The given differential equation is $x \frac{d^2y} {dx^2} - (2x-1)\frac{dy}{dx} + (x-1)y = 0$. This is a second-order linear homogeneous differential equation. Let's rewrite it in the standard form $y'' + P(x)y' + Q(x)y = 0$. Divide by $x$: $\frac{d^2y} {dx^2} - \left(2 - \frac{1}{x}\right)\frac{dy}{dx} + \left(1 - \frac{1}{x}\right)y = 0$. So, $P(x) = -(2 - \frac{1}{x}) = \frac{1}{x} - 2$, and $Q(x) = 1 - \frac{1}{x}$.This is a form of the confluent hypergeometric equation or related equation. Let's try to guess a solution. Common solutions involve powers of $x$, $e^x$, $\log x$, etc. The coefficients involve $x$ and constants, making it tricky.Let's check if $y=e^x$ is a solution. If $y=e^x$, then $y'=e^x$, $y''=e^x$. Substitute into the equation: $x e^x - (2x-1)e^x + (x-1)e^x = e^x [x - (2x-1) + (x-1)]$ $= e^x [x - 2x + 1 + x - 1] = e^x [0] = 0$. So, $y=e^x$ is a solution.Now that we have one solution $y_1(x) = e^x$, we can use the method of reduction of order to find a second linearly independent solution $y_2(x)$. The formula is $y_2(x) = y_1(x) \int \frac{e^{-\int P(x) dx}}{y_1(x)^2} dx$. In the standard form, $P(x) = -(2 - \frac{1} {x}) = \frac{1}{x} - 2$. So, $-\int P(x) dx = -\int (\frac{1}{x} - 2) dx = -(\ln|x| - 2x) = 2x - \ln|x|$. $e^{-\int P(x) dx} = e^{2x - \ln|x|} = e^{2x} e^{- \ln|x|} = e^{2x} \frac{1}{|x|}$. Assuming $x>0$, $e^{-\int P(x) dx} = \frac{e^{2x}}{x}$.Now substitute into the formula for $y_2$: $y_2(x) = e^x \int \frac{e^{2x}/x}{(e^x)^2} dx = e^x \int \frac{e^{2x}/x}{e^{2x}} dx$ $y_2(x) = e^x \int \frac{1}{x} dx$ $y_2(x) = e^x (\ln|x| + C_3)$. Assuming $x>0$, $y_2(x) = e^x (\ln x + C_3)$. We can choose $C_3=0$ to get a simple linearly independent solution. So, $y_2(x) = e^x \ln x$.The general solution is $y(x) = C_1 y_1(x) + C_2 y_2(x)$. $y(x) = C_1 e^x + C_2 e^x \ln x$. $y(x) = e^x (C_1 + C_2 \ln x)$.This matches Option B: $y = e^x (C_1 \log_e x + C_2)$.