Solve the following differential equations
`sin x(dy)/(dx)+y cos x= 2 sin^2 x cos x " if "y(pi/2)=1`.

Answer

To solve the differential equation given by

\[ \sin x \frac{dy}{dx} + y \cos x = 2 \sin^2 x \cos x \]

with the initial condition \( y\left(\frac{\pi}{2}\right) = 1 \), we will follow these steps:

### Step 1: Rewrite the Equation
First, we divide the entire equation by \(\sin x\) to simplify it:

\[ \frac{dy}{dx} + y \frac{\cos x}{\sin x} = 2 \frac{\sin^2 x \cos x}{\sin x} \]

This simplifies to:

\[ \frac{dy}{dx} + y \cot x = 2 \sin x \cos x \]

### Step 2: Identify \(p\) and \(q\)
In the standard form of a linear differential equation:

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

we can identify:

- \(p = \cot x\)
- \(q = 2 \sin x \cos x\)

### Step 3: Find the Integrating Factor
The integrating factor \(IF\) is given by:

\[ IF = e^{\int p \, dx} = e^{\int \cot x \, dx} \]

The integral of \(\cot x\) is:

\[ \int \cot x \, dx = \ln |\sin x| \implies IF = e^{\ln |\sin x|} = |\sin x| \]

Since \(\sin x\) is positive in the interval we are considering, we can simply use:

\[ IF = \sin x \]

### Step 4: Multiply the Equation by the Integrating Factor
Now, we multiply the entire differential equation by the integrating factor:

\[ \sin x \frac{dy}{dx} + y \sin x \cot x = 2 \sin^2 x \cos x \]

This simplifies to:

\[ \sin x \frac{dy}{dx} + y \cos x = 2 \sin^2 x \cos x \]

### Step 5: Write the Left Side as a Derivative
The left side can be expressed as the derivative of a product:

\[ \frac{d}{dx}(y \sin x) = 2 \sin^2 x \cos x \]

### Step 6: Integrate Both Sides
Now we integrate both sides:

\[ \int \frac{d}{dx}(y \sin x) \, dx = \int 2 \sin^2 x \cos x \, dx \]

The left side simplifies to:

\[ y \sin x \]

For the right side, we can use the substitution \(u = \sin x\), \(du = \cos x \, dx\):

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

### Step 7: Combine the Results
Thus, we have:

\[ y \sin x = \frac{2 \sin^3 x}{3} + C \]

### Step 8: Solve for \(y\)
Now, we can solve for \(y\):

\[ y = \frac{2 \sin^3 x}{3 \sin x} + \frac{C}{\sin x} = \frac{2 \sin^2 x}{3} + \frac{C}{\sin x} \]

### Step 9: Apply the Initial Condition
Now we apply the initial condition \(y\left(\frac{\pi}{2}\right) = 1\):

\[ 1 = \frac{2 \sin^2\left(\frac{\pi}{2}\right)}{3} + \frac{C}{\sin\left(\frac{\pi}{2}\right)} \]

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

\[ 1 = \frac{2 \cdot 1^2}{3} + C \implies 1 = \frac{2}{3} + C \implies C = 1 - \frac{2}{3} = \frac{1}{3} \]

### Step 10: Write the Particular Solution
Substituting \(C\) back into the equation for \(y\):

\[ y = \frac{2 \sin^2 x}{3} + \frac{1/3}{\sin x} \]

This can be simplified to:

\[ y = \frac{2 \sin^2 x + 1}{3 \sin x} \]

### Final Solution
Thus, the particular solution of the differential equation is:

\[ y = \frac{2 \sin^2 x + 1}{3 \sin x} \]