Key Concepts and Formulas

• First-Order Linear Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.
• Integrating Factor (I.F.): For a first-order linear differential equation, the integrating factor is $e^{\int P(x) dx}$.
• General Solution: The general solution to a first-order linear differential equation is $y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C$.

Step-by-Step Solution

Step 1: Identify the type of Differential Equation and Standardize it.

We are given the differential equation $\frac{dy}{dx} + \frac{3}{\cos^2 x} y = \frac{1}{\cos^2 x}, \quad x \in \left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$ This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$. Using the identity $\sec^2 x = \frac{1}{\cos^2 x}$, we can rewrite the equation as: $\frac{dy}{dx} + 3\sec^2 x \cdot y = \sec^2 x$ Comparing with the standard form, we identify $P(x) = 3\sec^2 x$ and $Q(x) = \sec^2 x$.

Step 2: Calculate the Integrating Factor (I.F.).

The integrating factor is given by $e^{\int P(x) dx}$. Substituting $P(x) = 3\sec^2 x$, we have: $\text{I.F.} = e^{\int 3\sec^2 x \, dx} = e^{3\int \sec^2 x \, dx}$ Since $\int \sec^2 x \, dx = \tan x$, the integrating factor is: $\text{I.F.} = e^{3\tan x}$

Step 3: Find the General Solution.

The general solution is given by $y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C$. Substituting $Q(x) = \sec^2 x$ and $\text{I.F.} = e^{3\tan x}$, we get: $y \cdot e^{3\tan x} = \int \sec^2 x \cdot e^{3\tan x} \, dx + C$ To evaluate the integral, let $u = 3\tan x$. Then $\frac{du}{dx} = 3\sec^2 x$, so $du = 3\sec^2 x \, dx$, which gives $\sec^2 x \, dx = \frac{1}{3} du$. Substituting, we have: $\int \sec^2 x \cdot e^{3\tan x} \, dx = \int e^u \cdot \frac{1}{3} \, du = \frac{1}{3} \int e^u \, du = \frac{1}{3} e^u + \text{constant} = \frac{1}{3} e^{3\tan x} + \text{constant}$ Therefore, the general solution is: $y \cdot e^{3\tan x} = \frac{1}{3} e^{3\tan x} + C$

Step 4: Use the Initial Condition to find C.

We are given that $y\left(\frac{\pi}{4}\right) = \frac{4}{3}$. Substituting $x = \frac{\pi}{4}$ and $y = \frac{4}{3}$ into the general solution, we get: $\frac{4}{3} \cdot e^{3\tan(\frac{\pi}{4})} = \frac{1}{3} e^{3\tan(\frac{\pi}{4})} + C$ Since $\tan\left(\frac{\pi}{4}\right) = 1$, we have: $\frac{4}{3} e^3 = \frac{1}{3} e^3 + C$ Solving for $C$, we find: $C = \frac{4}{3} e^3 - \frac{1}{3} e^3 = \frac{3}{3} e^3 = e^3$

Step 5: Find y(-π/4).

The particular solution is: $y \cdot e^{3\tan x} = \frac{1}{3} e^{3\tan x} + e^3$ We want to find $y\left(-\frac{\pi}{4}\right)$. Substituting $x = -\frac{\pi}{4}$, we have: $y\left(-\frac{\pi}{4}\right) \cdot e^{3\tan(-\frac{\pi}{4})} = \frac{1}{3} e^{3\tan(-\frac{\pi}{4})} + e^3$ Since $\tan\left(-\frac{\pi}{4}\right) = -1$, we have: $y\left(-\frac{\pi}{4}\right) \cdot e^{-3} = \frac{1}{3} e^{-3} + e^3$ Multiplying both sides by $e^3$, we get: $y\left(-\frac{\pi}{4}\right) = \frac{1}{3} + e^6$

Common Mistakes & Tips

• Remember the correct formula for the integrating factor. It is easy to forget the exponential.
• Be careful with trigonometric identities and signs, especially when dealing with $\tan(-x)$.
• Do not forget the constant of integration, $C$, when evaluating indefinite integrals.

Summary

We solved the first-order linear differential equation by finding the integrating factor, obtaining the general solution, and then using the given initial condition to determine the constant of integration. Finally, we evaluated the particular solution at the desired point $x = -\frac{\pi}{4}$ to find the value of $y\left(-\frac{\pi}{4}\right)$.

The final answer is $\boxed{\frac{1}{3} + e^6}$, which corresponds to option (A).