Question

The solution of $\left(\frac{1}{x}-\frac{1}{{\left(x-y\right)}^2}\right)dx+\left(\frac{1}{{\left(x-y\right)}^2}-\frac{1}{y}\right)dy=0$.

Answer 1

$\bullet$ Given the difference equation is $\{\frac{1}{x}-\frac{1}{{(x-y)}^2}\}dx+\{\frac{1}{{(x-y)}^2}-\frac{1}{y}\}dy=0$

on, $\frac{dy}{dx}-\frac{dy}{y}+\frac{dy-dx}{(x-y{)}^2}=0$

on, $\frac{dx}{x}-\frac{dy}{y}+\frac{d(y-x)}{(y-x{)}^2}=0$

Now, integrating both sides we have,

$\log |x|-\log |y|+(-\frac{1}{y-x})=c$

on, $\log |x|-\log |y|=\frac{1}{(y-x)}+c$

where ${}^{\prime }{c}^{\prime }$ is integrating constant.