`((dy)/(dx))^(2) + 2y cot x (dy)/(dx) = y^(2)` has the solution

Options

A. `y + (c)/(1+cos x) = 0`
B. `y = (c)/(1-cos x)`
C. `x = 2 sin^(-1) sqrt((c)/(2y))`
D. `x = 2 cos^(-1) ((c)/(2y))`

Explanation

To solve the differential equation

\[ \left(\frac{dy}{dx}\right)^2 + 2y \cot x \frac{dy}{dx} = y^2, \]

we will follow these steps:

### Step 1: Substitute \( \frac{dy}{dx} \) with \( t \)

Let

\[ \frac{dy}{dx} = t. \]

Then the equation becomes:

\[ t^2 + 2y \cot x \cdot t = y^2. \]

### Step 2: Rearrange the equation

Rearranging gives us a quadratic equation in \( t \):

\[ t^2 + 2y \cot x \cdot t - y^2 = 0. \]

### Step 3: Apply the quadratic formula

Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 1 \), \( b = 2y \cot x \), and \( c = -y^2 \). Thus,

\[ t = \frac{-2y \cot x \pm \sqrt{(2y \cot x)^2 - 4 \cdot 1 \cdot (-y^2)}}{2 \cdot 1}. \]

### Step 4: Simplify the expression

Calculating the discriminant:

\[ (2y \cot x)^2 + 4y^2 = 4y^2 (\cot^2 x + 1) = 4y^2 \csc^2 x. \]

So,

\[ t = \frac{-2y \cot x \pm 2y \csc x}{2} = -y \cot x \pm y \csc x. \]

### Step 5: Substitute back for \( \frac{dy}{dx} \)

Now substituting back for \( t \):

\[ \frac{dy}{dx} = -y \cot x \pm y \csc x. \]

### Step 6: Separate variables

We can separate variables:

\[ \frac{dy}{y} = (-\cot x \pm \csc x) dx. \]

### Step 7: Integrate both sides

Integrating both sides:

\[ \int \frac{dy}{y} = \int (-\cot x \pm \csc x) dx. \]

The left side integrates to:

\[ \ln |y| = \int (-\cot x \pm \csc x) dx. \]

### Step 8: Solve the integral on the right

The integral of \( -\cot x \) is \( -\ln |\sin x| \) and the integral of \( \csc x \) is \( -\ln |\csc x + \cot x| \). Therefore, we have:

\[ \ln |y| = -\ln |\sin x| \pm (-\ln |\csc x + \cot x|) + C. \]

### Step 9: Combine logarithmic terms

This can be simplified to:

\[ \ln |y| = \ln \left(\frac{1}{\sin x}\right) + \ln |\csc x + \cot x| + C. \]

### Step 10: Exponentiate to solve for \( y \)

Exponentiating both sides gives:

\[ y = e^C \cdot \frac{\csc x + \cot x}{\sin x}. \]

Let \( e^C = k \) (a constant), we have:

\[ y = k \cdot \frac{\csc x + \cot x}{\sin x}. \]

### Final Result

Thus, the solution to the differential equation is:

\[ y = k \cdot \frac{1 + \cos x}{\sin x}. \]