xy = cx + cy

Discussion in 'Math' started by KCHARROIS, Oct 13, 2012.

  1. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
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    1
    Hello,

    I need help making the following formula equal, y= ... instead of xy=...

    I know that I could send the x under cx + cy but then what do I do with the cy?
     
  2. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
    292
    1
    Heres the whole question

    show that the given equation is a solution of the given differential equation

    xy' + y^2 = 0 , xy = cx + cy

    Please help me solve this.
     
  3. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    [​IMG]

    Now you can substitute for 'c' in the equation to solve it in terms of x or y
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    How about just solving for y?

    Move all terms with y onto one side of the equation and all other terms to the other. Factor out y, divide both sides by the resulting coefficient of y.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    Are you sure that these equations are the given equations? Because they are not dimensionally consistent. The right hand equation requires that c, x, and y all have the same units (call them 'fred'). This means that y'=dy/dx is dimensionless. This in turn means that the left hand equation has units of 'fred' for the first term and 'fred'^2 for the second term.

    Are you sure the diffy-Q isn't

    (x^2)y' + y^2 = 0
     
  6. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
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    Yes look at c as if they were constants.
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    I have no choice but to assume that 'c' is a constant. But what do you mean "as if *they* were constants," there's only one 'c'. If you mean to say that the two occurances are not the same constant, then you need to call them something else.

    But none of that addresses the dimensional inconsistency.
     
  8. BillO

    Well-Known Member

    Nov 24, 2008
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    I have to agree with WBahn, or the answer is that xy=xc + cy it is not a solution of xy' +y^2=0


    BTW, y=cx/(x-c)
     
    Last edited: Oct 15, 2012
  9. panic mode

    Senior Member

    Oct 10, 2011
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    xy = cx + cy
    xy - cy = cx + cy - cy
    xy -cy = cx +0
    xy - cy = cx
    y(x-c) = cx
    y(x-c)/(x-c) = cx / (x-c)
    y * 1 = cx/(x-c)
    y=cx/(x-c)

    differentiate to get
    y' = [c(x-c) - cx]/(x-c)^2
    y'= -c^2/(x-c)^2

    also calculate y^2

    y^2=[cx/(x-c)]^2

    then try to verify
    xy' + y^2 = 0

    x*[-c^2/(x-c)^2]+[cx/(x-c)]^2 = 0

    which is not true except for special case where x=1 because

    x <> x^2
     
  10. WBahn

    Moderator

    Mar 31, 2012
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    And hopefully the OP (and others?) will take note that I didn't even have to try to solve the problem to spot (1) that the given equation could not be a solution of the given differential equation, and that (2) what the form of the diffy-Q would need to be in order for it to be a solution. That's part of the power of dimensional analysis -- even when there are no dimensions, per se.
     
  11. Wendy

    Moderator

    Mar 24, 2008
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    Actually this problem is simple algebra, which is bread and butter to all electronics one way or another. I didn't even consider the dimensional aspects, treating it as pure math.

    It wasn't until college chemistry I ran into the importance of units, and making sure all the i's were dotted and t's were crossed. Funny thing is it is every bit as true for electronics.
     
  12. WBahn

    Moderator

    Mar 31, 2012
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    Actually, the problem isn't pure algebra, since it involves a differential equation. But if you treat it as a "pure math problem", then you have little choice but to crank through the math and solve for y, take the derivative, square y, and plug all of that into the differential equation. Then, when you discover that it doesn't work, you are left to wonder whether it really doesn't work, or whether you just made a mistake along the way. So now you spend more time redoing it and/or checking your work to make sure that you didn't make any mistakes. For a simple problem like this, all of that doesn't take too long. But this could easily have been something that took a couple pages to work out (without looking much more complicated at face value than this one). But in either case, if you look for dimensional consistency, even in a 'pure math problem', you can spot things like this without doing any math at all and KNOW that it won't work out.

    Perhaps I'm just silly, but I prefer to spot problems early whenever possible.
     
    1chance likes this.
  13. 1chance

    Member

    Nov 26, 2011
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    I totally agree with WBahn. I teach my students to use dimensional analysis whenever possible, especially to do a quick check of answers for reasonableness. By the way, I teach dimensional analysis to my honors Algebra I students in September so they have yet one more tool in their toolbox. As students mature in their math knowledge, the applications are endless.
     
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