Question

Solve the differential equation
\[ \cos x \frac{dy}{dx} - \cos 2x = \cos 3x \]

• A. \(y = \sin 2x - x + 2x \sin x - \log(\sec x + \tan x) + C\)
• B. \(y = \sin 2x - x + 2 \sin x - \log(\sec x - \tan x) + C\)
• C. \(y = \sin x - x + 2 \sin x - \log(\cos x + \tan x) + C\)
• D. \(y = \sin 2x - x + 2 \sin x - \log(\sec x + \tan x) + C\)

Hint:

Recall the standard integral: \(\int \sec x \, dx = \log|\sec x + \tan x|\). Identifying this specific term in the options can immediately help narrow down the correct choice.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a first-order ordinary differential equation that can be solved by separating the variables \(y\) and \(x\), followed by direct integration using trigonometric identities.

Step 2: Detailed Explanation:
Let us rearrange the given differential equation:
\[ \cos x \frac{dy}{dx} = \cos 3x + \cos 2x \]
Divide both sides by \(\cos x\):
\[ \frac{dy}{dx} = \frac{\cos 3x + \cos 2x}{\cos x} \]
Recall the standard trigonometric identity for \(\cos 3x\):
\[ \cos 3x = 4\cos^3 x - 3\cos x \]
Substitute this into our expression:
\[ \frac{dy}{dx} = \frac{(4\cos^3 x - 3\cos x) + \cos 2x}{\cos x} \]
\[ \frac{dy}{dx} = 4\cos^2 x - 3 + \frac{\cos 2x}{\cos x} \]
Now write \(\cos 2x = 2\cos^2 x - 1\):
\[ \frac{dy}{dx} = 4\cos^2 x - 3 + \frac{2\cos^2 x - 1}{\cos x} \]
\[ \frac{dy}{dx} = 4\cos^2 x - 3 + 2\cos x - \sec x \]
Use the identity \(\cos^2 x = \frac{1 + \cos 2x}{2}\):
\[ 4\cos^2 x = 2(1 + \cos 2x) = 2 + 2\cos 2x \]
Substitute this back:
\[ \frac{dy}{dx} = (2 + 2\cos 2x) - 3 + 2\cos x - \sec x \]
\[ \frac{dy}{dx} = 2\cos 2x + 2\cos x - 1 - \sec x \]
Now, integrate both sides with respect to \(x\):
\[ y = \int (2\cos 2x + 2\cos x - 1 - \sec x) dx \]
Perform the integration term-by-term:
\[ \int 2\cos 2x \, dx = \sin 2x \]
\[ \int 2\cos x \, dx = 2\sin x \]
\[ \int -1 \, dx = -x \]
\[ \int -\sec x \, dx = -\log|\sec x + \tan x| \]
Combining these results and adding the constant of integration \(C\):
\[ y = \sin 2x + 2\sin x - x - \log(\sec x + \tan x) + C \]
Rearranging terms gives:
\[ y = \sin 2x - x + 2\sin x - \log(\sec x + \tan x) + C \]
This matches Option D.

Step 3: Final Answer:
The solution is \(y = \sin 2x - x + 2\sin x - \log(\sec x + \tan x) + C\).