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Homogeneous Differential Equation

  1. May 6, 2009 #1

    Air

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    I read that the 'Homogeneous Differential Equation' is one which has form [tex]\frac{\mathrm{d}y}{\mathrm{d}x} = f\left(\frac{y}{x}\right)[/tex] but I came across one example which was [tex]\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x+y}{x+5y}[/tex] and said that is 'Homogeneous Differential Equation' Too which has 'x' and 'y' in both numerator and denominator. Does that mean that 'x' and 'y' exist in numerator and denomintor, then it's 'Homogeneous Differential Equation'?
     
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  3. May 6, 2009 #2

    rock.freak667

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    If I remember correctly, the easiest way to see if an equation is homegenous is to add up the powers of the individual terms, and if they are all the same, then the equation is homogenous and y=Vx substitution can be used.
     
  4. May 6, 2009 #3

    HallsofIvy

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    No, it doesn't. The important thing here is that every term in both numerator and denominator has x or y to the same power. One test for homogeneous equations, that probably is in your text book, is to replace x and y by "ax" and "ay" respectively. If the "a" cancels out, then the equation is homogenous.
    That is, dy/dx= f(x,y) is "homogeneous" if f(ax,ay)= f(x,y). In your case,
    [tex]f(x,y)= \frac{x+ y}{x+ 5y}[/tex]
    [tex]f(ax,ay)= \frac{ax+ ay}{ax+ 5ay}= \frac{a(x+ y)}{a(x+ 5y}= \frac{x+ y}{x+ 5y}= f(x, y)[/tex]

    That means the if you divide both numerator and denominator by x, you get
    [tex]dy/dx= \frac{1+ \frac{y}{x}}{1+ \frac{5y}{x}}[/tex]
    Now, the substitution [tex]u= y/x[/tex] leads to y= xu so dy/dx= x du/dx+ u and the equation becomes
    [tex]x du/dx+ u= \frac{1+ u}{1+ 5u}[/tex]
    so
    [tex]x du/dx= \frac{1+ u}{1+ 5u}- u= \frac{1+ u- u- 5u^2}{1+ 5u}= \frac{1- 5u^2}{1+ 5u}[/tex]
    a separable equation in x and u.

    But
    [tex]dy/dx= \frac{y+ 1}{x- 2}[/tex]
    for example, is not homogeneous. (It is, by the way, separable.)

    [tex]dy/dx= \frac{x+ y}{x+ y+ 1}[/tex]
    is neither homogenous nor separable.
     
  5. May 6, 2009 #4

    Air

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    The 'a' test, you can only put 'a' infront of 'x' and 'y' but not in front of the constant? Am I correct?

    So...[tex]dy/dx= \frac{ay+ 1}{ax- 2}[/tex]

    And Not...[tex]dy/dx= \frac{ay+ a1}{ax- a2}[/tex] which would lead to 'a' cancelling out.
     
  6. May 6, 2009 #5

    Cyosis

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    The requirement is f(x,y)=f(ax,ay), so let f(x,y)=x+y+1, then f(ax,ay)=ax+ay+1. So you don't put a in front of terms, but you replace every x and y with ax and ay.
     
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