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.