If curve `y=f(x)` satisfying differential equation `dy/dx+(1/(x^2-1))y=((x-1)/(x+1))^(1/2)` such that `f(2)=sqrt3` then the value of `sqrt7 f(8)` is

Options

A. `7-3ln3`
B. `21-6ln2`
C. `21-6ln3`
D. `7+3ln2`

Answer

To solve the given differential equation and find the value of \(\sqrt{7} f(8)\), we will follow these steps:

### Step 1: Write down the differential equation
The given differential equation is:
\[ \frac{dy}{dx} + \frac{1}{x^2 - 1} y = \sqrt{\frac{x - 1}{x + 1}} \]

### Step 2: Identify \(p(x)\) and \(q(x)\)
From the standard form of a linear differential equation \(\frac{dy}{dx} + p(x)y = q(x)\), we identify:
- \(p(x) = \frac{1}{x^2 - 1}\)
- \(q(x) = \sqrt{\frac{x - 1}{x + 1}}\)

### Step 3: Find the integrating factor
The integrating factor \(I(x)\) is given by:
\[ I(x) = e^{\int p(x) \, dx} = e^{\int \frac{1}{x^2 - 1} \, dx} \]
To compute this integral, we can use partial fractions:
\[ \frac{1}{x^2 - 1} = \frac{1}{(x - 1)(x + 1)} = \frac{1/2}{x - 1} - \frac{1/2}{x + 1} \]
Thus,
\[ \int \frac{1}{x^2 - 1} \, dx = \frac{1}{2} \ln |x - 1| - \frac{1}{2} \ln |x + 1| + C = \frac{1}{2} \ln \left|\frac{x - 1}{x + 1}\right| + C \]
Therefore, the integrating factor is:
\[ I(x) = e^{\frac{1}{2} \ln \left|\frac{x - 1}{x + 1}\right|} = \left(\frac{x - 1}{x + 1}\right)^{1/2} \]

### Step 4: Multiply through by the integrating factor
Multiply the entire differential equation by \(I(x)\):
\[ \left(\frac{x - 1}{x + 1}\right)^{1/2} \frac{dy}{dx} + \left(\frac{x - 1}{x + 1}\right)^{1/2} \frac{1}{x^2 - 1} y = \left(\frac{x - 1}{x + 1}\right)^{1/2} \sqrt{\frac{x - 1}{x + 1}} \]
This simplifies to:
\[ \left(\frac{x - 1}{x + 1}\right)^{1/2} \frac{dy}{dx} + \left(\frac{x - 1}{x + 1}\right)^{1/2} \frac{1}{x^2 - 1} y = \frac{x - 1}{x + 1} \]

### Step 5: Integrate both sides
The left side can be integrated as:
\[ \frac{d}{dx}\left(y \left(\frac{x - 1}{x + 1}\right)^{1/2}\right) = \frac{x - 1}{x + 1} \]
Integrating both sides gives:
\[ y \left(\frac{x - 1}{x + 1}\right)^{1/2} = \int \frac{x - 1}{x + 1} \, dx \]
Calculating the integral:
\[ \int \frac{x - 1}{x + 1} \, dx = x - 2 \ln|x + 1| + C \]
Thus, we have:
\[ y \left(\frac{x - 1}{x + 1}\right)^{1/2} = x - 2 \ln|x + 1| + C \]

### Step 6: Solve for \(y\)
Rearranging gives:
\[ y = \frac{x - 2 \ln|x + 1| + C}{\left(\frac{x - 1}{x + 1}\right)^{1/2}} \]

### Step 7: Use the initial condition
We know \(f(2) = \sqrt{3}\):
\[ \sqrt{3} = \frac{2 - 2 \ln(3) + C}{\sqrt{\frac{1}{3}}} \]
Solving for \(C\) gives:
\[ C = \sqrt{3} \cdot \sqrt{3} + 2 \ln(3) - 2 = 3 + 2 \ln(3) - 2 = 1 + 2 \ln(3) \]

### Step 8: Find \(f(8)\)
Substituting \(x = 8\):
\[ f(8) = \frac{8 - 2 \ln(9) + (1 + 2 \ln(3))}{\sqrt{\frac{7}{9}}} \]
This simplifies to:
\[ f(8) = \frac{8 - 2 \ln(9) + 1 + 2 \ln(3)}{\frac{\sqrt{7}}{3}} = 3 \cdot (9 - 2 \ln(9) + 2 \ln(3)) \]

### Step 9: Calculate \(\sqrt{7} f(8)\)
Finally, we compute:
\[ \sqrt{7} f(8) = \sqrt{7} \cdot 3 \cdot (9 - 2 \ln(9) + 2 \ln(3)) \]

### Final Result
The value of \(\sqrt{7} f(8)\) is:
\[ \sqrt{7} f(8) = 6 \ln(3) \]