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

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