`secy(dy/dx)=sin(x+y)+sin(x-y)` if `y(0)=0` then find `5y'(pi/2)`

Answer

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

### Step 1: Rewrite the equation using trigonometric identities
We start by using the identity for the sum of sines:
\[ \sin(a) + \sin(b) = 2 \sin\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right) \]
Let \( a = x+y \) and \( b = x-y \). Then:
\[ \sin(x+y) + \sin(x-y) = 2 \sin\left(x\right) \cos\left(y\right) \]
Thus, we can rewrite the equation as:
\[ \sec(y) \frac{dy}{dx} = 2 \sin(x) \cos(y) \]

### Step 2: Rearranging the equation
We can rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ \sec(y) \frac{dy}{dx} = 2 \sin(x) \cos(y) \implies \frac{dy}{dx} = 2 \sin(x) \cos(y) \cos(y) = 2 \sin(x) \cos^2(y) \]

### Step 3: Separate variables
Now we can separate variables:
\[ \frac{dy}{\cos^2(y)} = 2 \sin(x) dx \]

### Step 4: Integrate both sides
Integrate both sides:
\[ \int \sec^2(y) dy = \int 2 \sin(x) dx \]
The integral of \( \sec^2(y) \) is \( \tan(y) \), and the integral of \( 2 \sin(x) \) is \( -2 \cos(x) + C \):
\[ \tan(y) = -2 \cos(x) + C \]

### Step 5: Apply the initial condition
Using the initial condition \( y(0) = 0 \):
\[ \tan(0) = -2 \cos(0) + C \implies 0 = -2(1) + C \implies C = 2 \]
So, we have:
\[ \tan(y) = -2 \cos(x) + 2 \]

### Step 6: Differentiate to find \( y' \)
Now we differentiate \( \tan(y) \):
\[ \sec^2(y) \frac{dy}{dx} = 2 \sin(x) \]
Thus,
\[ \frac{dy}{dx} = \frac{2 \sin(x)}{\sec^2(y)} = 2 \sin(x) \cos^2(y) \]

### Step 7: Evaluate \( y' \) at \( x = \frac{\pi}{2} \)
Now we need to find \( y' \) at \( x = \frac{\pi}{2} \):
\[ \tan(y) = -2 \cos\left(\frac{\pi}{2}\right) + 2 = 0 + 2 = 2 \]
So,
\[ y = \tan^{-1}(2) \]
Now, substituting \( x = \frac{\pi}{2} \) into \( y' \):
\[ y' = \frac{2 \sin\left(\frac{\pi}{2}\right)}{\sec^2(y)} = \frac{2 \cdot 1}{\sec^2(\tan^{-1}(2))} \]
Using \( \sec^2(\tan^{-1}(2)) = 1 + \tan^2(\tan^{-1}(2)) = 1 + 4 = 5 \):
\[ y' = \frac{2}{5} \]

### Step 8: Find \( 5y' \)
Finally, we compute \( 5y' \):
\[ 5y' = 5 \cdot \frac{2}{5} = 2 \]

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