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: Find derivatives of $y = v\sec x$

$$y' = v'\sec x + v\sec x\tan x$$

$$y'' = v''\sec x + 2v'\sec x\tan x + v\sec x\tan^2 x + v\sec^3 x$$

Step 2: Substitute into the differential equation

$$v''\sec x + 2v'\sec x\tan x + v\sec x\tan^2 x + v\sec^3 x$$ $$- 2\tan x(v'\sec x + v\sec x\tan x) + 5v\sec x = 0$$

Simplifying: $$v''\sec x + 2v'\sec x\tan x + v\sec x\tan^2 x + v\sec^3 x$$ $$- 2v'\sec x\tan x - 2v\sec x\tan^2 x + 5v\sec x = 0$$

$$v''\sec x + v\sec x\tan^2 x + v\sec^3 x - 2v\sec x\tan^2 x + 5v\sec x = 0$$

$$v''\sec x + v\sec x(\sec^2 x - \tan^2 x) + 5v\sec x = 0$$

Using $\sec^2 x - \tan^2 x = 1$: $$v''\sec x + v\sec x + 5v\sec x = 0$$

$$v'' + 6v = 0$$

Step 3: Solve the simplified equation

$$v'' + 6v = 0$$

Characteristic equation: $r^2 + 6 = 0 \Rightarrow r = \pm i\sqrt{6}$

General solution: $$v(x) = C_1\cos(\sqrt{6}x) + C_2\sin(\sqrt{6}x)$$

Step 4: Apply initial conditions

From $y(0) = 0$: $$y(0) = v(0)\sec(0) = v(0) \cdot 1 = 0$$ $$\Rightarrow v(0) = 0$$

$$v(0) = C_1 = 0$$

So: $v(x) = C_2\sin(\sqrt{6}x)$

From $y'(0) = \sqrt{6}$: $$y'(0) = v'(0)\sec(0) + v(0)\sec(0)\tan(0) = v'(0)$$

$$v'(x) = C_2\sqrt{6}\cos(\sqrt{6}x)$$ $$v'(0) = C_2\sqrt{6} = \sqrt{6}$$ $$C_2 = 1$$

Therefore: $v(x) = \sin(\sqrt{6}x)$

Step 5: Calculate $v\left(\frac{\pi}{6\sqrt{6}}\right)$

$$v\left(\frac{\pi}{6\sqrt{6}}\right) = \sin\left(\sqrt{6} \cdot \frac{\pi}{6\sqrt{6}}\right) = \sin\left(\frac{\pi}{6}\right) = 0.5$$

Answer: 0.5