# How to solve the following differential equation: $xy'-y=(x+y)(\ln(x+y)-\ln(x))$?

I'd really love your help with solving the following differential equation: $xy'-y=(x+y)(\ln(x+y)-\ln(x))$.

I transformed the equation to $((x+y)(\ln x-\ln(x+y))-y)dx+xdy=0$ and then checked if $\frac{(x+y)(\ln x-\ln(x+y))-y}{dy}=\frac{x}{dy}$ in order to find $\xi$ such that $\frac{d\xi}{dx}$=$\frac{d\xi}{dy}$, but it doesn't. Then I tried to multiply the equation with both $\mu(x)$ and $\mu(y)$ to find a integration factor in order to find $\xi$ but both $\frac {\mu'(x)}{\mu(x)}$ and $\frac {\mu'(y)}{\mu(y)}$ were depended on both $x$ and $y$.

Any suggestion for how should I solve this one?

Thanks a lot!

-
Just to clarify, is $y$ a function of $x$? – yohBS Apr 8 '12 at 12:39
Yes, $y$ is function of $x$. – Jozef Apr 8 '12 at 12:43
According to WA you should use substitution $y(x)=x\cdot v(x)$ , therefore $y'(x)=v(x)+x\cdot v'(x)$ – pedja Apr 8 '12 at 13:23

$xy' - y = (x+y)(\log(x+y) - \log(x))$ upon dividing by $x^2$ takes the form $$\left[1+\frac{y}{x}\right]' =\frac 1x \left(1+\frac{y}x\right) \log\left(1+\frac yx\right)$$
Now substitute $z = 1+y/x$. The rest should be clear.