Question

If \( \cos x \frac{dy}{dx} = y \sin x - 1, x \neq (2n+1)\frac{\pi}{2}, n \in \mathbb{Z} \) is the differential equation corresponding to the curve \( y = f(x) \) and \( f(0)=1 \) then \( f(x)= \)

• A. \( (1-x)\sec x + \tan x \)
• B. \( (1+x)\sec x \)
• C. \( (1-x)\sec x \)
• D. \( \sec x - x \)

Hint:

Recognize the standard form of linear differential equations. The integral of \( -\tan x \) leads to \( \ln|\cos x| \), which simplifies the IF nicely to \( \cos x \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

We rewrite the given equation as a first-order linear differential equation \( \frac{dy}{dx} + P(x)y = Q(x) \) and solve it using an Integrating Factor.
Step 2: Key Formula or Approach:

1. Standard form: \( y' - y \tan x = -\sec x \). 2. Integrating Factor: \( IF = e^{\int P dx} \). 3. Solution: \( y \cdot IF = \int Q \cdot IF dx + c \).
Step 3: Detailed Explanation:

Divide equation by \( \cos x \): \( \frac{dy}{dx} = y \tan x - \sec x \) \( \frac{dy}{dx} - (\tan x)y = -\sec x \). Here \( P(x) = -\tan x \), \( Q(x) = -\sec x \). IF \( = e^{\int -\tan x dx} = e^{\ln|\cos x|} = \cos x \). Multiply equation by IF: \( \frac{d}{dx}(y \cos x) = -\sec x \cdot \cos x = -1 \). Integrate: \( y \cos x = \int -1 dx = -x + c \). Use initial condition \( f(0) = 1 \implies y=1 \) when \( x=0 \). \( 1 \cdot \cos 0 = -0 + c \implies c = 1 \). Substitute c: \( y \cos x = 1 - x \). \( y = (1-x) \sec x \).
Step 4: Final Answer:

The function is \( f(x) = (1-x) \sec x \).