Step 1: Understanding the Concept:
The given equation is a first-order linear differential equation.
We need to rearrange it into the standard form $\frac{dy}{dx} + P y = Q$ to find the integrating factor.
Step 2: Key Formula or Approach:
The standard form is $\frac{dy}{dx} + P y = Q$.
The Integrating Factor (IF) is given by $e^{\int P dx}$.
The general solution is $y \cdot (\text{IF}) = \int Q \cdot (\text{IF}) dx + c$.
Step 3: Detailed Explanation:
Given differential equation is:
\[ x(x - 1)\frac{dy}{dx} = x^3(2x - 1) + (x - 2)y \]
Rearranging terms, we get:
\[ x(x - 1)\frac{dy}{dx} - (x - 2)y = x^3(2x - 1) \]
Dividing the entire equation by $x(x - 1)$, we obtain the standard linear form:
\[ \frac{dy}{dx} - \frac{x - 2}{x(x - 1)}y = \frac{x^3(2x - 1)}{x(x - 1)} \]
\[ \frac{dy}{dx} + \left( \frac{2 - x}{x(x - 1)} \right)y = \frac{x^2(2x - 1)}{x - 1} \]
Here, $P = \frac{2 - x}{x(x - 1)}$ and $Q = \frac{x^2(2x - 1)}{x - 1}$.
To integrate $P$, we use partial fractions:
\[ \frac{2 - x}{x(x - 1)} = \frac{A}{x} + \frac{B}{x - 1} \]
\[ 2 - x = A(x - 1) + Bx \]
Putting $x = 0$, we get $2 = A(-1) \Rightarrow A = -2$.
Putting $x = 1$, we get $1 = B(1) \Rightarrow B = 1$.
So, $P = -\frac{2}{x} + \frac{1}{x - 1}$.
Now, calculate the Integrating Factor (IF):
\[ \text{IF} = e^{\int P dx} = e^{\int \left( -\frac{2}{x} + \frac{1}{x - 1} \right) dx} \]
\[ \text{IF} = e^{-2\ln|x| + \ln|x - 1|} = e^{\ln(x^{-2}) + \ln(x - 1)} = e^{\ln\left(\frac{x - 1}{x^2}\right)} = \frac{x - 1}{x^2} \]
The general solution is given by:
\[ y \cdot (\text{IF}) = \int Q \cdot (\text{IF}) dx + c \]
\[ y \left( \frac{x - 1}{x^2} \right) = \int \left( \frac{x^2(2x - 1)}{x - 1} \right) \left( \frac{x - 1}{x^2} \right) dx + c \]
\[ y \left( \frac{x - 1}{x^2} \right) = \int (2x - 1) dx + c \]
\[ y \left( \frac{x - 1}{x^2} \right) = x^2 - x + c = x(x - 1) + c \]
Multiplying both sides by $x^2$, we get:
\[ y(x - 1) = x^3(x - 1) + cx^2 \]
Step 4: Final Answer:
The general solution is $y(x - 1) = x^3(x - 1) + cx^2$.