Suppose `cos^(2)y.(dy)/(dx)=sin(x+y)+sin(x-y), |x| le (pi)/(2) and |y|le(pi)/(2).` If `y((pi)/(3))=-(pi)/(2),` then `y((pi)/(2))` is

Options

A. 0
B. `(pi)/(4)`
C. `-(pi)/(4)`
D. `(pi)/(2)`

Answer

To solve the given differential equation step by step, we start with the equation:

\[ \cos^2 y \frac{dy}{dx} = \sin(x+y) + \sin(x-y) \]

### Step 1: Simplify the Right Side
Using the sine addition and subtraction formulas, we can rewrite the right side:

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

Thus, our equation becomes:

\[ \cos^2 y \frac{dy}{dx} = 2 \sin x \cos y \]

### Step 2: Separate Variables
We can separate the variables by dividing both sides by \(\cos^2 y\) and multiplying by \(dx\):

\[ \frac{dy}{\cos y} = 2 \sin x \, dx \]

### Step 3: Integrate Both Sides
Now we integrate both sides:

\[ \int \frac{dy}{\cos y} = \int 2 \sin x \, dx \]

The left side integrates to:

\[ \int \sec y \, dy = \ln |\sec y + \tan y| + C_1 \]

The right side integrates to:

\[ \int 2 \sin x \, dx = -2 \cos x + C_2 \]

Thus, we have:

\[ \ln |\sec y + \tan y| = -2 \cos x + C \]

where \(C = C_2 - C_1\).

### Step 4: Exponentiate to Remove the Logarithm
Exponentiating both sides gives:

\[ |\sec y + \tan y| = e^{-2 \cos x + C} = K e^{-2 \cos x} \]

where \(K = e^C\).

### Step 5: Solve for y
We can express this as:

\[ \sec y + \tan y = K e^{-2 \cos x} \]

### Step 6: Use Initial Condition
We are given the initial condition \(y\left(\frac{\pi}{3}\right) = -\frac{\pi}{2}\). We need to find \(K\):

At \(x = \frac{\pi}{3}\):

\[ \sec\left(-\frac{\pi}{2}\right) + \tan\left(-\frac{\pi}{2}\right) \text{ is undefined.} \]

However, we can evaluate \(K\) using the limit as \(y\) approaches \(-\frac{\pi}{2}\):

As \(y \to -\frac{\pi}{2}\):

\[ \sec y \to 0 \quad \text{and} \quad \tan y \to -\infty \]

This means:

\[ K e^{-2 \cos\left(\frac{\pi}{3}\right)} = K e^{-2 \cdot \frac{1}{2}} = K e^{-1} \]

### Step 7: Find \(y\left(\frac{\pi}{2}\right)\)
Now we need to find \(y\left(\frac{\pi}{2}\right)\):

At \(x = \frac{\pi}{2}\):

\[ \sec y + \tan y = K e^{-2 \cdot 0} = K \]

As \(y\) approaches \(0\):

\[ \sec(0) + \tan(0) = 1 + 0 = 1 \]

Thus, we find:

\[ K = 1 \]

So we have:

\[ \sec y + \tan y = e^{-2 \cos x} \]

### Final Step: Conclusion
Finally, substituting \(x = \frac{\pi}{2}\):

\[ \sec y + \tan y = 1 \]

This implies:

\[ y = 0 \]

Thus, the value of \(y\left(\frac{\pi}{2}\right)\) is:

\[ \boxed{0} \]