Question

The general solution of the differential equation \( \frac{dy}{dx} + \frac{\sec x}{\cos x + \sin x}y = \frac{\cos x}{1+\tan x} \) is

• A. \( (\cos x + \sin x)y = \sin x + c \)
• B. \( (\cos x + \sin x)y = \cos x + c \)
• C. \( (1+\tan x)y = \cos x + c \)
• D. \( \sec x(\cos x + \sin x)y = \sin x + c \)

Hint:

For a linear differential equation \( \frac{dy}{dx} + P(x)y = Q(x) \): 1. Calculate the Integrating Factor (IF): \( \text{IF} = e^{\int P(x)dx} \). 2. The general solution is \( y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C \). Simplify \(P(x)\) and \(Q(x)\) first. For \( \int \frac{f'(x)}{f(x)} dx = \log|f(x)| \).

Answer 1

The Correct Option is D

The given differential equation is \( \frac{dy}{dx} + P(x)y = Q(x) \), which is a linear first-order differential equation.
Here, \( P(x) = \frac{\sec x}{\cos x + \sin x} = \frac{1}{\cos x(\cos x + \sin x)} \).
And \( Q(x) = \frac{\cos x}{1+\tan x} = \frac{\cos x}{1+\frac{\sin x}{\cos x}} = \frac{\cos x}{\frac{\cos x + \sin x}{\cos x}} = \frac{\cos^2 x}{\cos x + \sin x} \).
Let's simplify \( P(x) \) and \( Q(x) \).
\( P(x) = \frac{\sec x}{\cos x + \sin x} \).
\( Q(x) = \frac{\cos^2 x}{\cos x + \sin x} \).
Integrating Factor (IF) = \( e^{\int P(x)dx} \).
\[ \int P(x)dx = \int \frac{\sec x}{\cos x + \sin x} dx = \int \frac{1}{\cos^2 x + \sin x \cos x} dx \] \[ = \int \frac{\sec^2 x}{1 + \tan x} dx \] Let \( u = 1+\tan x \).
Then \( du = \sec^2 x dx \).
\[ \int \frac{1}{u} du = \log|u| = \log|1+\tan x| = \log\left|\frac{\cos x + \sin x}{\cos x}\right| = \log|\cos x + \sin x| - \log|\cos x| \] So, IF = \( e^{\log|\cos x + \sin x| - \log|\cos x|} = e^{\log\left|\frac{\cos x + \sin x}{\cos x}\right|} = \left|\frac{\cos x + \sin x}{\cos x}\right| \).
Assuming \( \cos x>0 \) and \( \cos x + \sin x>0 \) (typical for general solution context if not specified), IF = \( \frac{\cos x + \sin x}{\cos x} = 1 + \tan x = \sec x (\cos x + \sin x) \).
The solution is \( y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) dx + C \).
\[ y \cdot \frac{\cos x + \sin x}{\cos x} = \int \frac{\cos^2 x}{\cos x + \sin x} \cdot \frac{\cos x + \sin x}{\cos x} dx + C \] \[ y \cdot \frac{\cos x + \sin x}{\cos x} = \int \frac{\cos^2 x}{\cos x} dx + C = \int \cos x dx + C \] \[ y \cdot \frac{\cos x + \sin x}{\cos x} = \sin x + C \] Multiply by \( \cos x \): \( y(\cos x + \sin x) = \sin x \cos x + C \cos x \).
This does not match any simple option.
Let's check the IF form \( \sec x (\cos x + \sin x) \).
Then \( y \cdot \sec x (\cos x + \sin x) = \int \frac{\cos^2 x}{\cos x + \sin x} \cdot \sec x (\cos x + \sin x) dx + C \) \[ y \cdot \sec x (\cos x + \sin x) = \int \cos^2 x \cdot \sec x dx + C \] \[ y \cdot \sec x (\cos x + \sin x) = \int \cos^2 x \cdot \frac{1}{\cos x} dx + C = \int \cos x dx + C \] \[ y \cdot \sec x (\cos x + \sin x) = \sin x + C \] This matches option (4).
The IF calculation was \( e^{\log|\sec x(\cos x + \sin x)|} = |\sec x (\cos x + \sin x)| \).
If we take \( \sec x (\cos x + \sin x) \) as the IF.
Let's review \( P(x) \).
\( P(x) = \frac{\sec x}{\cos x + \sin x} \).
\( \int P(x) dx = \int \frac{\sec x}{\cos x + \sin x} dx = \int \frac{1}{\cos x(\cos x + \sin x)} dx = \int \frac{1}{\cos^2 x (1+\tan x)} dx \) \( = \int \frac{\sec^2 x}{1+\tan x} dx \).
Let \( u = 1+\tan x \), so \( du = \sec^2 x dx \).
\( \int \frac{du}{u} = \log|u| = \log|1+\tan x| \).
So, IF = \( e^{\log|1+\tan x|} = |1+\tan x| \).
Let's assume \( 1+\tan x>0 \).
IF = \( 1+\tan x \).
Then the solution is \( y(1+\tan x) = \int Q(x)(1+\tan x) dx + C \).
\( Q(x) = \frac{\cos x}{1+\tan x} \).
So \( Q(x)(1+\tan x) = \cos x \).
\[ y(1+\tan x) = \int \cos x dx + C = \sin x + C \] This is \( (1+\tan x)y = \sin x + C \).
Let's check the options again.
Option (4) is \( \sec x(\cos x + \sin x)y = \sin x + c \).
\( \sec x(\cos x + \sin x) = \frac{1}{\cos x}(\cos x + \sin x) = 1 + \frac{\sin x}{\cos x} = 1+\tan x \).
So option (4) is identical to \( (1+\tan x)y = \sin x + c \).
This matches my result.