Question

In Problems 19-26 solve the given differential equation.

$\mathit{y}\mathbf{(}\mathit{l}\mathit{n}\mathit{x}\mathbf{-}\mathit{l}\mathit{n}\mathit{y}\mathbf{)}\mathit{d}\mathit{x}\mathbf{=}\mathbf{(}\mathit{x}\mathit{l}\mathit{n}\mathit{x}\mathbf{-}\mathit{x}\mathit{l}\mathit{n}\mathit{y}\mathbf{-}\mathit{y}\mathbf{)}\mathit{d}\mathit{y}$

Short answer

The solution of the given differential equation is $x(\ln x-\ln y)-x=c_{1}y-y\ln \left|y\right|$.

Step-by-step solution

Note the given data

Given the initial value problem $y(\ln x-\ln y)dx=(x\ln x-x\ln y-y)dy$.

We know that, $\ln u-\ln v=\ln \left(\frac{u}{v}\right)$.

Simplifying the given differential equation

Rewrite the differential equation as follows:

$\begin{array}{c}y(\ln x-\ln y)dx=(x\ln x-x\ln y-y)dy\\ y\ln \left(\frac{x}{y}\right)dx=(x\ln \left(\frac{x}{y}\right)-y)dy\end{array}$

Let $x=qy$.

Differentiate $x=qy$both sides as:

$\begin{array}{c}dx=qdy+ydq\\ y\ln q(qy+ydq)-(qy\ln q-y)=0\\ qy\ln qdy+y^2\ln qdq-(qy\ln q-y)dy=0\\ (qy\ln q-qy\ln q+y)dy+y^2\ln qdq=0\end{array}$

Further simplify the equation as follows:

$\begin{array}{c}ydy+y^2\ln qdq=0\\ \frac{dy}{y}+\ln qdq=0\\ \frac{dy}{y}=-\ln qdq\end{array}$

Find the required solution

Integrating$\frac{dy}{y}=-\ln qdq$ as follows:

$\begin{array}{c}\int \frac{dy}{y}=-\int \ln qdq\\ \ln \left|y\right|=-q\ln \left|q\right|+q+c_1\end{array}$

Substitute $q=\frac{y}{x}$into $\ln \left|y\right|=-q\ln \left|q\right|+q+c_1$as:

$\begin{array}{c}\ln \left|y\right|=-\frac{x}{y}\ln \left|\frac{x}{y}\right|+\frac{x}{y}+c_1\\ y\ln \left|y\right|=-x(\ln x-\ln y)+x+c_{1}y\\ x(\ln x-\ln y)-x=c_{1}y-y\ln \left|y\right|\end{array}$

Hence, the solution of the given differential equation is $x(\ln x-\ln y)-x=c_{1}y-y\ln \left|y\right|$.