Solve the differential equation :
`(dy)/(dx) -3ycotx = sin 2x`, given y=2, when `x=pi/2`

Answer

To solve the differential equation

\[ \frac{dy}{dx} - 3y \cot x = \sin 2x \]

with the initial condition \( y = 2 \) when \( x = \frac{\pi}{2} \), we will use the integrating factor method.

### Step 1: Identify \( p(x) \) and \( q(x) \)

The given equation can be rewritten in the standard form:

\[ \frac{dy}{dx} + p(x)y = q(x) \]

where \( p(x) = -3 \cot x \) and \( q(x) = \sin 2x \).

**Hint:** Identify the functions \( p(x) \) and \( q(x) \) from the standard form of the differential equation.

### Step 2: Calculate the integrating factor

The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int p(x) \, dx} = e^{\int -3 \cot x \, dx} \]

The integral of \( \cot x \) is \( \ln |\sin x| \), so:

\[ \int -3 \cot x \, dx = -3 \ln |\sin x| = \ln |\sin x|^{-3} \]

Thus, the integrating factor becomes:

\[ \mu(x) = e^{\ln |\sin x|^{-3}} = \frac{1}{\sin^3 x} \]

**Hint:** Remember that the integrating factor is derived from the integral of \( p(x) \).

### Step 3: Multiply the entire equation by the integrating factor

Now, multiply the entire differential equation by \( \mu(x) \):

\[ \frac{1}{\sin^3 x} \frac{dy}{dx} - 3y \frac{\cot x}{\sin^3 x} = \frac{\sin 2x}{\sin^3 x} \]

This simplifies to:

\[ \frac{d}{dx} \left( y \cdot \frac{1}{\sin^3 x} \right) = \frac{\sin 2x}{\sin^3 x} \]

**Hint:** Multiplying by the integrating factor allows us to express the left side as the derivative of a product.

### Step 4: Integrate both sides

Next, we integrate both sides:

\[ \int \frac{d}{dx} \left( y \cdot \frac{1}{\sin^3 x} \right) \, dx = \int \frac{\sin 2x}{\sin^3 x} \, dx \]

The left side simplifies to:

\[ y \cdot \frac{1}{\sin^3 x} = \int \frac{\sin 2x}{\sin^3 x} \, dx + C \]

**Hint:** Remember to integrate the right side carefully, possibly using trigonometric identities.

### Step 5: Solve the integral on the right side

The integral \( \int \frac{\sin 2x}{\sin^3 x} \, dx \) can be simplified using the identity \( \sin 2x = 2 \sin x \cos x \):

\[ \int \frac{2 \sin x \cos x}{\sin^3 x} \, dx = 2 \int \frac{\cos x}{\sin^2 x} \, dx = -2 \cot x + C \]

Thus, we have:

\[ y \cdot \frac{1}{\sin^3 x} = -2 \cot x + C \]

**Hint:** Use trigonometric identities to simplify the integral.

### Step 6: Solve for \( y \)

Now, multiply through by \( \sin^3 x \):

\[ y = -2 \cot x \sin^3 x + C \sin^3 x \]

**Hint:** Isolate \( y \) to express it in terms of \( x \) and \( C \).

### Step 7: Apply the initial condition

We are given \( y = 2 \) when \( x = \frac{\pi}{2} \):

\[ 2 = -2 \cot \left( \frac{\pi}{2} \right) \sin^3 \left( \frac{\pi}{2} \right) + C \sin^3 \left( \frac{\pi}{2} \right) \]

Since \( \cot \left( \frac{\pi}{2} \right) = 0 \) and \( \sin \left( \frac{\pi}{2} \right) = 1 \):

\[ 2 = C \]

**Hint:** Substitute the values of \( x \) and \( y \) to find \( C \).

### Step 8: Final solution

Substituting \( C = 2 \) back into the equation for \( y \):

\[ y = -2 \cot x \sin^3 x + 2 \sin^3 x \]

This can be simplified to:

\[ y = 2 \sin^3 x (1 - \cot x) \]

Thus, the final solution to the differential equation is:

\[ y = 2 \sin^3 x - 2 \sin^2 x \cos x \]

**Hint:** Ensure all terms are correctly simplified and check for any possible factorizations.