Key Concepts and Formulas

• Fundamental Theorem of Calculus: If $F(x) = \int_{a}^{x} f(t) \, dt$, then $F'(x) = f(x)$.
• First-Order Linear Differential Equation: $\frac{dy}{dx} + P(x)y = Q(x)$, with integrating factor $I.F. = e^{\int P(x) \, dx}$ and solution $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) \, dx + C$.
• Integration: $\int \tan x \, dx = \ln|\sec x| + C$

Step-by-Step Solution

Step 1: Differentiate the integral equation with respect to $x$.

• Why this step? We want to eliminate the integral and find an expression for $g(x)$. Differentiating both sides using the Fundamental Theorem of Calculus will achieve this.

We are given: $\int_0^x g(t) dt = x - \int_0^x tg(t) dt$ Differentiating both sides with respect to $x$, we have: $\frac{d}{dx} \left( \int_0^x g(t) dt \right) = \frac{d}{dx} \left( x - \int_0^x tg(t) dt \right)$ Applying the Fundamental Theorem of Calculus: $g(x) = 1 - xg(x)$

Step 2: Solve for $g(x)$.

• Why this step? We need to isolate $g(x)$ to find its explicit form.

Rearranging the equation from Step 1: $g(x) + xg(x) = 1$ $g(x)(1 + x) = 1$ $g(x) = \frac{1}{1 + x}$

Step 3: Substitute $g(x)$ into the differential equation.

• Why this step? We now have an expression for $g(x)$, so we substitute it into the given differential equation to get an equation involving only $y$ and $x$.

The differential equation is: $\frac{dy}{dx} - y \tan x = 2(x+1) \sec x g(x)$ Substituting $g(x) = \frac{1}{1 + x}$: $\frac{dy}{dx} - y \tan x = 2(x+1) \sec x \left( \frac{1}{1 + x} \right)$ $\frac{dy}{dx} - y \tan x = 2 \sec x$

Step 4: Find the integrating factor (I.F.).

• Why this step? The differential equation is now in the standard form for a first-order linear differential equation. We need to find the integrating factor to solve it.

The equation is in the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = -\tan x$ and $Q(x) = 2 \sec x$. The integrating factor is: $I.F. = e^{\int P(x) dx} = e^{\int -\tan x dx} = e^{-\int \tan x dx}$ Since $\int \tan x dx = \ln|\sec x| + C$, we have: $I.F. = e^{-\ln|\sec x|} = e^{\ln|\sec x|^{-1}} = e^{\ln|\cos x|} = \cos x$ We can drop the absolute value since $x \in [0, \frac{\pi}{2})$.

Step 5: Multiply the differential equation by the integrating factor and integrate.

• Why this step? Multiplying by the integrating factor makes the left-hand side a perfect derivative, which we can then integrate.

Multiplying the differential equation by $\cos x$: $\cos x \frac{dy}{dx} - y \cos x \tan x = 2 \sec x \cos x$ $\cos x \frac{dy}{dx} - y \sin x = 2$ The left side is the derivative of $y \cos x$: $\frac{d}{dx}(y \cos x) = 2$ Integrating both sides with respect to $x$: $\int \frac{d}{dx}(y \cos x) dx = \int 2 dx$ $y \cos x = 2x + C$

Step 6: Apply the initial condition $y(0) = 0$ to find $C$.

• Why this step? We need to find the particular solution that satisfies the initial condition.

Substituting $x = 0$ and $y = 0$ into the equation $y \cos x = 2x + C$: $0 \cdot \cos(0) = 2(0) + C$ $0 = 0 + C$ $C = 0$

Step 7: Find $y(x)$.

• Why this step? We now have the particular solution, so we need to isolate $y$ to find $y(x)$.

Substituting $C = 0$ into $y \cos x = 2x + C$: $y \cos x = 2x$ $y(x) = \frac{2x}{\cos x} = 2x \sec x$

Step 8: Evaluate $y(\frac{\pi}{3})$.

• Why this step? This is the final step, where we find the value of the solution at the specified point.

$y\left(\frac{\pi}{3}\right) = 2\left(\frac{\pi}{3}\right) \sec\left(\frac{\pi}{3}\right)$ Since $\sec\left(\frac{\pi}{3}\right) = \frac{1}{\cos(\frac{\pi}{3})} = \frac{1}{1/2} = 2$: $y\left(\frac{\pi}{3}\right) = 2\left(\frac{\pi}{3}\right)(2) = \frac{4\pi}{3}$

Common Mistakes & Tips

• Remember to use the Fundamental Theorem of Calculus correctly when differentiating integrals.
• Be careful with signs when finding the integrating factor.
• Don't forget the constant of integration when integrating, and use the initial condition to find its value.

Summary

We found $g(x)$ by differentiating the integral equation and then solved the resulting first-order linear differential equation using the integrating factor method. Applying the initial condition allowed us to find the particular solution and then evaluate it at $x = \frac{\pi}{3}$. The final result is $\frac{4\pi}{3}$.

The final answer is $\boxed{\frac{4 \pi}{3}}$, which corresponds to option (A).