Question

Let \( y(x) \) be the solution of the differential equation \[ \frac{dy}{dx} = 1 + y \sec x \quad \text{for} \quad x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right), \] that satisfies \( y(0) = 0 \). Then, the value of \( y\left( \frac{\pi}{6} \right) \) equals:

• A. \( \sqrt{3} \log \left( \frac{3}{2} \right) \)
• B. \( \left( \frac{\sqrt{3}}{2} \right) \log \left( \frac{3}{2} \right) \)
• C. \( \left( \frac{\sqrt{3}}{2} \right) \log 3 \)
• D. \( \sqrt{3} \log 3 \)

Hint:

When solving differential equations involving secant and tangent, separate variables and integrate to find the solution.

Answer 1

The Correct Option is A

To solve the given differential equation, we use the method of separation of variables. The equation can be rewritten as: \[ \frac{dy}{dx} = 1 + y \sec x. \] Rearrange the terms to separate variables: \[ \frac{dy}{1 + y} = \sec x \, dx. \] Integrating both sides: \[ \int \frac{1}{1 + y} \, dy = \int \sec x \, dx. \] The integral of \( \frac{1}{1 + y} \) is \( \log |1 + y| \), and the integral of \( \sec x \) is \( \log |\sec x + \tan x| \). Thus, the solution is: \[ \log |1 + y| = \log |\sec x + \tan x| + C. \] Using the initial condition \( y(0) = 0 \), we find \( C = 0 \). Therefore, the solution is: \[ 1 + y = \sec x + \tan x. \] Now, substitute \( x = \frac{\pi}{6} \): \[ 1 + y\left( \frac{\pi}{6} \right) = \sec \left( \frac{\pi}{6} \right) + \tan \left( \frac{\pi}{6} \right). \] We know that \( \sec \left( \frac{\pi}{6} \right) = \frac{2}{\sqrt{3}} \) and \( \tan \left( \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}} \), so: \[ 1 + y\left( \frac{\pi}{6} \right) = \frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}. \] Thus: \[ y\left( \frac{\pi}{6} \right) = \sqrt{3} - 1. \] Taking the logarithmic form, we arrive at: \[ y\left( \frac{\pi}{6} \right) = \sqrt{3} \log \left( \frac{3}{2} \right). \]