Question

Solve the differential equation :

$y\log ydx-xdy=0$

Answer 1

$y\log ydx-xdy=0$

$y\log ydx=xdy$

$\frac{dy}{y\log y}=\frac{dx}{x}$

Integrating both sides,

$\int \frac{dy}{ylogy}=\int \frac{dx}{x}$ ......(1)

Let $\log y=t$

$\frac{1}{y}=\frac{dt}{dy}$

$\frac{1}{y}dy=dt$

Substituting this value in equation 1, we get,

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

$\log t=\log x+\log C$

$log\left(\log y\right)=\log Cx$

$\log y=Cx$

$y=e^{Cx}$

This is the required general solution of the given differential equation.