Homogeneous Differential Equation

[Dr.Doom]

Homework Statement

Find the general solution of the differential equation:

(x+y)y'=x-y

Homework Equations

I want to solve this as a homogeneous differential equation, so our equations are:
v=$\frac{y}{x}$, y=vx, $\frac{dy}{dx}$=v+x$\frac{dv}{dx}$

The Attempt at a Solution

I need to get this into the form $\frac{dy}{dx}$=F($\frac{y}{x}$), so I rewrite it as y'=$\frac{x-y}{x+y}$. Dividing by x I get, y'=$\frac{1-\frac{y}{x}}{1+\frac{y}{x}}$. From here, I substitute to get v+x$\frac{dv}{dx}$=$\frac{1-v}{1+v}$. When looking at the solution manual, however, it says that it should be in the form x(v+1)v'=-(v²+2v-1) before I integrate. I don't see how i can get it into this form. Also, it gives the answer as y²+2xy-x²=C. I can correctly solve this differential equation using different methods, but I would really like to know how to solve this using the homogeneous method. I would really appreciate any help with this. Thanks!

 

[hunt_mat]

You are almost there...

You have to compute:
$$\frac{1-v}{1+v}-v=x\frac{dv}{dx}$$
This will be a separable equation which is solved in the usual fashion.

 

[xxhizors]

you just have to take subtract v on both sides of the equation you got man...(xdv/dx=(1-v)/(1+v) - v..

 

[Dr.Doom]

Ok, so I rewrite it as $\frac{1}{x}$dx=$\frac{1+v}{1-v}$-$\frac{1}{v}$dv. Do I need to do an additional substitution on $\frac{1+v}{1-v}$?

 

[xxhizors]

you solved it wrong on right side... you will get dv/{(1-v)/(1+V) - v}...which comes out to be (1+v)/(1-v^2-2v)...just substitute 1-v^2-2v as t...

 

[Dr.Doom]

I'm having trouble seeing how it comes out to (1+v)/(1-v²-2v)

 

[hunt_mat]

Do the addition as I said to get:
$$\frac{1-v}{1+v}-v=\frac{1-v-v(1+v)}{1+v}=\frac{1-v-v-v^{2}}{1+v}=\frac{1-2v-v^{2}}{1+v}$$