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Simple differential equation - Finding the solution

  1. Dec 1, 2015 #1
    1. The problem statement, all variables and given/known data
    We are given the following differential equation:

    $$ y+yy'-xy'=0 $$

    Let's find the solution.
    2. Relevant equations


    3. The attempt at a solution
    So in my course usually we have to do some sort of substitution by the lines of x/y=z or y/x=z. This equation has proven difficult. I have taken a look at wolfram alpha and the y(x)=xv(x) substitution is not one I am familiar with.
     
  2. jcsd
  3. Dec 1, 2015 #2

    RUber

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    ##y + (y - x ) y' = 0##
    Can be written as:
    ##y' = -\frac{y}{y-x} = -\frac{1}{1 - x/y} ##
    So, if you were to use ## z = x/y,## you would get ##z' = \frac{yx' - xy'}{y^2}##.
    Then substituting into the original equation would give yy' + y^2 z' = 0.
    Assume y is not zero, and you get:
    ##\frac{y'}{y} = -z' ##
    Integrate both sides.
     
  4. Dec 1, 2015 #3
    Could you give me a more chew down process of the substitution? What goes where to get that lovely equation down there at the bottom of your answer?
     
  5. Dec 1, 2015 #4

    SammyS

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    I think he did show how he got it.

    Work through a bit of algebra.
     
  6. Dec 2, 2015 #5

    RUber

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    Here is a bit more information to guide your algebra.
    ##z' = \frac{yx' - xy'}{y^2} = \frac{1}{y^2}( y - xy') ##. So ##y^2 z' = y - xy'## which matches 2 of the 3 terms in your original equation.
    So, substituting into the original equation would give ##yy' + y^2 z' = 0##.
    Assume y is not zero, and divide the whole thing by ##y^2##, you get:
    ##\frac{y'}{y} + z'=0 ## which is the same as ##\frac{y'}{y} = -z' ##.
     
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