If `(dy)/(dx)+3/(cos^(2)x)y=1/(cos^(2)x),xin((-pi)/3,(pi)/3),` and `y((pi)/4)=4/3`, then `y(-(pi)/4)` equals `1/3+e^(k)`. The value of k is __________.

Explanation

To solve the given differential equation and find the value of \( k \), we will follow these steps:

### Step 1: Write the Differential Equation
The given differential equation is:
\[ \frac{dy}{dx} + \frac{3}{\cos^2 x} y = \frac{1}{\cos^2 x} \]

### Step 2: Identify \( p(x) \) and \( q(x) \)
From the standard form of a linear first-order differential equation \( \frac{dy}{dx} + p(x) y = q(x) \), we identify:
- \( p(x) = \frac{3}{\cos^2 x} \)
- \( q(x) = \frac{1}{\cos^2 x} \)

### Step 3: Find the Integrating Factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int p(x) \, dx} = e^{\int \frac{3}{\cos^2 x} \, dx} \]
The integral of \( \frac{1}{\cos^2 x} \) is \( \tan x \), so:
\[ \int \frac{3}{\cos^2 x} \, dx = 3 \tan x \]
Thus, the integrating factor is:
\[ \mu(x) = e^{3 \tan x} \]

### Step 4: Multiply the Differential Equation by the Integrating Factor
Multiplying the entire differential equation by \( e^{3 \tan x} \):
\[ e^{3 \tan x} \frac{dy}{dx} + 3 e^{3 \tan x} \frac{y}{\cos^2 x} = e^{3 \tan x} \frac{1}{\cos^2 x} \]

### Step 5: Rewrite the Left Side
The left side can be rewritten as the derivative of a product:
\[ \frac{d}{dx} \left( e^{3 \tan x} y \right) = e^{3 \tan x} \sec^2 x \]

### Step 6: Integrate Both Sides
Integrating both sides:
\[ \int \frac{d}{dx} \left( e^{3 \tan x} y \right) \, dx = \int e^{3 \tan x} \sec^2 x \, dx \]
The right side can be solved using substitution \( u = \tan x \), \( du = \sec^2 x \, dx \):
\[ \int e^{3u} \, du = \frac{1}{3} e^{3u} + C = \frac{1}{3} e^{3 \tan x} + C \]

### Step 7: Solve for \( y \)
Thus, we have:
\[ e^{3 \tan x} y = \frac{1}{3} e^{3 \tan x} + C \]
Dividing by \( e^{3 \tan x} \):
\[ y = \frac{1}{3} + C e^{-3 \tan x} \]

### Step 8: Use the Initial Condition
We are given \( y\left(\frac{\pi}{4}\right) = \frac{4}{3} \):
\[ \frac{4}{3} = \frac{1}{3} + C e^{-3 \cdot 1} \]
This simplifies to:
\[ \frac{4}{3} - \frac{1}{3} = C e^{-3} \implies 1 = C e^{-3} \implies C = e^{3} \]

### Step 9: Write the Final Solution
Substituting \( C \) back into the equation:
\[ y = \frac{1}{3} + e^{3} e^{-3 \tan x} \]

### Step 10: Find \( y\left(-\frac{\pi}{4}\right) \)
Now, we need to find \( y\left(-\frac{\pi}{4}\right) \):
\[ y\left(-\frac{\pi}{4}\right) = \frac{1}{3} + e^{3} e^{-3 \cdot (-1)} = \frac{1}{3} + e^{3} e^{3} = \frac{1}{3} + e^{6} \]

### Step 11: Compare with the Given Format
The problem states that \( y\left(-\frac{\pi}{4}\right) = \frac{1}{3} + e^{k} \). Therefore, we can equate:
\[ e^{k} = e^{6} \implies k = 6 \]

### Final Answer
The value of \( k \) is:
\[ \boxed{6} \]