Question

Let \( y = y(x) \) be the solution of the differential equation \( \frac{dy}{dx} + 3(\tan^2 x) y + 3y = \sec^2 x \), with \( y(0) = \frac{1}{3} + e^3 \). Then \( y\left(\frac{\pi}{4}\right) \) is equal to

• A. \( \frac{2}{3} \)
• B. \( \frac{4}{3} \)
• C. \( \frac{4}{3} + e^3 \)
• D. \( \frac{2}{3} + e^3 \)

Hint:

Recognize the linear differential equation form \( \frac{dy}{dx} + P(x) y = Q(x) \). Find the integrating factor \( IF = e^{\int P(x) dx} \). The solution is \( y \cdot IF = \int Q(x) \cdot IF \, dx + c \). Use the initial condition to find the value of the constant \( c \), and then substitute the required value of \( x \) to find \( y \). Remember trigonometric identities to simplify the equation and integration.

Answer 1

The Correct Option is B

The differential equation is given as: \[ \frac{dy}{dx} + 3(\tan^2 x) y + 3y = \sec^2 x \] This can be rewritten as: \[ \frac{dy}{dx} + 3(\tan^2 x + 1) y = \sec^2 x \] Using the identity \( \tan^2 x + 1 = \sec^2 x \): \[ \frac{dy}{dx} + 3\sec^2 x \cdot y = \sec^2 x \] This is a linear differential equation of the form \( \frac{dy}{dx} + P(x) y = Q(x) \), with \( P(x) = 3\sec^2 x \) and \( Q(x) = \sec^2 x \). The integrating factor (IF) is calculated as \( e^{\int P(x) dx} \): \[ IF = e^{\int 3\sec^2 x dx} = e^{3\tan x} \] The general solution for a linear differential equation is \( y \cdot IF = \int Q(x) \cdot IF \, dx + c \): \[ y \cdot e^{3\tan x} = \int \sec^2 x \cdot e^{3\tan x} \, dx + c \] To solve the integral, let \( u = 3\tan x \). Then \( du = 3\sec^2 x \, dx \), which implies \( \sec^2 x \, dx = \frac{1}{3} du \). Substituting these into the equation: \[ y \cdot e^{3\tan x} = \int e^u \cdot \frac{1}{3} \, du + c \] \[ y \cdot e^{3\tan x} = \frac{1}{3} e^u + c \] Substituting back \( u = 3\tan x \): \[ y \cdot e^{3\tan x} = \frac{1}{3} e^{3\tan x} + c \] Using the initial condition \( y(0) = \frac{1}{3} + e^3 \). At \( x = 0 \), \( \tan(0) = 0 \). \[ \left(\frac{1}{3} + e^3\right) e^{3(0)} = \frac{1}{3} e^{3(0)} + c \] \[ \frac{1}{3} + e^3 = \frac{1}{3} (1) + c \] \[ \frac{1}{3} + e^3 = \frac{1}{3} + c \] This gives \( c = e^3 \). The particular solution is: \[ y \cdot e^{3\tan x} = \frac{1}{3} e^{3\tan x} + e^3 \] We need to find \( y\left(\frac{\pi}{4}\right) \). At \( x = \frac{\pi}{4} \), \( \tan\left(\frac{\pi}{4}\right) = 1 \). \[ y \cdot e^{3(1)} = \frac{1}{3} e^{3(1)} + e^3 \] \[ y \cdot e^3 = \frac{1}{3} e^3 + e^3 \] Dividing by \( e^3 \): \[ y = \frac{1}{3} + 1 = \frac{4}{3} \] Therefore, \( y\left(\frac{\pi}{4}\right) = \frac{4}{3} \).