The solution of dy = cos x `(2 - y " cosec " x) dx` , where ` y = sqrt(2) , " when " x = pi//4 ` is

Options

A. `y=sin x + (1)/(2)cosec x`
B. `y = tan (x//2)+cot (x//2)`
C. `y=(1//sqrt(2))sec(x//2)+sqrt(2)cos (x//2)`
D. None of the above

Answer

To solve the differential equation \( dy = \cos x (2 - y \csc x) dx \) with the initial condition \( y = \sqrt{2} \) when \( x = \frac{\pi}{4} \), we can follow these steps:

### Step 1: Rewrite the equation
Starting with the given equation:
\[ dy = \cos x (2 - y \csc x) dx \]
We can rearrange it to:
\[ \frac{dy}{dx} = 2 \cos x - y \cos x \csc x \]
Since \( \csc x = \frac{1}{\sin x} \), we can rewrite the equation as:
\[ \frac{dy}{dx} = 2 \cos x - \frac{y \cos x}{\sin x} = 2 \cos x - y \cot x \]

### Step 2: Identify the form of the equation
This can be rewritten in the standard linear form:
\[ \frac{dy}{dx} + y \cot x = 2 \cos x \]
where \( P = \cot x \) and \( Q = 2 \cos x \).

### Step 3: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int P \, dx} = e^{\int \cot x \, dx} \]
The integral of \( \cot x \) is \( \ln |\sin x| \), so:
\[ \mu(x) = e^{\ln |\sin x|} = |\sin x| \]
Since \( \sin x \) is positive in the interval we are considering, we can take:
\[ \mu(x) = \sin x \]

### Step 4: Multiply through by the integrating factor
Multiplying the entire differential equation by \( \sin x \):
\[ \sin x \frac{dy}{dx} + y \sin x \cot x = 2 \sin x \cos x \]
This simplifies to:
\[ \sin x \frac{dy}{dx} + y \cos x = 2 \sin x \cos x \]

### Step 5: Recognize the left side as a derivative
The left-hand side can be recognized as the derivative of a product:
\[ \frac{d}{dx}(y \sin x) = 2 \sin x \cos x \]

### Step 6: Integrate both sides
Integrating both sides:
\[ \int \frac{d}{dx}(y \sin x) \, dx = \int 2 \sin x \cos x \, dx \]
The right-hand side can be simplified using the identity \( \sin 2x = 2 \sin x \cos x \):
\[ y \sin x = \int \sin 2x \, dx = -\frac{1}{2} \cos 2x + C \]

### Step 7: Solve for \( y \)
Thus, we have:
\[ y \sin x = -\frac{1}{2} \cos 2x + C \]
Solving for \( y \):
\[ y = \frac{-\frac{1}{2} \cos 2x + C}{\sin x} \]

### Step 8: Apply the initial condition
Using the initial condition \( y = \sqrt{2} \) when \( x = \frac{\pi}{4} \):
\[ \sqrt{2} \sin\left(\frac{\pi}{4}\right) = -\frac{1}{2} \cos\left(\frac{\pi}{2}\right) + C \]
Since \( \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \) and \( \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 0 + C \implies C = 1 \]

### Step 9: Substitute \( C \) back into the equation
Substituting \( C \) back into the equation:
\[ y = \frac{-\frac{1}{2} \cos 2x + 1}{\sin x} \]

### Final Answer
Thus, the solution to the differential equation is:
\[ y = \csc x - \frac{\cos 2x}{2 \sin x} \]