Categorize first order differential equation

BlueSocks
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Homework Statement


I'm trying to determine which categories various first order differential equations fall into (and once they're categorized they're nice and easy to solve). My list of categories is the following; linear equations, homogenous equations, bernoulli equations, exact equations, exact equations with special integrating factors, separable equations, equations with linear coefficients, and equations that fit the form y'=g(ax+by).

However, I can't seem to transform the following equation into anything that would fit any of the above.

dy/dx=-y2/(x2 + 4xy)


The Attempt at a Solution



I've tried transforming the equation in pretty much every way I can think of, but I'm not finding any readily solvable form. the closest I've gotten (I think) is y' =(x2/y2 - 4x(1/y))-1 but other than isolating y' (in a different way), that doesn't seem to be much good. Am I missing something obvious here? Sorry, I feel a little silly. If that's not enough attempt at a solution I can show my work for all the other forms of the equation that I've worked to.

Thank you very much for any direction/assistance you might provide.


Cheers!
 
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BlueSocks said:

Homework Statement


I'm trying to determine which categories various first order differential equations fall into (and once they're categorized they're nice and easy to solve). My list of categories is the following; linear equations, homogenous equations, bernoulli equations, exact equations, exact equations with special integrating factors, separable equations, equations with linear coefficients, and equations that fit the form y'=g(ax+by).

However, I can't seem to transform the following equation into anything that would fit any of the above.

dy/dx=-y2/(x2 + 4xy)

Doesn't that fit your category M and N are homogeneous of degree 2?
 
...yup. Well, I feel incredibly silly all of a sudden. Thank you, I really appreciate you pointing that out. I think I'll be good from here on in.
 
Indeed, the ODE can be written in the form \frac{dy}{dx}=f\left(\frac{y}{x}\right)
 
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