The solution of the differential equation `sin 2y (dy)/(dx) +2 tan x cos ^(2) y=2 sec x cos ^(3) y` is: (where C is arbitary constant)

Options

A. `cos y sec x= tan x+C`
B. ` sec y cos x = tan x +C`
C. `sec y sec x = tan x +C`
D. `tan y sec x = sec x +C`

Answer

To solve the differential equation given by:

\[ \sin 2y \frac{dy}{dx} + 2 \tan x \cos^2 y = 2 \sec x \cos^3 y \]

we will follow these steps:

### Step 1: Rearranging the Equation
First, we divide both sides by \(\cos^3 y\):

\[ \frac{\sin 2y}{\cos^3 y} \frac{dy}{dx} + \frac{2 \tan x \cos^2 y}{\cos^3 y} = 2 \sec x \]

This simplifies to:

\[ \frac{\sin 2y}{\cos^3 y} \frac{dy}{dx} + 2 \tan x \sec y = 2 \sec x \]

### Step 2: Using Trigonometric Identity
Using the identity \(\sin 2y = 2 \sin y \cos y\), we rewrite the equation:

\[ \frac{2 \sin y \cos y}{\cos^3 y} \frac{dy}{dx} + 2 \tan x \sec y = 2 \sec x \]

This simplifies to:

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

### Step 3: Simplifying the Equation
We can cancel the factor of 2 from all terms:

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

### Step 4: Substituting \( \sec y \)
Let \( \sec y = t \). Then, differentiating both sides with respect to \( x \):

\[ \sec y \tan y \frac{dy}{dx} = \frac{dt}{dx} \]

Substituting this into our equation gives:

\[ t \tan y \frac{dt}{dx} + \tan x t = \sec x \]

### Step 5: Linear Differential Equation Form
This can be rearranged to:

\[ \frac{dt}{dx} + \tan x t = \sec x \]

This is a linear first-order differential equation in \( t \).

### Step 6: Finding the Integrating Factor
The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int \tan x \, dx} = e^{-\log(\cos x)} = \sec x \]

### Step 7: Applying the Integrating Factor
Multiplying the entire equation by the integrating factor:

\[ \sec x \frac{dt}{dx} + \sec x \tan x t = \sec^2 x \]

### Step 8: Integrating Both Sides
Integrating both sides with respect to \( x \):

\[ \int \sec x \frac{dt}{dx} \, dx + \int \sec x \tan x t \, dx = \int \sec^2 x \, dx \]

This simplifies to:

\[ t \sec x = \tan x + C \]

### Step 9: Substituting Back for \( t \)
Recall \( t = \sec y \):

\[ \sec y \sec x = \tan x + C \]

### Final Solution
Thus, the solution to the differential equation is:

\[ \sec y \sec x = \tan x + C \]