Understanding Leibnitz Notation for Differential Equations

[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.

 

[tiny-tim]

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

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

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 ?

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!

 

[rock.freak667]

Because

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

 

[HallsofIvy]

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.

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.

 

[Mark44]

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.

 

[tiny-tim]

… This is saying that the derivative wrt x of y/x² = 1, which can only be true if y/x² = x.

plus a constant!

 

[James889]

Hello,

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

 

[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/e^x is the same as (1/e^x)dx

 

[HallsofIvy]

Hello,

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

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 $e^{-x}dx$.

 

[James889]

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 $e^{-x}dx$.

Yes yes, I am not the sharpest knife in the box

 

[Mark44]

plus a constant!

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