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

Options

A. `y((pi)/(4))=(pi^(2))/(8sqrt2),y'((pi)/(3))=(4pi)/(3)+(2pi^(2))/(3sqrt3)`
B. `y((pi)/(4))=(pi^(2))/(4sqrt2),y'((pi)/(4))=(pi^(2))/(18)`
C. `y((pi)/(3))=(pi^(2))/(9),y'((pi)/(3))=(4pi)/(3)+(pi^(2))/(3sqrt3)`
D. `y((pi)/(3))=(pi^(2))/(4sqrt2),y'((pi)/(3))=(pi^(2))/(18)`

Explanation

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

### Step 1: Rewrite the Differential Equation
We start with the given equation:
\[ y' - y \tan x = 2x \sec x \]
This can be rewritten as:
\[ \frac{dy}{dx} - y \tan x = 2x \sec x \]

### Step 2: Identify \( p(x) \) and \( q(x) \)
In the standard form of a linear differential equation \( \frac{dy}{dx} + p(x)y = q(x) \), we identify:
- \( p(x) = -\tan x \)
- \( q(x) = 2x \sec x \)

### Step 3: Compute 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:
\[ -\ln |\cos x| = \ln |\sec x| \]
Thus, the integrating factor is:
\[ \mu(x) = e^{\ln |\sec x|} = \sec x \]

### Step 4: Multiply the Equation by the Integrating Factor
We multiply the entire differential equation by \( \sec x \):
\[ \sec x \frac{dy}{dx} - y \sec x \tan x = 2x \]

### Step 5: Rewrite the Left Side as a Derivative
The left-hand side can be expressed as the derivative of a product:
\[ \frac{d}{dx}(y \sec x) = 2x \]

### Step 6: Integrate Both Sides
Integrating both sides with respect to \( x \):
\[ \int \frac{d}{dx}(y \sec x) \, dx = \int 2x \, dx \]
This gives:
\[ y \sec x = x^2 + C \]

### Step 7: Solve for \( y \)
Now, we can express \( y \) as:
\[ y = x^2 \cos x + C \cos x \]

### Step 8: Apply the Initial Condition
Using the initial condition \( y(0) = 0 \):
\[ 0 = 0^2 \cos(0) + C \cos(0) \implies 0 = C \]
Thus, \( C = 0 \).

### Final Solution
The solution simplifies to:
\[ y = x^2 \cos x \]

### Step 9: Evaluate \( y \) at Specific Points
Now, we can evaluate \( y \) at \( x = \frac{\pi}{4} \) and \( x = \frac{\pi}{3} \):
1. For \( x = \frac{\pi}{4} \):
\[ y\left(\frac{\pi}{4}\right) = \left(\frac{\pi}{4}\right)^2 \cos\left(\frac{\pi}{4}\right) = \frac{\pi^2}{16} \cdot \frac{1}{\sqrt{2}} = \frac{\pi^2}{16\sqrt{2}} \]

2. For \( x = \frac{\pi}{3} \):
\[ y\left(\frac{\pi}{3}\right) = \left(\frac{\pi}{3}\right)^2 \cos\left(\frac{\pi}{3}\right) = \frac{\pi^2}{9} \cdot \frac{1}{2} = \frac{\pi^2}{18} \]

### Step 10: Compute the Derivative \( y' \)
To find \( y' \), we differentiate \( y = x^2 \cos x \):
Using the product rule:
\[ y' = 2x \cos x - x^2 \sin x \]
Now evaluate at \( x = \frac{\pi}{4} \) and \( x = \frac{\pi}{3} \).

1. For \( x = \frac{\pi}{4} \):
\[ y'\left(\frac{\pi}{4}\right) = 2 \cdot \frac{\pi}{4} \cdot \cos\left(\frac{\pi}{4}\right) - \left(\frac{\pi}{4}\right)^2 \sin\left(\frac{\pi}{4}\right) = \frac{\pi}{2} \cdot \frac{1}{\sqrt{2}} - \frac{\pi^2}{16} \cdot \frac{1}{\sqrt{2}} = \frac{\pi - \frac{\pi^2}{8}}{\sqrt{2}} \]

2. For \( x = \frac{\pi}{3} \):
\[ y'\left(\frac{\pi}{3}\right) = 2 \cdot \frac{\pi}{3} \cdot \cos\left(\frac{\pi}{3}\right) - \left(\frac{\pi}{3}\right)^2 \sin\left(\frac{\pi}{3}\right) = \frac{2\pi}{3} \cdot \frac{1}{2} - \frac{\pi^2}{9} \cdot \frac{\sqrt{3}}{2} = \frac{\pi}{3} - \frac{\pi^2 \sqrt{3}}{18} \]