Question

For the differential equations, find the general solution:$y\ log\ y \ dx-x\ dy=0$

Answer 1

The given differential equation is:

$y\ log\ y \ dx-x\ dy=0$

$⇒y \ log∫y\ dx=x\ dy$

$⇒\frac {dy}{y }log\ y=\frac {dx}{x}$

Integrating both sides, we get:

$∫\frac {dy}{y }log\ y=∫\frac {dx}{x}$ ...(1)

$Let \ log\ y=t$

$∴\frac {d}{d}(log\ y)=\frac {dt}{dy}$

$⇒\frac 1y=\frac {dt}{dy}$

$⇒\frac {1}{y} dy=dt$

Substituting this value in equation(1), we get:

$∫\frac {dt}{t}=∫\frac {dx}{x}$

$⇒log\ t =log\ x+log\ C$

$⇒log\ (log\ y)=log\ Cx$

$⇒log\ y=Cx$

$⇒y=e^{Cx}$

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