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

proto

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

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

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 cant see a mathematical way of figuring it out
its all intuition.and i my intuition is very bad

Fightfish

[tex]\frac{dz}{dx} = -\frac{1}{x^2}[/tex]
[tex]dz = -\frac{1}{x^2} dx[/tex]
[tex]\frac{dz}{dy} = -\frac{1}{x^2} \frac{dx}{dy}[/tex]

proto

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

i cant undestand how you cameup with the first step
[tex]
\frac{dz}{dx} = -\frac{1}{x^2}
[/tex]

i cant understand what are you doing here?

proto

and your conclution
doesnt 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 [tex]\frac{x'}{x^2}[/tex] necessary equals [tex]-\frac{dz}{dy}[/tex]. The 1st step was obtained from your substitution variable [tex]z = x^{-1}[/tex].

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 [tex]z' = \frac{dz}{dx}[/tex], then how are you going to solve [tex]-z'+\frac{z}{y}=2ln y[/tex]? You now have what, three variables?

D H

Because you defined it that way, right here:
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.