# 1st order homogenous ODE: (x+y)dy/dx=(x-y)

2h2o

## Homework Statement

Find a general solution

## Homework Equations

$$(x+y)\frac{dy}{dx} = x-y$$

## The Attempt at a Solution

$$\frac{dy}{dx} = \frac{x-y}{x+y}$$

let v=y/x
y=xv

$$\frac{dy}{dx} = v+x\frac{dv}{dx}$$

now,

$$v+x\frac{dv}{dx} = \frac{x-xv}{x+xv}$$

$$= \frac{1-v}{1+v}$$

$$= \frac{1}{1+v} - \frac{v}{1+v}$$

$$=\frac{1}{1+\frac{y}{x}}-\frac{\frac{y}{x}}{1+\frac{y}{x}}$$

Which takes me back to where I started if I clear the denominators, so I'm spinning the wheels. This looks very familiar to me, but I don't recall what it is or what to do with it. Separate the variables, then integrate? Could do it, wrt v, by parts, but isn't there a more efficient way?

Last edited:

## Answers and Replies

Gold Member
Well, first, v/(1+v) = ((v+1)-1)/(1+v) = 1 - 1/(v+1). So that will help.

2h2o
Ok, that gets me a little further, but I still run into trouble:

$$= \frac{1}{1+v} - \frac{v}{1+v}$$

$$= \frac{1}{1+v} - \frac{(1+v)-1}{1+v}$$

$$= \frac{1}{1+v} - 1 + \frac{1}{1+v}$$

$$= \frac{2}{1+v} - 1$$

$$= \frac{2-(1+v}{1+v}$$

$$= \frac{2}{1+v} - 1$$

Which will lead me again, back to my starting point. When do I actually start back-substituting to get something useful?

2h2o
Problem solved: way back in one of the first steps, I found it to be much easier if I collect all the v's on the RHS (but leaving x*dv/dx on the LHS) and onwards from there. Thanks Char, for trying to help anyway :)