Question

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

• A. \(\cosec(x + y) + \tan(x + y) = x + C\)
• B. \(x + \cosec(x + y) = C\)
• C. \(x + \tan(x + y) = C\)
• D. \(x + \sec(x + y) = C\)

Hint:

The substitution \(v = x + y\) is useful when the right-hand side is a function of \(x+y\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Substitute \(v = x + y\) to reduce to separable form.

Step 2: Detailed Explanation:
Let \(v = x + y\). Then \(\frac{dv}{dx} = 1 + \frac{dy}{dx}\). So \(\frac{dy}{dx} = \frac{dv}{dx} - 1\). Given \(\frac{dy}{dx} = \sin v \tan v - 1\). Thus \(\frac{dv}{dx} - 1 = \sin v \tan v - 1 \Rightarrow \frac{dv}{dx} = \sin v \tan v = \frac{\sin^2 v}{\cos v}\). Separate: \(\frac{\cos v}{\sin^2 v} dv = dx\). Integrate: \(\int \cot v \cosec v \, dv = x + C\). \(\int \cot v \cosec v \, dv = -\cosec v\). So \(-\cosec v = x + C\), or \(x + \cosec v = C'\). Thus \(x + \cosec(x+y) = C\).

Step 3: Final Answer:
\(x + \cosec(x + y) = C\), which corresponds to option (B).