Resolving a differential equation manually It is my first week dealing with Differential Equations, and I am stuck at the following question:
Find all the solutions of the following equation:
$2y^2dx-(x+y)^2dy=0$
I have consulted an online calculator, which gave the solution $log\frac{y(x)}{x}+2arctan\frac{y}{x}=c_1-log(x)$.
However, I would like to be able to do this manually.
Could anybody help out?
 A: Hint:
$$\dfrac{dy}{dx}=\dfrac{2y^2}{(x+y)^2}=\dfrac{2\left(\dfrac yx\right)^2}{\left(1+\dfrac yx\right)^2}$$
As the denominator & numerator are Homogeneous polynomials
Set $y=vx,\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$
A: The differential equation
$$2y^2dx-(x+y)^2dy=0$$
can be rewriten as 
$$2\dfrac{dx}{dy}=\left(\dfrac{x+y}{y}\right)^2$$ 
or
$$2\dfrac{dx}{dy}=\left(\dfrac{x}{y}+1\right)^2\ \ \ ...(1)$$ 
Now substituting $\dfrac{x}{y}$ as t, or $x=yt$ 
If you differentiate $x=yt$ w.r.t. $y$ on both sides you'll get $\dfrac{dx}{dy}=t+y\dfrac{dt}{dy}$
Substituting the value of $\dfrac{dx}{dy} \text{ and } \dfrac{x}{y}$ in equation (1)
$$2t+2y\dfrac{dt}{dy}=(t+1)^2$$
$$2t+2y\dfrac{dt}{dy}=t^2+1+2t$$
$$2y\dfrac{dt}{dy}=t^2+1$$
$$2\dfrac{dt}{t^2+1}=\dfrac{dy}{y}$$
Now integrating both sides
$$\displaystyle \int 2\dfrac{dt}{t^2+1}=\int \dfrac{dy}{y}$$
$$2\tan ^{-1} t=\ln(y)+C$$
Substituting the value of t in this expression
$$2\tan ^{-1} \dfrac{x}{y}=\ln(y)+C$$
In this question integrating $2\dfrac{dx}{dy}=\left(\dfrac{x+y}{y}\right)^2$ was much easier than integrating $\dfrac{dy}{dx}=\dfrac{2y^2}{(x+y)^2}$, that's why I preferred the former.
Although my solution doesn't match yours it doesn't mean its incorrect, its just a matter of integration constant.
