8. $left1+ln(x)+\frac{y}{x}\rightdx = 1 - ln(x)dy$

Explanation:

Step1: Rearrange the differential equation

We rewrite the given equation ([1 + \ln(x)+\frac{y}{x}]dx=[1 - \ln(x)]dy) as (\frac{dy}{dx}=\frac{1+\ln(x)+\frac{y}{x}}{1 - \ln(x)}). This is a first - order linear differential equation of the form (\frac{dy}{dx}+P(x)y = Q(x)). First, we rewrite it in the standard form. (\frac{dy}{dx}-\frac{1}{x(1 - \ln(x))}y=\frac{1+\ln(x)}{x(1 - \ln(x))})

Step2: Find the integrating factor

The integrating factor (I(x)=e^{\int P(x)dx}), where (P(x)=-\frac{1}{x(1 - \ln(x))}). Let (u = 1-\ln(x)), then (du=-\frac{1}{x}dx). So, (\int P(x)dx=\int\frac{1}{u}du=\ln|u|=\ln|1 - \ln(x)|) The integrating factor (I(x)=1 - \ln(x))

Step3: Multiply the differential equation by the integrating factor

((1 - \ln(x))\frac{dy}{dx}-\frac{1}{x}y=\frac{1+\ln(x)}{x}) The left - hand side is the derivative of ((1 - \ln(x))y) with respect to (x) by the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = 1 - \ln(x)) and (v = y).

Step4: Integrate both sides

(\int\left[(1 - \ln(x))\frac{dy}{dx}-\frac{1}{x}y\right]dx=\int\frac{1+\ln(x)}{x}dx) (\int d((1 - \ln(x))y)=\int\frac{1+\ln(x)}{x}dx) Let (t=\ln(x)), then (dt=\frac{1}{x}dx). So, (\int\frac{1+\ln(x)}{x}dx=\int(1 + t)dt=t+\frac{t^{2}}{2}+C=\ln(x)+\frac{(\ln(x))^{2}}{2}+C) ((1 - \ln(x))y=\ln(x)+\frac{(\ln(x))^{2}}{2}+C) (y=\frac{\ln(x)+\frac{(\ln(x))^{2}}{2}+C}{1 - \ln(x)})

Answer:

(y=\frac{\ln(x)+\frac{(\ln(x))^{2}}{2}+C}{1 - \ln(x)})