Question

solve the given differential equation. y(ln(x) - ln(y)) dx=(x ln(x) - x ln(y) - y) dy

Explanation:

Step1: Rearrange the terms

First, rewrite the given differential equation \(y(\ln(x)-\ln(y))dx=(x\ln(x)-x\ln(y)-y)dy\) into the form \(\frac{dy}{dx}\). We get \(\frac{dy}{dx}=\frac{y(\ln(x)-\ln(y))}{x\ln(x)-x\ln(y)-y}\). Let \(u = \frac{y}{x}\), then \(y = ux\) and \(\frac{dy}{dx}=u + x\frac{du}{dx}\).

Step2: Substitute \(y = ux\) into the equation

Substitute \(y = ux\) into the differential - equation. The left - hand side is \(u + x\frac{du}{dx}\), and the right - hand side:
\[

$$\begin{align*} &\frac{ux(\ln(x)-\ln(ux))}{x\ln(x)-x\ln(ux)-ux}\\ =&\frac{ux(\ln(x)-\ln(x)-\ln(u))}{x\ln(x)-x(\ln(x)+\ln(u))-ux}\\ =&\frac{-ux\ln(u)}{x\ln(x)-x\ln(x)-x\ln(u)-ux}\\ =&\frac{-u\ln(u)}{-\ln(u)-u} \end{align*}$$

\]
So, \(u + x\frac{du}{dx}=\frac{-u\ln(u)}{- \ln(u)-u}\).

Step3: Simplify the equation

\[

$$\begin{align*} u + x\frac{du}{dx}&=\frac{u\ln(u)}{\ln(u)+u}\\ x\frac{du}{dx}&=\frac{u\ln(u)}{\ln(u)+u}-u\\ x\frac{du}{dx}&=\frac{u\ln(u)-u(\ln(u)+u)}{\ln(u)+u}\\ x\frac{du}{dx}&=\frac{u\ln(u)-u\ln(u)-u^{2}}{\ln(u)+u}\\ x\frac{du}{dx}&=\frac{-u^{2}}{\ln(u)+u}\\ \frac{\ln(u)+u}{u^{2}}du&=-\frac{1}{x}dx \end{align*}$$

\]

Step4: Integrate both sides

Integrate \(\int\frac{\ln(u)+u}{u^{2}}du=-\int\frac{1}{x}dx\).
For \(\int\frac{\ln(u)+u}{u^{2}}du=\int\frac{\ln(u)}{u^{2}}du+\int\frac{1}{u}du\).
Use integration by parts for \(\int\frac{\ln(u)}{u^{2}}du\). Let \(v=\ln(u)\) and \(dw = \frac{1}{u^{2}}du\), then \(dv=\frac{1}{u}du\) and \(w =-\frac{1}{u}\).
\(\int\frac{\ln(u)}{u^{2}}du=-\frac{\ln(u)}{u}+\int\frac{1}{u^{2}}du=-\frac{\ln(u)}{u}-\frac{1}{u}+C_1\). And \(\int\frac{1}{u}du=\ln|u|+C_2\).
So, \(-\frac{\ln(u)}{u}-\frac{1}{u}+\ln|u|=-\ln|x| + C\).
Substitute back \(u=\frac{y}{x}\):
\[

$$\begin{align*} -\frac{\ln(\frac{y}{x})}{\frac{y}{x}}-\frac{x}{y}+\ln(\frac{y}{x})&=-\ln|x| + C\\ -\frac{x\ln(\frac{y}{x})}{y}-\frac{x}{y}+\ln(\frac{y}{x})+\ln|x|&=C\\ -\frac{x(\ln(y)-\ln(x))}{y}-\frac{x}{y}+\ln(y)-\ln(x)+\ln(x)&=C\\ -\frac{x\ln(y)}{y}+\frac{x\ln(x)}{y}-\frac{x}{y}+\ln(y)&=C \end{align*}$$

\]

Answer:

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