Working the differential equation dy/dx = x-y/x+y

[bitrex]

Homework Statement

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

Homework Equations

Homogeneous differential equation rules = $$v = \frac{y}{x}$$$$\frac{1}{y} = \frac{x}{y}$$ $$\frac{dy}{dx} = v + x\frac{dv}{dx}$$

The Attempt at a Solution

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

$$x+\frac{dv}{dx} = (1+v)^-1-(1+1/v)^-1$$


I'd like to know if what I've done here looks good so far? I'm not getting the right answer when I complete the integration, so I'm curious to know if I'm making an error after this point or if I've just completely set the problem up wrong. Thanks for any help!

 

[djeitnstine]

Check your last step its supposed to be $$v+x\frac{dv}{dx}$$

 

[HallsofIvy]

Homework Statement

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

Homework Equations

Homogeneous differential equation rules = $$v = \frac{y}{x}$$$$\frac{1}{y} = \frac{x}{y}$$ $$\frac{dy}{dx} = v + x\frac{dv}{dx}$$

The Attempt at a Solution

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

$$x+\frac{dv}{dx} = (1+v)^-1-(1+1/v)^-1$$


I'd like to know if what I've done here looks good so far? I'm not getting the right answer when I complete the integration, so I'm curious to know if I'm making an error after this point or if I've just completely set the problem up wrong. Thanks for any help!

As djeitnstine pointed out, your left side should be v+ x dv/dx. The right side is
$$\frac{1}{1+ v}- \frac{1}{1+ \frac{1}{v}}$$
Multiplying the numerator and denominator of the last fraction by v, this is
$$\frac{1}{1+ v}- \frac{v}{v+ 1}$$
which is equal to ?