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

#### bitrex

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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

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

#### HallsofIvy

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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 ?

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