Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dividing differentials

  1. Nov 4, 2013 #1
    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. jcsd
  3. Nov 4, 2013 #2
    Hi !
    You suppose that y/x = const.
    This supposition is false. So, all consequences of the supposition are false.
     
  4. Nov 4, 2013 #3
    And why the supposition is wrong?
     
  5. Nov 4, 2013 #4

    UltrafastPED

    User Avatar
    Science Advisor
    Gold Member

    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.
     
  6. Nov 4, 2013 #5
    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.
     
  7. Nov 4, 2013 #6
    You suppose x=y/a , then :
    dx = dy/a
    y(dy/a)+ydy=0
    Mistake : you wrote (-a/y)dy+ydy=0
     
  8. Nov 4, 2013 #7

    UltrafastPED

    User Avatar
    Science Advisor
    Gold Member

    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.
     
  9. Nov 4, 2013 #8
    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.
     
  10. Nov 4, 2013 #9
    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
     
  11. Nov 5, 2013 #10

    UltrafastPED

    User Avatar
    Science Advisor
    Gold Member

    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.
     
  12. Nov 5, 2013 #11
    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.
     
  13. Nov 5, 2013 #12

    UltrafastPED

    User Avatar
    Science Advisor
    Gold Member

    Good bye!
     
  14. Nov 5, 2013 #13
    Thats a good answer indeed.
     
  15. Nov 5, 2013 #14
    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
  16. Nov 5, 2013 #15
    I just see the message #11 from inglezakis !!!
    Good bye inglezakis !
    That is also my good answer indeed.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook