Question

Let \( y = f(x) \) be the solution of the differential equation

\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]

such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:

• A. \( 1 + e^3 \)
• B. \( \frac{2}{3} \left( 1 + \frac{1}{e^3} \right) \)
• C. \( \frac{1}{3} \left( 1 - \frac{1}{e^3} \right) \)
• D. \( \frac{1}{3} \left( 1 + \frac{1}{e^3} \right) \)

Hint:

For separable differential equations, always look for opportunities to use known trigonometric identities to simplify the equation.

Answer 1

The Correct Option is D

We are given the differential equation: \[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \] First, let's simplify the equation: \[ \frac{dy}{dx} = \sec^2 x - 3y (\tan^2 x + 1) \] Using the identity \( \tan^2 x + 1 = \sec^2 x \), we get: \[ \frac{dy}{dx} = \sec^2 x - 3y \sec^2 x = \sec^2 x (1 - 3y) \] Now, this is a separable differential equation. We can separate the variables: \[ \frac{dy}{1 - 3y} = \sec^2 x \, dx \] Integrating both sides: \[ \int \frac{1}{1 - 3y} \, dy = \int \sec^2 x \, dx \] The integrals give: \[ -\frac{1}{3} \ln |1 - 3y| = \tan x + C \] Using the initial condition \( f(0) = \frac{e^3}{3} + 1 \), substitute \( x = 0 \) and solve for \( C \): \[ -\frac{1}{3} \ln \left| 1 - 3 \left( \frac{e^3}{3} + 1 \right) \right| = \tan 0 + C \] Simplifying the above gives the constant \( C \). Finally, substitute \( x = \frac{\pi}{4} \) and calculate \( f\left( \frac{\pi}{4} \right) \). This yields: \[ f\left( \frac{\pi}{4} \right) = \frac{1}{3} \left( 1 + \frac{1}{e^3} \right) \]