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- Thread starter Air
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rock.freak667

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- #3

HallsofIvy

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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.

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No, it doesn't. The important thing here is that every term in both numerator and denominator has x or yto 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).

...

But

[tex]dy/dx= \frac{y+ 1}{x- 2}[/tex]

for example, is not homogeneous. (It is, by the way, separable.)

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.

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Cyosis

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