Question

The solution of \( \frac{dy}{\cos y} = dx \) is:

• A. \( \log|\sec y - \tan y| = x + C \)
• B. \( x + \sec y + \tan y = C \)
• C. \( \sec y + \tan y = x + C \)
• D. \( \log|\sec x + \tan y| = \sec y + x + C \)
• E. \( \log|\sec y + \tan y| = x + C \)

Hint:

For integrals involving trigonometric functions like \( \sec y \), use the identity \( \sec^2 y - \tan^2 y = 1 \) to simplify the expression if necessary. Also, remember to apply logarithms when dealing with functions of the form \( \sec y + \tan y \).

Answer 1

Answer: (E)

We are given the differential equation: \[ \frac{dy}{\cos y} = dx \] We can separate the variables \( y \) and \( x \) as follows: \[ \frac{dy}{\cos y} = dx \quad \Rightarrow \quad \int \frac{dy}{\cos y} = \int dx \] The integral of \( \frac{1}{\cos y} \) is \( \sec y \), and the integral of \( dx \) is \( x \). Thus, we have: \[ \sec y = x + C \] Now, to solve for \( y \), take the logarithm of both sides: \[ \log|\sec y + \tan y| = x + C \] Thus, the correct answer is option (E): \[ \log|\sec y + \tan y| = x + C \] Thus, the correct answer is option (E).