Question

If \( y(x) = v(x) \sec x \) is the solution of \[ y'' - (2 \tan x) y' + 5y = 0, -\frac{\pi}{2} < x < \frac{\pi}{2}, \text{ satisfying } y(0) = 0 \text{ and } y'(0) = \sqrt{6}, \] then \( v \left( \frac{\pi}{6 \sqrt{6}} \right) \) is .............. (correct up to two decimal places). 
 

Hint:

When solving second-order differential equations, always use the initial conditions to determine the constants in the solution.

Answer 1

Correct Answer: 0.5

Step 1: Solve the given second-order differential equation.
We are given the second-order linear differential equation \( y'' - (2 \tan x) y' + 5y = 0 \). The general solution to this equation can be found using standard methods for solving linear ODEs, such as undetermined coefficients or variation of parameters.

Step 2: Use the given initial conditions.
Using the initial conditions \( y(0) = 0 \) and \( y'(0) = \sqrt{6} \), we determine the constants in the solution. The solution for \( v(x) \) is then found as \( v(x) = \sec x \).

Step 3: Evaluate \( v \left( \frac{\pi}{6 \sqrt{6}} \right) \).
Now, evaluate \( v \left( \frac{\pi}{6 \sqrt{6}} \right) \). Since \( v(x) = \sec x \), we compute: \[ v \left( \frac{\pi}{6 \sqrt{6}} \right) = \sec \left( \frac{\pi}{6 \sqrt{6}} \right). \] Using a calculator, we find that \( v \left( \frac{\pi}{6 \sqrt{6}} \right) \approx \boxed{1.155} \).