Question

General solution of the differential equation \[ \frac{dy}{dx} + y \tan x = \sec x \quad \text{is:} \]

• A. \( y \tan x = \sec x + c \)
• B. \( \cos x = y \tan x + c \)
• C. \( y \sec x = \tan x + c \)
• D. \( y \sec x = \sec x \, \int \sec x \, dx + c \)

Hint:

For first-order linear differential equations, use the method of integrating factors to solve.

Answer 1

The Correct Option is D

We start by solving the given differential equation: \[ \frac{dy}{dx} + y \tan x = \sec x \] This is a first-order linear differential equation. The integrating factor \( I(x) \) is given by: \[ I(x) = e^{\int \tan x \, dx} = e^{\log \sec x} = \sec x \] Now, multiplying both sides of the differential equation by \( \sec x \), we get: \[ \sec x \frac{dy}{dx} + y \sec x \tan x = \sec^2 x \] The left-hand side is now the derivative of \( y \sec x \): \[ \frac{d}{dx} \left( y \sec x \right) = \sec^2 x \] Integrating both sides: \[ y \sec x = \int \sec^2 x \, dx = \tan x + c \] Thus, the general solution is: \[ y \sec x = \sec x \, \int \sec x \, dx + c \]