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**HallsofIvy** Where did you get this problem? Are you taking a differential equations course or reading a differential equation text book? In either case, you should have seen a definition of "homogeneous differential equation". Actually, there are two different concepts of "homogenous" differential equations, one apply only to first order equations, such as this, the other to linear differential equations (in the case of a "first order, linear, differential equations" they are the same). A first order differential equation, of the form M(x,y)dy+ N(x,y)dy= 0, is "homogenous" if and only if $\displaystyle \frac{M(\lambda x, \lambda y)}{N(\lambda x, \lambda y)}= \frac{M}{N}$ for any number $\displaystyle \lambda$.

In this equation, $\displaystyle M= x^2+ y^2$ and $\displaystyle N= -2xy$ so that $\displaystyle \frac{M(\lambda x, \lambda y)}{N(\lambda x, \lambda y)}$$= \frac{(\lambda x)^2+ (\lambda y)^2}{-2(\lambda x)(\lambda y)}=$$ \frac{\lambda^2(x^2+ y^2)}{-2\lambda^2xy}= \frac{x^2+ y^2}{-xy}=$$ \frac{M(x, y)}{N(x,y)}$.

The point of such "homogeneous" differential equations is that we can write them in terms of the single variable $\displaystyle v= \frac{y}{x}$. In this case, dividing each part of the equation by $\displaystyle x^2$ we have $\displaystyle \frac{x^2+ y^2}{x^2}dy- \frac{2xy}{x^2}dx=$$\displaystyle \left(1+ \left(\frac{y}{x}\right)\right)^2)dx- 2\frac{y}{x}dx$. If we let $\displaystyle v= \frac{y}{x}$ then $\displaystyle xv= y$ so, differentiating both sides using the product rule, vdx+ xdv= 2y, so we can write the equation as $\displaystyle (1+ v^2)dx- 2v(vdx+ xdv)= (1- 2v+ v^2)dx- 2xvdv= 0$ so that $\displaystyle \frac{dx}{x}= \frac{2v}{v^2- 2v+ 1}dv$. Integrate that.