The solution of $x (x - 1) \frac{dy}{dx} - (x - 2) y = x^{3} (2 x - 1)$ is

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$y \frac{(x - 1)}{x^{2}} = x^{2} - x + c$

$y \frac{(x - 1)}{x^{2}} = x^{2} + x + c$

$y \frac{(x + 1)}{x^{2}} = x^{2} - x + c$

$y \frac{(x + 1)}{x^{2}} = x^{2} + x + c$

answer is A.

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Detailed Solution

$x (x - 1) \frac{dy}{dx} - (x - 2) y = x^{3} (2 x - 1)$

Divide by $x (x - 1)$

$\frac{dy}{dx} - \frac{(x - 2)}{x (x - 1)} y = \frac{x^{3} \cdot (2 x - 1)}{x (x - 1)}$

then y the form $\frac{dy}{dx} + Py = Q$

$\begin{array}{rcl} P & = & \frac{- (x - 2)}{x (x - 1)} = \frac{2 - x}{x (x - 1)} \\ I . F & = & e^{\int \frac{2 - x}{x (x - 1)}} dx \\ \frac{2 - x}{x (x - 1)} & = & \frac{A}{x} + \frac{B}{x - 1} \\ \Rightarrow & \frac{2 - x}{x (x - 1)} = \frac{A (x - 1) + Bx}{x (x - 1)} \end{array}$

$\Rightarrow 2 - x = Ax - A + Bx \Rightarrow 2 - x = - A + x (A + B) \Rightarrow A = - 2, A + B = - 1 \Rightarrow B = 1$

$\begin{array}{rcl} \Rightarrow & \frac{2 - x}{x (x - 1)} = \frac{- 2}{x} + \frac{1}{x - 1} \\ \int \frac{2 - x}{x (x - 1)} & = & \int \frac{- 2}{x} dx + \int \frac{1}{x - 1} dx \\ = & - 2 \log x + \log (x - 1) \\ \Rightarrow & \log (x - 1) - \log x^{2} \\ \Rightarrow & \log (x - 1) - \log x^{2} = \log \frac{x - 1}{x^{2}} \\ I . F & = & e^{\log (\frac{x - 1}{x^{2}})} = \frac{x - 1}{x^{2}} \end{array}$

multiply with $\frac{x - 1}{x^{2}}$ and integrate B.S

$\begin{array}{rcl} y \times \frac{x - 1}{x^{2}} & = & \int \frac{x^{3} (2 x - 1)}{x (x - 1)} \times \frac{x - 1}{x^{2}} \\ = & \int 2 x dx - \int dx \\ y (\frac{x - 1}{x^{2}}) & = & x^{2} - x + c \end{array}$