Solve the following differential equations :
`tan x (dy)/(dx)+2y= cos x`.

Answer

To solve the differential equation \( \tan x \frac{dy}{dx} + 2y = \cos x \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \tan x \frac{dy}{dx} + 2y = \cos x \]
We can rewrite this equation by dividing through by \( \tan x \):
\[ \frac{dy}{dx} + \frac{2y}{\tan x} = \frac{\cos x}{\tan x} \]
Since \( \tan x = \frac{\sin x}{\cos x} \), we can express \( \frac{\cos x}{\tan x} \) as:
\[ \frac{\cos x}{\tan x} = \frac{\cos^2 x}{\sin x} \]
Thus, the equation becomes:
\[ \frac{dy}{dx} + \frac{2y \cos x}{\sin x} = \frac{\cos^2 x}{\sin x} \]

### Step 2: Identify \( p(x) \) and \( q(x) \)
From the standard form of a linear differential equation \( \frac{dy}{dx} + p(x)y = q(x) \), we identify:
\[ p(x) = \frac{2 \cos x}{\sin x} = 2 \cot x \]
\[ q(x) = \frac{\cos^2 x}{\sin x} \]

### Step 3: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int p(x) \, dx} = e^{\int 2 \cot x \, dx} \]
The integral of \( 2 \cot x \) is:
\[ \int 2 \cot x \, dx = 2 \ln |\sin x| = \ln |\sin^2 x| \]
Thus, the integrating factor is:
\[ \mu(x) = e^{\ln |\sin^2 x|} = \sin^2 x \]

### Step 4: Multiply through by the integrating factor
Now we multiply the entire differential equation by the integrating factor:
\[ \sin^2 x \frac{dy}{dx} + 2y \sin x \cos x = \cos^2 x \]

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

### Step 6: Integrate both sides
Integrating both sides gives:
\[ y \sin^2 x = \int \cos^2 x \, dx \]
Using the identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \), we have:
\[ \int \cos^2 x \, dx = \frac{1}{2} \int (1 + \cos 2x) \, dx = \frac{1}{2}(x + \frac{1}{2} \sin 2x) + C \]
Thus,
\[ y \sin^2 x = \frac{x}{2} + \frac{1}{4} \sin 2x + C \]

### Step 7: Solve for \( y \)
Finally, we solve for \( y \):
\[ y = \frac{\frac{x}{2} + \frac{1}{4} \sin 2x + C}{\sin^2 x} \]

### Final Solution
The solution to the differential equation is:
\[ y = \frac{x/2 + \frac{1}{4} \sin 2x + C}{\sin^2 x} \]