`((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 \( t = \frac{dy}{dx} \)

Rewriting the equation in terms of \( t \):

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

### Step 2: Rearranging the equation

This is a quadratic equation in \( t \). We can rearrange it as:

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

### 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,

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

### Step 4: Simplifying the expression

Calculating the discriminant:

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

Since \( \cot^2 x + 1 = \csc^2 x \):

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

### Step 5: Expressing \( \frac{dy}{dx} \)

Thus, we have:

\[ \frac{dy}{dx} = -y \cot x + y \quad \text{or} \quad \frac{dy}{dx} = -y \cot x - y. \]

### Step 6: Separating variables

For the first case:

\[ \frac{dy}{dx} = y(1 - \cot x). \]

This can be rewritten as:

\[ \frac{dy}{y} = (1 - \cot x) dx. \]

### Step 7: Integrating both sides

Integrating both sides:

\[ \int \frac{dy}{y} = \int (1 - \cot x) dx. \]

The left side gives:

\[ \ln |y| = \int (1 - \cot x) dx. \]

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

The integral of \( 1 \) is \( x \) and the integral of \( -\cot x \) is \( -\ln |\sin x| \):

\[ \ln |y| = x - \ln |\sin x| + C. \]

### Step 9: Exponentiating both sides

Exponentiating both sides gives:

\[ y = e^{C} \cdot \frac{e^x}{|\sin x|}. \]

Let \( e^{C} = C' \) (a constant), we have:

\[ y = \frac{C' e^x}{|\sin x|}. \]

### Step 10: General solution

Thus, the general solution of the differential equation is:

\[ y = \frac{C e^x}{\sin x}, \]

where \( C \) is a constant.

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