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## Main Question or Discussion Point

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:

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: