u=xy(x)$$

The equation becomes

$$ u'=xu^3$$

So $$ \frac{u'(x)}{u^3(x)}=x $$

You guess .. :)

So $$ \int \frac{u'(x)}{u^3(x)} dx = \frac{x^2}{2} + \varpi $$ $$ u^2(x)=\dfrac{-1}{\varpi+\frac{x^2}{2}} $$

Then discuss values of definition.