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.

  • 3
    $\begingroup$ These are called homogeneous equations: equations $y'=F(x,y)$ where $F$ depends only on $y/x$. There is a nice discussion of the general theory, along with your particular example, here. $\endgroup$ Mar 9, 2017 at 23:18

2 Answers 2


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?


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.


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