# Leibnitz notation

1. Jul 17, 2009

### James889

Hi,

I was looking at a solution to a differential equation problem. And there are some parts that i don't understand.

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

And the solution looks like this:

$$e^\int{ \frac{-2}{x}} = \frac{1}{x^2}$$

$$\frac{1}{x^2}\frac{dy}{dx} -\frac{2}{x^3}y=1$$

$$\frac{d}{dx}\frac{y}{x^2}=1$$

This is the part i don't understand, what does $$\frac{d}{dx}\frac{y}{x^2}$$ mean, and why is it equal to 1 ?

To be honest i find this notation pretty confusing.

2. Jul 17, 2009

### tiny-tim

Hi James889!

It just means d/dx of y/x2

use the product rule, and that's (dy/dx)(1/x2) + y(-2/x3) …

which is the LHS of the previous line!

3. Jul 17, 2009

### rock.freak667

Because

$$\frac{d}{dx}( \frac{y}{x^2}) = \frac{1}{x^2}\frac{dy}{dx}- \frac{2}{x^3}y$$

4. Jul 17, 2009

### HallsofIvy

What that means is that you differentiate $yx^2$ with respect to x! That's pretty basic terminology. If you going to study calculus, you had better learn it:
$$\frac{df}{dx}$$ means "differentiate f with respect to x".

In this case, y is a function of x so we need to use both the product rule and the chain rule.
$$\frac{d yx^2}{dx}= \frac{dy}{dx}(x^2)+ y\frac{dx^2}{dx}$$
exactly what is on the left side of your equation above. It is equal to one because it is equal to the left side of the equation whose right side is 1: If A= B and B= C then A= C.

5. Jul 17, 2009

### Staff: Mentor

James,
You're not asking the right question. The question is not why that derivative is 1, but what does it mean that that derivative is 1. The equation is
$$\frac{d}{dx} \frac{y}{x^2} = 1$$

This is saying that the derivative wrt x of y/x2 = 1, which can only be true if y/x2 = x. This means that y/x2 = x ==> y = x3. Notice that this is the solution of the given differential equation.

6. Jul 18, 2009

### tiny-tim

plus a constant!

7. Jul 18, 2009

### James889

Hello,

What is the difference in meaning between these notations?
$$\frac{dx}{e^x}$$ and $$e^x dx$$

8. Jul 18, 2009

### tiny-tim

Hello James889!

(try using the X2 tag just above the Reply box )

There's nothing special about dx "over" something …

(unless the "something" also starts with "d", of course!)

dx/ex is the same as (1/ex)dx

9. Jul 18, 2009

### HallsofIvy

That's a peculiar question!

$e^x dx$ is dx times $e^x$ and $dx/e^x$ is dx divided by $e^x$

You could also write $dx/e^x$ as $(1/e^x)dx$ or $e^{-x}dx$.

10. Jul 18, 2009

### James889

Yes yes, im not the sharpest knife in the box

11. Jul 18, 2009

### Staff: Mentor

Right, but in the DE the constant was zero, so I omitted it in that step.