Question

If \[ \cos x \frac{dy}{dx} - y \sin x = 6x, \quad (0 < x < \frac{\pi}{2}) \quad \text{and} \quad y(\frac{\pi}{3}) = 0, \quad \text{then} \quad y(\frac{\pi}{6}) = \]

• A. \( \frac{-\pi^2}{4\sqrt{3}} \)
• B. \( \frac{-\pi^2}{2} \)
• C. \( \frac{-\pi^2}{2\sqrt{3}} \)
• D. \( \frac{\pi^2}{2\sqrt{3}} \)

Hint:

For solving differential equations with initial conditions, apply standard methods like separation of variables or integrating factors, and always check the consistency with the given conditions.

Answer 1

The Correct Option is C

Step 1: Solve the differential equation We are given the differential equation: \[ \cos x \frac{dy}{dx} - y \sin x = 6x \] and the initial condition \( y\left(\frac{\pi}{3}\right) = 0 \). Rearrange the equation: \[ \frac{dy}{dx} = \frac{6x + y \sin x}{\cos x} \] Step 2: Solve using the given initial condition Using the appropriate method (such as integrating or applying standard solutions), we can solve the equation. After solving and applying the initial conditions, we get: \[ y\left(\frac{\pi}{6}\right) = \frac{-\pi^2}{2\sqrt{3}} \]

Answer 2

Given the differential equation:
\[ \cos x \frac{dy}{dx} - y \sin x = 6x, \quad 0 < x < \frac{\pi}{2} \] with the initial condition:
\[ y\left(\frac{\pi}{3}\right) = 0 \] Find \( y\left(\frac{\pi}{6}\right) \).

Step 1: Rewrite the differential equation:
\[ \cos x \frac{dy}{dx} = y \sin x + 6x \] \[ \frac{dy}{dx} = y \tan x + \frac{6x}{\cos x} \]

Step 2: This is a linear differential equation of the form:
\[ \frac{dy}{dx} - y \tan x = \frac{6x}{\cos x} \] with integrating factor (IF):
\[ \mu(x) = e^{-\int \tan x \, dx} = e^{\ln \cos x} = \cos x \]

Step 3: Multiply both sides by IF:
\[ \cos x \frac{dy}{dx} - y \sin x = 6x \] which matches the original equation, confirming IF.

Step 4: The left side is the derivative of \( y \cos x \):
\[ \frac{d}{dx} (y \cos x) = 6x \] Integrate both sides:
\[ y \cos x = \int 6x \, dx + C = 3x^2 + C \]

Step 5: Solve for \( y \):
\[ y = \frac{3x^2 + C}{\cos x} \]

Step 6: Apply initial condition \( y(\frac{\pi}{3}) = 0 \):
\[ 0 = \frac{3 \left(\frac{\pi}{3}\right)^2 + C}{\cos \frac{\pi}{3}} = \frac{3 \times \frac{\pi^2}{9} + C}{\frac{1}{2}} = 2 \left( \frac{\pi^2}{3} + C \right) \] \[ \implies \frac{\pi^2}{3} + C = 0 \implies C = -\frac{\pi^2}{3} \]

Step 7: Find \( y(\frac{\pi}{6}) \):
\[ y\left(\frac{\pi}{6}\right) = \frac{3 \left(\frac{\pi}{6}\right)^2 - \frac{\pi^2}{3}}{\cos \frac{\pi}{6}} = \frac{3 \times \frac{\pi^2}{36} - \frac{\pi^2}{3}}{\frac{\sqrt{3}}{2}} = \frac{\frac{\pi^2}{12} - \frac{\pi^2}{3}}{\frac{\sqrt{3}}{2}} = \frac{-\frac{\pi^2}{4}}{\frac{\sqrt{3}}{2}} = -\frac{\pi^2}{4} \times \frac{2}{\sqrt{3}} = -\frac{\pi^2}{2 \sqrt{3}} \]

Therefore,
\[ \boxed{ y\left(\frac{\pi}{6}\right) = -\frac{\pi^2}{2 \sqrt{3}} } \]