Question

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

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

Hint:

When dealing with trigonometric functions in differential equations, remember to apply sum-to-product identities for simplification, which can make the equation easier to solve.

Answer 1

The Correct Option is C

Given the differential equation:

\(\frac{dy}{dx} \tan y = \sin(x + y) + \sin(x - y)\)

First, simplify the right side using the sum-to-product identities:

\(\sin(x + y) + \sin(x - y) = 2 \sin x \cos y\)

Thus, the equation becomes:

\(\frac{dy}{dx} \tan y = 2 \sin x \cos y\)

Rewrite \(\tan y\) as \(\frac{\sin y}{\cos y}\):

\(\frac{dy}{dx} \frac{\sin y}{\cos y} = 2 \sin x \cos y\)

Multiply both sides by \(\cos y\):

\(\frac{dy}{dx} \sin y = 2 \sin x \cos^2 y\)

Separate the variables:

\(\frac{\sin y}{\cos^2 y} dy = 2 \sin x dx\)

This can be rewritten as:

\(\sin y \sec^2 y dy = 2 \sin x dx\)

Integrate both sides:

\(\int \sin y \sec^2 y \, dy = \int 2 \sin x \, dx\)

Substitute \(\sin y \sec^2 y\) as \(\tan y \sec y\) and integrate:

\(\int \tan y \sec y \, dy = \sec y + C_1\)

On the right side:

\(\int 2 \sin x \, dx = -2 \cos x + C_2\)

Equate both expressions:

\(\sec y + C_1 = -2 \cos x + C_2\)

Simplify to find the general solution:

\(\sec y = -2 \cos x + C\)

where \(C=C_2-C_1\).

Hence, the correct answer is:

\(\sec y = -2 \cos x + C\)

Answer 2

To solve the differential equation \( \frac{dy}{dx} \tan y = \sin(x + y) + \sin(x - y) \), we follow these steps:

1. Recall the trigonometric identity: \(\sin A + \sin B = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\). Applying it to \(\sin(x+y) + \sin(x-y)\), we get:
   \(\sin(x+y) + \sin(x-y) = 2 \sin(x) \cos(y)\).
2. Substitute this back into the differential equation:\
   \(\frac{dy}{dx} \tan y = 2 \sin(x) \cos(y)\).
3. Rewrite \(\tan y\) as \(\frac{\sin y}{\cos y}\), resulting in:
   \(\frac{dy}{dx} \frac{\sin y}{\cos y} = 2 \sin(x) \cos(y)\).
4. Rearrange the terms to separate variables:
   \(\frac{\sin y}{\cos^2 y} dy = 2 \sin(x) dx\).
5. Integrate both sides. Integrate \(\frac{\sin y}{\cos^2 y} dy\) as \(\int \frac{\sin y}{\cos^2 y} dy = -\int \sec y \cdot d(\cos y) = -\sec y\).
   Integrate \(2 \sin(x) dx\) as \(-2 \cos(x)\):
   \(-\sec y = -2 \cos(x) + C\).
6. Rewriting gives the solution:
   \(\sec y = -2 \cos(x) + C\).

This matches the option: \(\sec y = -2 \cos x + C\).