If `y= y(x), y in [0, (pi)/(2))` is the solution of the differential equation `sec y (dy)/(dx)- sin (x+y) - sin (x-y)= 0`, with y(0)= 0, then `5y'((pi)/(2))` is equal to _____

Answer

To solve the given differential equation \( \sec y \frac{dy}{dx} - \sin(x+y) - \sin(x-y) = 0 \) with the initial condition \( y(0) = 0 \), we will follow these steps:

### Step 1: Rearranging the Differential Equation
We start by rearranging the differential equation:
\[ \sec y \frac{dy}{dx} = \sin(x+y) + \sin(x-y) \]

### Step 2: Using the Sine Addition Formula
Using the sine addition formula, we can express \( \sin(x+y) + \sin(x-y) \):
\[ \sin(x+y) + \sin(x-y) = 2 \sin x \cos y \]
Thus, we can rewrite the equation as:
\[ \sec y \frac{dy}{dx} = 2 \sin x \cos y \]

### Step 3: Multiplying by \(\cos y\)
Next, we multiply both sides by \(\cos y\):
\[ \frac{dy}{dx} = 2 \sin x \cos^2 y \]

### Step 4: Separating Variables
We can separate variables to integrate:
\[ \frac{dy}{\cos^2 y} = 2 \sin x \, dx \]

### Step 5: Integrating Both Sides
Now we integrate both sides:
\[ \int \sec^2 y \, dy = \int 2 \sin x \, dx \]
The left side integrates to:
\[ \tan y = -2 \cos x + C \]

### Step 6: Applying the Initial Condition
Using the initial condition \( y(0) = 0 \):
\[ \tan(0) = -2 \cos(0) + C \implies 0 = -2(1) + C \implies C = 2 \]
Thus, we have:
\[ \tan y = -2 \cos x + 2 \]

### Step 7: Finding \(y\) at \(x = \frac{\pi}{2}\)
Now we need to find \(y\) when \(x = \frac{\pi}{2}\):
\[ \tan y = -2 \cos\left(\frac{\pi}{2}\right) + 2 = 0 + 2 = 2 \]
Thus, we have:
\[ y = \tan^{-1}(2) \]

### Step 8: Differentiating to Find \(y'\)
Now we differentiate \( \tan y = -2 \cos x + 2 \):
\[ \sec^2 y \frac{dy}{dx} = 2 \sin x \]
Thus,
\[ \frac{dy}{dx} = \frac{2 \sin x}{\sec^2 y} = 2 \sin x \cos^2 y \]

### Step 9: Evaluating \(y'\) at \(x = \frac{\pi}{2}\)
Now we evaluate \(y'\) at \(x = \frac{\pi}{2}\):
\[ y' = 2 \sin\left(\frac{\pi}{2}\right) \cos^2 y = 2(1) \cos^2(\tan^{-1}(2)) \]
Using the identity \( \cos(\tan^{-1}(a)) = \frac{1}{\sqrt{1+a^2}} \):
\[ \cos(\tan^{-1}(2)) = \frac{1}{\sqrt{1+2^2}} = \frac{1}{\sqrt{5}} \]
Thus,
\[ \cos^2(\tan^{-1}(2)) = \frac{1}{5} \]
So,
\[ y' = 2 \cdot 1 \cdot \frac{1}{5} = \frac{2}{5} \]

### Step 10: Finding \(5y'\)
Finally, we calculate \(5y'\):
\[ 5y' = 5 \cdot \frac{2}{5} = 2 \]

Thus, the final answer is:
\[ \boxed{2} \]