- Thread starter
- #1

#### paulmdrdo

##### Active member

- May 13, 2013

- 386

please help me continue solving this,

$\displaystyle \frac{dy}{dx}=\ln(x)-\ln(y)+\frac{x-y}{x+y}$

this is where I can get to,

$\displaystyle \frac{dy}{dx}=\ln(\frac{x}{y})+\frac{x-y}{x+y}$

multiplying the 2nd term by $\frac{1}{x}$

$\displaystyle \frac{dy}{dx}=\ln(\frac{x}{y})+\frac{\frac{x}{y}-1}{\frac{x}{y}+1}$

using the substitution, $\displaystyle v=\frac{y}{x}$--->$\displaystyle y=vx$----->>$\displaystyle\frac{1}{v}=\frac{x}{y}$

$\displaystyle \frac{dy}{dx}=\ln(\frac{1}{v})+\frac{\frac{1}{v}-1}{\frac{1}{v}+1}$

until here I don't know how to proceed.. please help!