How to know here on what variable its a derivative of(diff)

[proto]

(2x^2ylny-x)y'=y
(2x^2ylny-x)dy=ydx
then i divide both sides by dy
(2x^2ylny-x)=yx'
then i divide both sides by y
(2x^2lny-\frac{x}{y})=x'
x'+\frac{x}{y}=2x^2lny
so i have here a bernuly foruma
i divide both sides by x^2
\frac{x'}{x^2}+\frac{1}{xy}=2lny
z=x^{-1}
z'=-1x^{-2}x'
-z'+\frac{z}{y}=2ln y

z is defined to be a function of x
so z'=\frac{dz}{dx}
why the book interprets z'=\frac{dz}{dy}
??

z is linked to y not in a direct way .
but z linked to x in a direct way
z and x are more close to each other.

i can't see a mathematical way of figuring it out
its all intuition.and i my intuition is very bad

 

[Fightfish]

\frac{dz}{dx} = -\frac{1}{x^2}
dz = -\frac{1}{x^2} dx
\frac{dz}{dy} = -\frac{1}{x^2} \frac{dx}{dy}

 

[proto]

z=x'
z'=-x^(-2)x'

i can't undestand how you cameup with the first step
<br /> \frac{dz}{dx} = -\frac{1}{x^2}<br />

i can't understand what are you doing here?

 

[proto]

and your conclution
doesn't show that z is a derivative by y
it shows that z' is a derivative by x

 

[Fightfish]

What I was doing up there was proving mathematically that \frac{x&#039;}{x^2} necessary equals -\frac{dz}{dy}. The 1st step was obtained from your substitution variable z = x^{-1}.

On a more intuitive basis, what are you attempting to do when performing the substitution? It's to simplify the DE via substituting x with a new variable z, and thus why would there be a dx lying around after you have completed your substitution? If z&#039; = \frac{dz}{dx}, then how are you going to solve -z&#039;+\frac{z}{y}=2ln y? You now have what, three variables?

 

[D H]

z is defined to be a function of x
so z&#039;=\frac{dz}{dx}
why the book interprets z&#039;=\frac{dz}{dy}

Because you defined it that way, right here:

z=x^{-1}
z&#039;=-1x^{-2}x&#039;

In the first line you defined z to be a function of x. In the second, you defined z' to be in a derivative of the same variable as used for x' -- which was y.