# Basic Differential Equations

1. Oct 3, 2012

### Physics-Pure

Hello all~

Given the equation:
dy/dx = (x/y)
I know we would initially go to:
∫dy =∫ (x/y) dx
then too:
∫(y)(dy) = ∫x dx
Until arriving at:
(y2/2) + C1 = (x2/2) + C2
(y2) - (x2) = C

My question is:
Where does the dy disappear to in step 4? Where the anti-derivative is taken.

Why does ∫dy become just y when solving an equation of the form
dy/dx = (x2 + 1), but it disappears in the first example?

Thank you~

2. Oct 3, 2012

### arildno

This is, basically, wrong all the way.
First off, if you have dy/dx=x/y(x), we may rewrite this, bu multiplying both sides with y(x) as:

y(x)dy/dx=x.

Then, up to an arbitrary constant of integration, we'll have.

int((y(x)dy/dx)dx)=int(xdx).
That is integrating BOTH sides with respect to the same variable, i.e, "x".

Now, on the left-hand side, we use the reverse of the chain rule of differentiation, that is, integration by substitution, letting "y" be our integration variable.
There is no magical disappearance of any variables or infinitesemals.

3. Oct 3, 2012

### Physics-Pure

Where did that y(x) come from?
And are you saying that we let y = y(x)dy/dx)dx?

P.S. I was simply following the "Introduction to differential equations" video, under calculus. Found here: http://www.hippocampus.org/Calculus & Advanced Math;jsessionid=BAEE0BB1E88F4A594768EEBE4D8FC1EA

4. Oct 3, 2012

### Studiot

Following the method presented (which is known as variables separable or separation of variables)

The answer is that ∫dy is not ∫dy it is the ∫1dy.
When you integrate this the integral of 1 is y and the dy drops out as it did in the previous example. You are not integrating the dy.
You asked why the dy drops out - well it is really a book- keeping symbol I see someone else already told you this in another thread.