# Dividing differentials

1. Nov 4, 2013

### inglezakis

Consider the differential equation

dx+ydy=0, the integration leads to (x2-x1)+(y2^2-y1^2)/2=0 (1)

Suppose we know that y/x = const.

Lest proceed to the following manipulation on the initial equation, by dividing by (x), then

dx/x+(y/x)dy=0, now the integration gives ln(x2/x1)+(y/x)*(y2-y1) (2)

Correct? Well solutions (1) and (2) are different, i.e. for the same set of x1,x2,y1 they give different values of y2.

Where is the mistake?

2. Nov 4, 2013

### JJacquelin

Hi !
You suppose that y/x = const.
This supposition is false. So, all consequences of the supposition are false.

3. Nov 4, 2013

### inglezakis

And why the supposition is wrong?

4. Nov 4, 2013

### UltrafastPED

The condition y/x=constant means y=ax.
Thus dx + ydy = dx + ax*adx = dx + a^2 xdx = (1+a^2 x)dx = 0

But this means that (1 + a^2 x)=0, so x is not a variable: your constraint converted a differential equation anto an algebraic equation.

5. Nov 4, 2013

### inglezakis

The point of my question is that when we divide a differential equation by a function or variable we result in different solution (not always). Take the example:

ydx+ydy=0, constaint: xy=a

By substituting x with y/a and after some manipulations we arrive to

(-a/y)dy+ydy=0 and on integration we have -aln(y2/y1)+(y2^2-y1^2)/2=0

Suppose we divide the original equation by y. We have

dx+dy=0 and then the solution is (x2-x1) + (y2-y1) =0 (the constraint is eliminated by eliminating y)

Or, we divide by (y^2), we have

dx/y+dy/y=0 and the solution is (x2^2-x1^2)/2a+ln(y2/y1)=0

Here the constraint is needed in order to substitute y for x in the dx term.

So, by dividing the original equation we arrive in different solutions. I dont have the space here, but it can be shown that the 3 solutions above converge for small (x1-x2). Anyway, the issue is when and under what condition we are allowed to divide the differential equation, the purpose being to arrive in the same solution.

6. Nov 4, 2013

### JJacquelin

You suppose x=y/a , then :
dx = dy/a
y(dy/a)+ydy=0
Mistake : you wrote (-a/y)dy+ydy=0

7. Nov 4, 2013

### UltrafastPED

But you are _not_ dividing by an arbitrary function, or even a specified function; you are imposing a _constraint_ on the solutions. In this case your original equation,

ydx + ydy=0 is the same as dx + dy=0 which integrates to x+y=c - a line with slope of -1.

But then you impose a constraint: xy=a, which is a hyperbolic constraint. This will give you a solution space with at most two points which vary with the parameter a and the boundary conditions which fix c.

You should sit down with your instructor; you have a misconception at work here which is detrimental to your progress.

8. Nov 4, 2013

### JJacquelin

Hi inglezakis !

One must not beat around the bush:
The solutions of dx+ydy=0 are x=-y²/2 +constant. You can check it by puting it back into dx+ydy=0.
If you suppose y/x=c then x=y/c which is not equal to -y²/2 + constant. The supposition is false. That's all.

9. Nov 4, 2013

### inglezakis

Hi JJacquelin,

i don't think the supposition is wrong. You can still solve the equation and get the result. By the way, this equation represents a physical system (fluid flow) and i know that is true and the solution is correct. The target of the question is to find out what happens to a differential equation when we divide it with y^2, or in general with a function. I know that this is permitted, there is no rule, unless the function is zero. But as you can see (you can check other examples) when you start dividing by a condition then you arrive with a different equation and so, the solutions are different. This is the point

10. Nov 5, 2013

### UltrafastPED

Yes, and I explained what is happening!

Of course you can divide/multiply by an arbitrary function ... that is what you do with integrating factors.

11. Nov 5, 2013

### inglezakis

The constraint in the example we discuss is, xy=a, so f(x)=y=a/x and thus it is a function. What you mean by arbitrary function?

Lets say you dont eliminate y and you use the constraint to solve the equation:

ydx+ydy=0, constaint: xy=a

By substituting x with a/y and after some manipulations we arrive to

(-a/y)dy+ydy=0 and on integration we have -aln(y2/y1)+(y2^2-y1^2)/2=0

Fine, this is straightforward and clear. But, if you first divide by y and then introduce the constraint you arrive in a different equation, by dividing by y^2 another and so on.

12. Nov 5, 2013

### UltrafastPED

Good bye!

13. Nov 5, 2013

### inglezakis

14. Nov 5, 2013

### JJacquelin

Of course one can divide by an arbitrary function.
For example, starting from dx+ydy=0 :
one can divide by y, leading to the correct equation : dx/y+dy=0
or one can divide by ax, leading to the correct equation : dx/(ax)+ydy/(ax)=0
But one cannot divide by two different functions, for example dividing the first term by ax and dividing the second term by y:
dx/(ax)+ydy/y=0 is false.
because supposing that y=ax is the big mistake !
Of course, one is allowed to say "I suppose that y=ax" when one doesn't know. But as soon as it is prouved that y is not equal to ax, one must admit that the supposition was false.

Last edited: Nov 5, 2013
15. Nov 5, 2013

### JJacquelin

I just see the message #11 from inglezakis !!!
Good bye inglezakis !
That is also my good answer indeed.