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 and find \( y\left(\frac{\pi}{2}\right) \), we will follow these steps:

### Step 1: Rewrite the Differential Equation
The given differential equation is:
\[ \cos^2 y \frac{dy}{dx} = \sin(x+y) + \sin(x-y) \]
Using the sine addition and subtraction formulas, we can rewrite the right-hand side:
\[ \sin(x+y) + \sin(x-y) = 2 \sin x \cos y \]
Thus, the equation becomes:
\[ \cos^2 y \frac{dy}{dx} = 2 \sin x \cos y \]

### Step 2: Separate Variables
We can separate the variables \( y \) and \( x \):
\[ \frac{dy}{\cos y} = \frac{2 \sin x}{\cos^2 y} dx \]
This simplifies to:
\[ \frac{dy}{\cos y} = 2 \sin x \, dx \]

### 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:
\[ \ln |\sec y + \tan y| + C_1 \]
The right side integrates to:
\[ -2 \cos x + C_2 \]
Thus, we have:
\[ \ln |\sec y + \tan y| = -2 \cos x + C \]

### Step 4: Exponentiate Both Sides
Exponentiating both sides gives:
\[ |\sec y + \tan y| = e^{-2 \cos x + C} = Ae^{-2 \cos x} \]
where \( A = e^C \).

### Step 5: Use Initial Condition
We are given the initial condition \( y\left(\frac{\pi}{3}\right) = -\frac{\pi}{2} \). We can find \( A \) using this condition:
\[ \sec\left(-\frac{\pi}{2}\right) + \tan\left(-\frac{\pi}{2}\right) \text{ is undefined.} \]
However, we can evaluate the limit as \( y \) approaches \(-\frac{\pi}{2}\):
\[ \sec y \to 0 \quad \text{and} \quad \tan y \to -\infty \]
Thus, we can assume \( A \) is a constant that we will determine later.

### Step 6: Find \( y\left(\frac{\pi}{2}\right) \)
Now we need to find \( y\left(\frac{\pi}{2}\right) \):
Using the equation:
\[ |\sec y + \tan y| = Ae^{-2\cos\left(\frac{\pi}{2}\right)} = A \]
As \( \cos\left(\frac{\pi}{2}\right) = 0 \), we have:
\[ |\sec y + \tan y| = A \]
Since \( y \) approaches \( 0 \) as \( x \) approaches \( \frac{\pi}{2} \), we find:
\[ \sec(0) + \tan(0) = 1 + 0 = 1 \]
Thus, \( A = 1 \).

### Conclusion
Therefore, we conclude:
\[ y\left(\frac{\pi}{2}\right) = 0 \]