Question

$\left\{xy\log \frac{x}{y}\right\}dx+\{y^2-x^2\log \frac{x}{y}\}dy=0$ given that, $y\left(1\right)=0$

Answer 1

Given,

$\left\{xy\log \frac{x}{y}\right\}dx+\{y^2-x^2\log \frac{x}{y}\}dy=0$

$\Rightarrow \frac{dx}{dy}=-\left(\frac{y^2-x^2\log \frac{x}{y}}{xy\log \frac{x}{y}}\right)$

Let $x=vy\to \frac{dx}{dy}=v+y\frac{dv}{dy}$

$v+y\frac{dv}{dy}=\frac{v^2\log v-1}{v\log v}$

$y\frac{dv}{dy}=-\frac{1}{v\log v}$

$\Rightarrow v\log vdv+\frac{1}{y}dy=0$

integrating on both sides, we get,

$\frac{v^2}{2}\log v-\frac{v^2}{4}+\log y=c$

$\frac{x^2}{2y^2}\log \frac{x}{y}-\frac{x^2}{4y^2}+\log y=c$

given, $y\left(1\right)=0$

$\frac{1^2}{2(0{)}^2}\log \frac{1}{0}-\frac{1^2}{4(0{)}^2}+\log 0=c$

$c=0$

$\therefore \frac{x^2}{2y^2}\log \frac{x}{y}-\frac{x^2}{4y^2}+\log y=0$