Question

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

Answer 1

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

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

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

$\frac{dx}{dy}=\frac{-y^2\left(1-\frac{x^2}{y^2}\log \frac{x}{y}\right)}{y^2\left(\frac{x}{y}\log \frac{x}{y}\right)}$

$\frac{dx}{dy}=\frac{-\left(1-\frac{x^2}{y^2}\log \frac{x}{y}\right)}{\frac{x}{y}\log \frac{x}{y}}$

$put\frac{x}{y}=v$

$x=vy$

$\frac{dx}{dy}=v+y\frac{dv}{dy}$

putting in u , we get

$v+y\frac{dv}{dy}=\frac{-\left(1-v^2\log \text{v}\right)}{v\text{log v}}$

$y\frac{dv}{dy}=\frac{-\left(1-v\text{log v}\right)}{v\text{log v}}-v$

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

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

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

Now integrating, we have

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

$\log v\frac{v^2}{2}-$ $\int \frac{1}{v}.\frac{v^2}{2}dv=-\log y+c$

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

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

Now substitute value of v from (2)

$\therefore \frac{{\left(x/y\right)}^2}{2}\log \frac{x}{y}-\frac{{\left(x/y\right)}^2}{4}=-\log y\text{+ c}$

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

$2x^2\log \frac{x}{y}-x^2=4y^2\left(c-\log y\right)$