If y(x) satisfies the differential equation `y'-y tan x=2x sec x and y(0)=0`, then

Options

A. `y((pi)/(4)) =(pi^(2))/(8sqrt(2))`
B. `y'((pi)/(4))=(pi^(2))/(18)`
C. `y((pi)/(3))=(pi^(2))/(9)`
D. `y'((pi)/(3))=(4pi)/(3)+(2pi^(2))/(3sqrt(3))`

Answer

To solve the differential equation \( y' - y \tan x = 2x \sec x \) with the initial condition \( y(0) = 0 \), we can follow these steps:

### Step 1: Identify the form of the differential equation
The given equation is of the form:
\[ y' + P(x)y = Q(x) \]
where \( P(x) = -\tan x \) and \( Q(x) = 2x \sec x \).

### 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:
\[ -\ln |\sec x| = \ln |\cos x| \]
Thus, the integrating factor becomes:
\[ \mu(x) = e^{\ln |\cos x|} = \cos x \]

### Step 3: Multiply the entire differential equation by the integrating factor
Multiplying the original equation by \( \cos x \):
\[ \cos x \cdot y' - y \cos x \tan x = 2x \cos x \sec x \]
This simplifies to:
\[ \cos x \cdot y' - y \sin x = 2x \]

### Step 4: Rewrite the left side as a derivative
The left-hand side can be expressed as the derivative of a product:
\[ \frac{d}{dx} (y \cos x) = 2x \]

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

### Step 6: Solve for \( y \)
Rearranging the equation gives:
\[ y = \frac{x^2 + C}{\cos x} \]

### Step 7: Apply the initial condition
Using the initial condition \( y(0) = 0 \):
\[ 0 = \frac{0^2 + C}{\cos(0)} \implies 0 = C \]
Thus, \( C = 0 \).

### Step 8: Write the final solution
The final solution for \( y \) is:
\[ y = \frac{x^2}{\cos x} \]

### Summary
The solution to the differential equation \( y' - y \tan x = 2x \sec x \) with the initial condition \( y(0) = 0 \) is:
\[ y = \frac{x^2}{\cos x} \]