# Substitution Methods and Exact Equations

1. Sep 18, 2006

### enviroengnhopeful

There is a problem in my book which wants us to find the general solution to the given equation. I understand most of the problem it is just the integral part that is tricky. Here is the problem:

x(x+y)y' = y(x-y)

In this problem I know that you need to divide the equation by x and you get (1+y/x)y' = y/x(1-y/x). Then I know that you can substitue y/x by saying v = y/x or y = vx. Therefore, dy/dx = v + x dv/dx. Next, I know you can substitue the equation with v to make it (1+v)(v +x dv/dx) = v(1-v). From that point on, if you divide both sides by v(1+v) you can get your equation down to
x v' = v(1-v)/v(1+v)
Then I know you are supposed to put like terms on each side so it makes the equation:

v(1+v) dv/v(1-v) = dx/x.

I know the right side is ln (x), but how do you integrate the left?

2. Oct 1, 2006

### Fermat

You are correct down to here:(1+v)(v +x dv/dx) = v(1-v).

Thereafter,

(v +x dv/dx) = v(1-v)/(1+v)
x dv/dx = v(1-v)/(1+v) - v = v(1-v)/(1+v) - v(1+v)/(1+v) = -2v²/(1+v)

giving,

{(1+v)/(-2v²)} dv = (1/x) dx