Let y=y(x) be the solution of the differential equation,
`dy/dx+y tan x=2x+x^(2)tanx, x in(-pi/2,pi/2),` such that
y(0)= 1. Then

Options

A. `y'((pi)/(4 ) ) - y' (-(pi)/(4) )=pi - sqrt(2)`
B. `Y((pi ) /(4)) - Y (-(pi)/(4) ) = sqrt(2)`
C. ` Y ((pi)/(4)) + Y (-(pi)/(4))=(pi^2)/(2) +2`
D. `y' ((pi)/(4))+Y'( -(pi)/( 4))= sqrt(2)`

Answer

To solve the given differential equation \( \frac{dy}{dx} + y \tan x = 2x + x^2 \tan x \) with the initial condition \( y(0) = 1 \), we will follow these steps:

### Step 1: Identify the standard form
The given equation is in the form of a linear first-order differential equation:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
where \( P(x) = \tan x \) and \( Q(x) = 2x + x^2 \tan x \).

### Step 2: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \tan x \, dx} \]
The integral of \( \tan x \) is \( -\log(\cos x) \), so:
\[ \mu(x) = e^{-\log(\cos x)} = \frac{1}{\cos x} = \sec x \]

### Step 3: Multiply the differential equation by the integrating factor
Multiplying the entire differential equation by \( \sec x \):
\[ \sec x \frac{dy}{dx} + y \sec x \tan x = (2x + x^2 \tan x) \sec x \]
This simplifies to:
\[ \sec x \frac{dy}{dx} + y \sec x \tan x = 2x \sec x + x^2 \]

### Step 4: Rewrite the left-hand side
The left-hand side can be rewritten as:
\[ \frac{d}{dx}(y \sec x) = 2x \sec x + x^2 \]

### Step 5: Integrate both sides
Integrate both sides with respect to \( x \):
\[ \int \frac{d}{dx}(y \sec x) \, dx = \int (2x \sec x + x^2) \, dx \]
This gives:
\[ y \sec x = \int (2x \sec x) \, dx + \int x^2 \, dx + C \]

### Step 6: Solve the integrals
1. The integral \( \int x^2 \, dx = \frac{x^3}{3} \).
2. For \( \int 2x \sec x \, dx \), we can use integration by parts:
- Let \( u = 2x \) and \( dv = \sec x \, dx \).
- Then \( du = 2 \, dx \) and \( v = \log(\sec x + \tan x) \).
- Thus, \( \int 2x \sec x \, dx = 2x \log(\sec x + \tan x) - \int 2 \log(\sec x + \tan x) \, dx \).

This integral can be complex, but we can denote it as \( I \) for simplicity.

### Step 7: Combine results
Combining the results:
\[ y \sec x = 2x \log(\sec x + \tan x) - I + \frac{x^3}{3} + C \]

### Step 8: Solve for \( y \)
Now, we can express \( y \):
\[ y = \frac{2x \log(\sec x + \tan x) - I + \frac{x^3}{3} + C}{\sec x} \]

### Step 9: Apply the initial condition
Using the initial condition \( y(0) = 1 \):
\[ 1 = \frac{2(0) \log(\sec(0) + \tan(0)) - I(0) + \frac{0^3}{3} + C}{\sec(0)} \]
Since \( \sec(0) = 1 \) and \( \log(1) = 0 \), we have:
\[ 1 = C \]

### Final Solution
Thus, the solution to the differential equation is:
\[ y = \frac{2x \log(\sec x + \tan x) - I + \frac{x^3}{3} + 1}{\sec x} \]