Question

Solve: $(x-y\ln y+y\ln x)dx+x(\ln y-\ln x)dy=0$

• A. $x\ln \left(\frac{y}{x}\right)-y+x\ln x+cy=0$
• B. $y\ln \left(\frac{x}{y}\right)-y+x\ln x+cx=0$
• C. $x\ln \left(\frac{x}{y}\right)-y+x\ln x+cy=0$
• D. $y\ln \left(\frac{y}{x}\right)-y+x\ln x+cx=0$

Answer 1

The correct option is D $y\ln \left(\frac{y}{x}\right)-y+x\ln x+cx=0$

$y\ln x-y\ln y+x\frac{dy}{dx}(\ln y-\ln x)+x=0$
Substituting $y=xv\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\Rightarrow x+x(-\ln x+\ln \left(xv\right))(x\frac{dv}{dx}+v)+x\left(\ln x\right)v-x\left(\ln xv\right)v=0\Rightarrow x(x\ln v\frac{dv}{dx}+1)=0\Rightarrow \frac{dv}{dx}=-\frac{1}{x\ln v}\Rightarrow \ln v\frac{dv}{dx}=-\frac{1}{x}$
Integrating both sides
$\int \ln v\frac{dv}{dx}dx=\int -\frac{1}{x}dx\Rightarrow -v+\left(\ln v\right)v=-\ln x+c\Rightarrow -\frac{y}{x}+\left(\ln \frac{y}{x}\right)\frac{y}{x}=-\ln x+c\Rightarrow y\left(\ln \frac{y}{x}\right)-y+x\ln x+cx=0$