Hi folks,

I am trying to solve the following ordinary differential equation. I am using the substitution y = vx to make it separable.

$\dfrac{dy}{dx} = \dfrac{(x^2 + y^2)}{x(x + y)} $ ...............(1)

so substituting $y = vx$ and $\dfrac{dy}{dx} = v + x \dfrac{dv}{dx} $ into (1) gives:

$v + x \dfrac{dv}{dx} = \dfrac{x^2 (1 + v^2)}{x^2 (1 + v)}$

such that

$\int \dfrac {(1 + v)}{(1 - v)} dv = \int \dfrac{dx}{x} $ ............... (2)

and

$- \int \dfrac{-1. dv}{(1 - v)} + \int \dfrac{v}{(1 - v)} dv = \int \dfrac{dx}{x} $

the left hand side can be split, giving:

$- \ln (1 - v) + \int \dfrac{v}{(1 - v)} dv = \ln |x| + c $ ..................(3) (c can be ln A)

Now, use substitution $ w = 1 - v, \dfrac{dw}{dv} = -1$

$\int \dfrac{v}{(1 - v)} dv = \int \dfrac{(w - 1)}{w} \dfrac{dv}{dw} dw = \int dw - \int \dfrac{1}{w} dw = w - \ln |w| $

such that

$\int \dfrac{v}{(1 - v)} dv = (1 - v) - \ln (1 - v)$

from (3) we get

$ - \ln (1 - v) + (1 - v) - \ln (1 - v) = \ln |x| + c $ ...................(4)

and

$- 2 \ln (1 - v) + 1 - v = \ln |Ax| $

Now using the substitution $v = \frac{y}{x} $

$ 1 - \dfrac{y}{x} = \ln |Ax| + 2 \ln (1 - \dfrac{y}{x} ) $

which gives

$ 1 - \dfrac{y}{x} = \ln |Ax| (\dfrac{x - y}{x})^2 $

and

$ y = x(1 - \ln |Ax| (\dfrac{x - y}{x})^2 ) $

this is the wrong answer. Correct is

$y = x \ln ( \dfrac{|Ax|}{(x - y)^2})$

I thought these might be equivalent forms, but I can't get them to match. Can anyone find the error?