Solving Non-Separable ODEs: A Beginner's Guide to Differential Equations

[sporff]

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:
$$(y^2 - x^2)dy + 2xydx = 0$$

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

or (2):
$$\frac{1}{2}(\frac{x}{y}-\frac{y}{x}) = \frac{dx}{dy}$$

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


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:

 

[lurflurf]

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.

 

[GCT]

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...


$$y=vx,~y'=v+v'x$$

$$\frac{2xy}{x^2-y^2} = \frac{dy}{dx}$$

$$\frac{2x^{2}v}{x^{2}(1-v^{2})} = \frac{dv}{dx}x+v$$

now try separating

 

[sporff]

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?

 

[HallsofIvy]

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, $$\frac{2xy}{x^2-y^2}$$ 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 $$\frac{2\frac{y}{x}}{1- (\frac{y}{x})^2}$$. Let v=y/x and that becomes $$\frac{2v}{1-v^2}$$.
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.

 

[sporff]

ahh. thank you HallsofIvy. it makes sense to me now.

 

[omenka]

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.