Solve the differential equation x(1-y) dy/dx = -2y given y=2 when x=e, and show that y = ln((x2y)/2).

The given differential equation is:

$$x(1-y)\frac{dy}{dx} = -2y$$

Divide both sides by $xy$:

$$\frac{1-y}{y} \frac{dy}{dx} = -\frac{2}{x}$$

Separate the variables $y$ and $x$:

$$\left(\frac{1}{y} - 1\right) dy = -\frac{2}{x} dx$$

Integrate both sides of the equation:

$$\int \left(\frac{1}{y} - 1\right) dy = \int -\frac{2}{x} dx$$ $$\ln|y| - y = -2\ln|x| + C$$

where $C$ is the constant of integration.

Given that $y=2$ when $x=e$:

$$\ln|2| - 2 = -2\ln|e| + C$$

Since $\ln e = 1$:

$$\ln 2 - 2 = -2(1) + C$$ $$\ln 2 - 2 = -2 + C$$ $$C = \ln 2$$

$$\ln|y| - y = -2\ln|x| + \ln 2$$

Rearrange the terms to match the required form. Move $y$ to one side and logarithmic terms to the other:

$$y = \ln|y| + 2\ln|x| - \ln 2$$

Using logarithm properties, $a\ln b = \ln(b^a)$, $\ln a + \ln b = \ln(ab)$, and $\ln a - \ln b = \ln\left(\frac{a}{b}\right)$:

$$y = \ln|y| + \ln(x^2) - \ln 2$$ $$y = \ln\left(\frac{|y|x^2}{2}\right)$$

Since the problem asks to show $y = \ln\left(\frac{x^2y}{2}\right)$, we can assume $y > 0$ and $x \neq 0$.

$$y = \ln\left(\frac{x^2y}{2}\right)$$

This matches the required form.