`xdx+secxdy=0, y=0" if " x=(pi)/(2)` A)`x+y cos y+cos y=(pi)/(2)` B)`y+x sin x-cosx=(pi)/(2)` C)`y+x sin x+cos x=(pi)/(2)` D)`x+y sin y+cos y=(pi)/(2)`

Options

A. `x+y cos y+cos y=(pi)/(2)`
B. `y+x sin x-cosx=(pi)/(2)`
C. `y+x sin x+cos x=(pi)/(2)`
D. `x+y sin y+cos y=(pi)/(2)`

Answer

To solve the differential equation \( xdx + \sec(x) dy = 0 \) with the initial conditions \( y = 0 \) and \( x = \frac{\pi}{2} \), we will follow these steps:

### Step 1: Rearrange the Differential Equation
We start with the given equation:
\[ xdx + \sec(x) dy = 0 \]
Rearranging gives us:
\[ \sec(x) dy = -xdx \]
Dividing both sides by \(\sec(x)\):
\[ dy = -x \cos(x) dx \]

### Step 2: Integrate Both Sides
Now we will integrate both sides:
\[ \int dy = \int -x \cos(x) dx \]
The left side integrates to:
\[ y \]
For the right side, we will use integration by parts where:
- Let \( u = x \) and \( dv = \cos(x) dx \)
- Then, \( du = dx \) and \( v = \sin(x) \)

Using integration by parts:
\[ \int x \cos(x) dx = x \sin(x) - \int \sin(x) dx \]
The integral of \(\sin(x)\) is \(-\cos(x)\), so:
\[ \int x \cos(x) dx = x \sin(x) + \cos(x) \]
Thus, we have:
\[ y = - (x \sin(x) + \cos(x)) + C \]

### Step 3: Solve for the Constant \(C\)
Now we apply the initial conditions \( y = 0 \) when \( x = \frac{\pi}{2} \):
\[ 0 = -\left(\frac{\pi}{2} \sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right)\right) + C \]
Calculating the values:
- \(\sin\left(\frac{\pi}{2}\right) = 1\)
- \(\cos\left(\frac{\pi}{2}\right) = 0\)

Substituting these values:
\[ 0 = -\left(\frac{\pi}{2} \cdot 1 + 0\right) + C \]
This simplifies to:
\[ C = \frac{\pi}{2} \]

### Step 4: Write the Final Solution
Substituting \(C\) back into the equation for \(y\):
\[ y = - (x \sin(x) + \cos(x)) + \frac{\pi}{2} \]
Rearranging gives:
\[ y + x \sin(x) + \cos(x) = \frac{\pi}{2} \]

### Conclusion
The final form of the solution is:
\[ y + x \sin(x) + \cos(x) = \frac{\pi}{2} \]