Solve a simple differential equation This is an equation I met while solving a probability theory problem:

$ dy = 2(m/x-y) \cdot dx/x $

m is a constant 

Also it is known that $y(m) =1$.

I have the function which satisfies the equation and I verified that it works. But I haven't solved a single differential equation in many years so... I just got curious how one can derive this solution. 

Is this equation of some particular type which is solvable through a well-known procedure? Any reference? 

Sorry... I am typing on my phone... I will try to improve the equation outlook now.
We have $$\\ \\ \frac { dy }{ dx } =\frac { 2\left( \frac { m }{ x } -y \right)  }{ x } =\frac { 2m-2xy }{ { x }^{ 2 } } \\ { y'x }^{ 2 }+2xy=2m\\ d\left( { yx }^{ 2 } \right) =2m\\ y{ x }^{ 2 }=2mx+C\\ y\left( x \right) =\frac { 2mx+C }{ { x }^{ 2 } } \\ \\ \\ \\  $$
and the fact  $y\left( m \right) =1$ gives us

$$y\left( m \right) =\frac { 2{ m }^{ 2 }+C }{ { m }^{ 2 } } =1\Rightarrow C=-{ m }^{ 2 }$$


  $$y\left( x \right) =\frac { 2mx-{ m }^{ 2 } }{ { x }^{ 2 } } $$
$$\dfrac{dy}{dx}=\dfrac{2m}{x(x-y)}\\\dfrac{dx}{dy}=\dfrac{x^2-xy}{2m}\\ x'+\frac{y}{2m}x=\frac1{2m}x^2$$
It's Bernoulli form, make $z=x^{-1}$ then $\frac{dz}{dy}=-x^{-2}\frac{dx}{dy}$ isolate x' and substitute in the ode, it reduces into a Linear Form do integrating factor and some algebra and you get it in the form $x(y)= F(y) +C$