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)