Question

If the solution curve \( y = y \, x \) of the differential equation \((1 + y^2) \left(1 + \log_e x\right) dx + x \, dy = 0, \quad x > 0\) passes through the point \( (1, 1) \) and\[y(e) = \frac{\alpha - \tan\left(\frac{3}{2}\right)}{\beta + \tan\left(\frac{3}{2}\right)},\]then \( \alpha + 2\beta \) is

Answer 1

Correct Answer: 3

Step 1: Separate Variables in the Differential Equation

\[ \int \left( \frac{1}{x} + \frac{\ln x}{x} \right) dx + \int \frac{dy}{1 + y^2} = 0 \]

Step 2: Integrate Both Sides

Integrating, we get:

\[ \ln x + \frac{(\ln x)^2}{2} + \tan^{-1} y = C \]

Step 3: Apply Initial Condition \((x, y) = (1, 1)\)

Substitute \(x = 1\) and \(y = 1\) to find \(C\):

\[ \ln 1 + \frac{(\ln 1)^2}{2} + \tan^{-1}(1) = C \Rightarrow C = \frac{\pi}{4} \]

Step 4: Rewrite the Solution with \(C = \frac{\pi}{4}\)

\[ \ln x + \frac{(\ln x)^2}{2} + \tan^{-1} y = \frac{\pi}{4} \]

Step 5: Evaluate \(y(e)\)

Substitute \(x = e\) into the equation:

\[ \ln e + \frac{(\ln e)^2}{2} + \tan^{-1} y = \frac{\pi}{4} \]

\[ 1 + \frac{1}{2} + \tan^{-1} y = \frac{\pi}{4} \]

Solving for \(y\), we get:

\[ y = \tan \left( \frac{\pi}{4} - \frac{3}{2} \right) = \frac{1 - \tan \frac{3}{2}}{1 + \tan \frac{3}{2}} \]

Step 6: Identify \(\alpha\) and \(\beta\)

Comparing with the given expression for \(y(e)\), we find \(\alpha = 1\) and \(\beta = 1\).

Step 7: Calculate \(\alpha + 2\beta\)

\[ \alpha + 2\beta = 1 + 2 \cdot 1 = 3 \]

So, the correct answer is: 3