- #1

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[tex] x*\frac{dy}{dx}=(y^2-y x^2)/(y^2+x^2)[/tex]

:tongue2:

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- Thread starter gvk
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- #1

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[tex] x*\frac{dy}{dx}=(y^2-y x^2)/(y^2+x^2)[/tex]

:tongue2:

- #2

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Maybe convert to polar coodinates? That xgvk said:

[tex] x*\frac{dy}{dx}=(y^2-y x^2)/(y^2+x^2)[/tex]

:tongue2:

Alex

- #3

saltydog

Science Advisor

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gvk said:

[tex] x*\frac{dy}{dx}=(y^2-y x^2)/(y^2+x^2)[/tex]

:tongue2:

I tend to look at it this way:

[tex](y^2-yx^2)dx-(xy^2+x^3)dy=0[/tex]

but are unable to proceed. It's not exact, can't make it exact, and can't come up with any differentials by inspection.

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