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

1. Aug 16, 2009

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

2. Aug 16, 2009

### Fightfish

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

3. Aug 16, 2009

### proto

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

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

i cant understand what are you doing here?

4. Aug 16, 2009

### proto

and your conclution
doesnt show that z is a derivative by y
it shows that z' is a derivative by x

5. Aug 16, 2009

### Fightfish

What I was doing up there was proving mathematically that $$\frac{x'}{x^2}$$ 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 $$z' = \frac{dz}{dx}$$, then how are you going to solve $$-z'+\frac{z}{y}=2ln y$$? You now have what, three variables?

6. Aug 16, 2009

### D H

Staff Emeritus
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.

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