To solve this problem, let's follow these steps systematically:

1. Integrate $f''(x)$ to find $f'(x)$: $f''(x) = -\frac{4}{(x-1)^2} - 2$ Integrate $f''(x)$ with respect to $x$.

2. Integrate $f'(x)$ to find $f(x)$: After obtaining $f'(x)$, integrate again to find $f(x)$.

3. Apply the initial conditions: Use $f'(2) = 0$ and $f(2) = 5$ to determine the constants of integration.

Step 1: Integrating $f''(x)$

We start with: $f''(x) = -\frac{4}{(x-1)^2} - 2$ The integral of $-\frac{4}{(x-1)^2}$ is: $\int -\frac{4}{(x-1)^2} dx = \frac{4}{x-1}$ The integral of $-2$ is: $\int -2 \, dx = -2x$ Thus: $f'(x) = \frac{4}{x-1} - 2x + C_1$

Step 2: Integrating $f'(x)$

Now integrate $f'(x)$: $f'(x) = \frac{4}{x-1} - 2x + C_1$ The integral of $\frac{4}{x-1}$ is: $\int \frac{4}{x-1} dx = 4 \ln|x-1|$ The integral of $-2x$ is: $\int -2x \, dx = -x^2$ The integral of $C_1$ is: $\int C_1 \, dx = C_1x$ Thus: $f(x) = 4 \ln|x-1| - x^2 + C_1x + C_2$

Step 3: Applying Initial Conditions

Using $f'(2) = 0$: $f'(x) = \frac{4}{x-1} - 2x + C_1$ Substitute $x = 2$: $0 = \frac{4}{2-1} - 2(2) + C_1$ $0 = 4 - 4 + C_1 \quad \implies \quad C_1 = 0$

Using $f(2) = 5$: $f(x) = 4 \ln|x-1| - x^2 + C_1x + C_2$ Substitute $x = 2$ and $C_1 = 0$: $5 = 4 \ln|2-1| - 2^2 + C_2$ $5 = 4 \ln(1) - 4 + C_2$ $5 = 0 - 4 + C_2 \quad \implies \quad C_2 = 9$

Final Solution

Substitute $C_1 = 0$ and $C_2 = 9$ into $f(x)$: $f(x) = 4 \ln|x-1| - x^2 + 9$

For $x > 1$, $\ln|x-1| = \ln(x-1)$. Therefore: $f(x) = 4 \ln(x-1) - x^2 + 9$

Let me know if you want further details or have any questions! Here are some related questions to consider:

1. How is the constant of integration determined in differential equations?
2. Why is $\ln|x-1|$ used when integrating $\frac{1}{x-1}$?
3. How can you confirm that this solution satisfies the original differential equation?
4. What happens if the initial conditions are different?
5. How does $x > 1$ simplify $\ln|x-1|$ to $\ln(x-1)$?

Tip: Always verify the solution by substituting it back into the original differential equation!