Is this ODE separable?

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I'm almost finished my calculus book (I'm self-teaching) and in the last 2 chapters it's giving a brief intro to differential equations. the second section is for "separable" and I'm stuck on this one halfway through the exercises. It doesn't seem to be separable by any means I can see unless there's some kind of substitution (which he's never mentioned anywhere yet).

my book gives:
[tex](y^2 - x^2)dy + 2xydx = 0[/tex]

the closest I can seem to get it is (1):
[tex]\frac{x^2-y^2}{2xy} = \frac{dx}{dy}[/tex] *or* [tex]\frac{2xy}{x^2-y^2} = \frac{dy}{dx}[/tex]

or (2):
[tex]\frac{1}{2}(\frac{x}{y}-\frac{y}{x}) = \frac{dx}{dy}[/tex]

or even (3):
[tex]y dy - \frac{x^2}{y}dy + 2xdx = 0[/tex]


Now... I have the Schaum's "3000 solved problems in calculus" and in it there's a problem which simplifies into form (1) I have up there and goes on to say it's a "homogeneous" so substitute in y=vx.
Is my problem even a separable one? Excuse my DE newbieness. :yuck:
 

Answers and Replies

  • #2
lurflurf
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This would make more sense in a broader view in which one would recognize it as homogeneous.
you could solve
y^2dy+xydx=0
and
-x^2dy+xydx=0
then realize the sum of the two solutions is the solution of the original problem.
or you could realize
d(x/y) and d(y/x) should be involved, then try to obtain the differential equation in terms of them.
 
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  • #3
GCT
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yes, it's homogenous, one way to test is to substitute ax and ay for every x and y, and see if the constants cancel out to give you the original form. In pertinence to the "schaum's" example you mentioned....


[tex]y=vx,~y'=v+v'x [/tex]

[tex]\frac{2xy}{x^2-y^2} = \frac{dy}{dx}[/tex]

[tex] \frac{2x^{2}v}{x^{2}(1-v^{2})} = \frac{dv}{dx}x+v[/tex]

now try separating
 
  • #4
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thanks guys. very strange that my book would do this. he doesn't go over homogeneous for another couple of sections but he goes and gives me a problem on it. at least now that I know what it is I can go about figuring it out.

but anyway...
what is the reasoning behind the substitution of y=vx? and how do you get y' = v + v'x from that?
 
  • #5
HallsofIvy
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The fact that function is "homogeneous" in x and y really means that it can be written as a function of y/x. For example, [tex]\frac{2xy}{x^2-y^2}[/tex] is homogeneous because it has the same "power" of the variables (2) in both numerator and denominator. If you divide both numerator and denominator by x2, you get [tex]\frac{2\frac{y}{x}}{1- (\frac{y}{x})^2}[/tex]. Let v=y/x and that becomes [tex]\frac{2v}{1-v^2}[/tex].
Taking v= y/x is the same as y= vx. And, of course, if y= vx then y'= v(x)'+ v'x by the product rule. And that is y'= v+ v'x since (x)'= 1.
 
  • #6
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ahh. thank you HallsofIvy. it makes sense to me now.
 
  • #7
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Just in case, there is a test for separability for any ODE in the form dy/dx = F(x,y). In other words, can we write F(x,y) as f(x)g(y). You can google this test.
 

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