how to solve this equation ODE

Question

I am asked to solved

Reduce the differential equatioin into linear form and hence solve it;

$\frac{dy}{dx}+ \frac{y\ln y}{x-\ln y}=0$

I put: $\ln y= t\Rightarrow \frac{1}{y} \frac{dy}{dt}=1$

could anyone tell me how to proceed?

Answer 1

$\frac{dy}{dx}+\frac{y\ln(y)}{x-\ln(y)}=0$

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

$\frac1y\frac{dy}{dx}=\frac{\ln(y)}{\ln(y)-x}$

let $\ln(y) = t\implies \frac1y\frac{dy}{dx} = \frac{dt}{dx}$

$\frac{dt}{dx}=\frac{t}{t-x}$

$\frac{dx}{dt}=1-\frac xt$

This is a homogenous equation, let $x = wt\implies \frac{dx}{dt}=w+t\frac{dw}{dt}$

$w+t\frac{dw}{dt}=1-w$

$t\frac{dw}{dt}=1-2w$

$\frac1{1-2w}dw=\frac1t\,dt$

integrate on both sides

$\int\frac1{1-2w}\,dw=\int\frac1t\,dt$

$\frac{\ln(1-2w)}{-2}=\ln(t)+C$

$\frac{\ln{(1-2\frac xt)}}{-2}=\ln(t)+C$

$-\ln(\frac{t-2x}{t})=\ln(t^2)+C$

$-\ln(\frac{\ln(y)-2x}{\ln(y)})=\ln({\ln^2{y}})+C$

$\ln(\ln(y)(\ln(y)-2x))=C$

$\ln^2(y)-2x\ln(y)=C$

Answer 2

Hint: consider the inverse function $x=x(y)$ and the equation for $dx/dy$.