How to solve differential equations that look like these y' = (x +y)/(x-y)
In my Finnish math book they classify this as a differential equation which is "reducible to equal degree". They present a way of solving this type of equation, but the explanation is really bad. I've been trying to find more information, but I just can't figure out what these are called in English. Even Wolfram Alpha doesn't know (it suggests a more general classification, "first-order nonlinear differential equation").
Does anyone know of a name for this type of differential equation? Can you direct me to a book, website or video which explains how to solve these? I'm really stuck. Any material at all would be appreciated.
 A: This is a first-order homogeneous differential equation since one can write it in the form:
$$y'=F\left(\frac{y}{x}\right)$$
By dividing the numerator and denominator by $x$.

A general method to solve these is to substitute $y=v\cdot x$ and $\frac{dy}{dx}=\frac{dv}{dx}x+v$. The derivative was obtained as a result of the product rule since $v$ is a function of $x$. Then you will obtain a separable ODE.
Thus, you obtain:
$$\frac{dy}{dx}=\frac{x+y}{x-y} \implies \frac{dv}{dx}x+v=\frac{x+vx}{x-vx} \implies \frac{dv}{dx}x=\frac{1+v}{1-v}-v$$
Putting it all into one fraction gives:
$$\frac{dv}{dx}x=\frac{v^2+1}{1-v}$$
Which is separable. Can you continue?
A: If a function $F(x,y)$ satisfies the condition
$$ F(tx,ty)=t^nF(x,y)$$
for some $n$ it is said to be homogeneous in degree $n$ and the substitution $y=ux$ (alternately $x=uy$) will result in a separable DE in $x$ and $u$ (alternately $y$ and $u$).
There are alternate ways of doing this but I will use the method which utilizes the form
$$ M(x,y)\,dx+N(x,y)\,dy=0$$
Using your example,
\begin{equation}
(x+y)\,dx-(x-y)\,dy=0\\
\end{equation}
Let $y=ux, dy=u\,dx+x\,du$. Then
\begin{eqnarray}
(x+ux)\,dx-(x-ux)(u\,dx+x\,du)&=&0\\
x(1+u^2)\,dx-x^2(1-u)\,du&=&0\\
\frac{1}{x}\,dx-\frac{1-u}{1+u^2}\,du&=&0
\end{eqnarray}
which is routine from this point. In the final answer you will replace $u$ with $\dfrac{y}{x}$ and simplify the expression if possible.
