Solution of $(x^2 + y^2)\ dx -2xy\ dy$ = 0 Solve $(x^2 + y^2)dx -2xydy = 0$
The answer is $x^2 - y^2 = Cx$
I've tried the following methods but I'm not getting the answer :


*

*Variable Separable (n/a)

*Homogenous Differential Equation (Coefficients not of degree 1)

*Linear Differential Equation; Not Linear due to $n \ne 1$

*Exact Differential Equation: Not Equal

*Unexact Differential Equaiton: Doesn't fit

*Bernoulli's: getting $y^2 + x^2/3 = C/x$


What else can I try? Any hint?
 A: $$x^2 + y^2 = 2xy \dfrac{dy}{dx} \implies x^2 + y^2 = x\dfrac{d(y^2)}{dx} .$$
Substitute $y^2 = z$ and you will get the nice form:
  $$x \dfrac{dz}{dx}-z = x^2.$$
A: Geometry also helps. $2xy$ looks like a diagonal term, rotation by $45^\circ$ will simplify the equation. Use
$$x=u+v$$
$$y=u-v$$
(substitution of new variables $u=(x+y)/2$ and $v=(x-y)/2$).
We get
$$2(u^2+v^2)(du+dv)-2(u^2-v^2)(du-dv)=0$$
$$v^2du+u^2dv=0$$
This is actually as good as it can possibly get. You can separate the variables!
$$\int \frac{du}{u^2}=-\int\frac{dv}{v^2}$$
$$1/u+1/v=C$$
$$u+v=Cuv$$
$$C'x=x^2-y^2$$
When I forgot the minus in the second term ($-2xy$) I got the same wrong solution you got with Bernoulli, so you should just check your method again and be careful with the signs.
A: Hint:
$${x^2} + {y^2} = 2xy.y' \Rightarrow y' = \frac{{{x^2} + {y^2}}}{{2xy}} = \frac{{1 + {{\left( {\frac{y}{x}} \right)}^2}}}{{2\frac{y}{x}}}.$$
Put $u=\frac{y}{x}$, we have $y=ux$, and so $y'=u'x+u$. The latter equation becomes
$$u'x + u = \frac{{1 + {u^2}}}
{u} = \frac{1}
{u} + \frac{u}
{2} \Rightarrow u'x = \frac{1}
{u} - \frac{u}
{2} = \frac{{2 - {u^2}}}
{{2u}}.$$
A: Switching to polar coordinates
$$r^2(\cos\theta\,dr-r\sin\theta\,d\theta)=2r^2\cos\theta\sin\theta(\sin\theta\,dr+r\cos\theta\,d\theta)$$
makes the equation separable:
$$\frac{dr}r=\frac{\sin\theta+2\cos^2\theta\sin\theta}{\cos\theta-2\cos\theta\sin^2\theta}d\theta=-\frac{\sin\theta\,(2\cos^2\theta+1)}{\cos\theta\,(2\cos^2\theta-1)}d\theta=\frac{2t^2+1}{t(2t^2-1)}dt.$$
Then
$$\ln(r)=\ln\left(\frac t{2t^2-1}\right)+C$$
$$r(2\cos^2\theta-1)=C\cos\theta,$$
$$2r^2\cos^2\theta-r^2=Cr\cos\theta,$$
$$x^2-y^2=Cx.$$
A: With the integrand factor $\dfrac1{x^2}$,
$$\frac{(x^2+y^2)\,dx-2xy\,dy}{x^2}=dx-d\frac{y^2}x,$$and $$\frac{x^2-y^2}x=C.$$
A: This seem to be the most simple answer:
Substituting $u = x^2$ and $v = y^2$ gives $(u+v)du - 2udv = 0$
which has the solution $v = u + Cu^{1/2}$ or  $y^2 - x^2 = Cx$.
A: $$
x' = \frac{2xy}{x^2+y^2} 
$$
Let $x=vy$ then we get
$$
v'y + v = \frac{2v}{v^2+1}
$$
Which is a nicer form don't you think?
A: $(x^2+y^2)dx−2xydy=0$
$\frac{dy}{dx}=\frac{x^2+y^2}{2xy}  $..(i)
This is a homogeneous differential equation because it has homogeneous functions of same degree 2.
homogeneous functions are:
$(x^2+y^2)$ and $2xy$, both functions have degree 2.
Solution of differential equation:
Equation (i) can be written as,
$\frac{dy}{dx}=\frac{1+\frac{y}{x}}{2(\frac{y}{x})}$..(ii) 
Substitute, $\frac{y}{x}=v \implies\frac{dy}{dx}=x\frac{dv}{dx}+v$
Now, equation(ii)becomes,
$x\frac{dv}{dx}+v=\frac{1+v^{2}}{2v}\implies x\frac{dv}{dx}=\frac{1+v^{2}}{2v}-v\implies x\frac{dv}{dx}=\frac{1-v^{2}}{2v} $
Integrate both side:
$\int {\frac{2v}{1-v^2}dv}=\int {\frac{1}{x}}dx$...(iii)
Substitute, $1-v^2 =t\implies2v\;dv=-dt$
$\int {\frac{-1}{t}dt}=\int {\frac{1}{x}}dx$
$ -lnt=lnx + lnC\implies t^{-1}=xC$
Put the value of t
$(1-v^{2})^{-1}=xC\implies \frac{1}{xC}=1-v^{2}$
Put the value of v,
$\frac{1}{xC}=1-\frac{y^{2}}{x^{2}}$ 
$C=\frac{1}{x(1-\frac{y^{2}}{x^{2}})}$
