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.): $I.F. = e^{\int P(x) dx}$. The solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.
• Definite Integral Properties: $\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is an odd function, and $\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx$ if $f(x)$ is an even function.

Step-by-Step Solution

Step 1: Identify $P(x)$ and $Q(x)$

We are given the differential equation $\frac{dy}{dx} + \frac{xy}{x^2-1} = \frac{x^4+2x}{\sqrt{1-x^2}}.$ Comparing this to the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we identify $P(x) = \frac{x}{x^2-1}, \quad Q(x) = \frac{x^4+2x}{\sqrt{1-x^2}}.$ This step is crucial for setting up the problem and applying the correct formulas.

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

The integrating factor is given by $I.F. = e^{\int P(x) dx}$. We need to calculate $\int P(x) dx = \int \frac{x}{x^2-1} dx$. Let $u = x^2-1$, then $du = 2x dx$, so $x dx = \frac{1}{2} du$. Therefore, $\int \frac{x}{x^2-1} dx = \int \frac{1}{u} \cdot \frac{1}{2} du = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln |u| + C_1 = \frac{1}{2} \ln |x^2-1| + C_1.$ Since $x \in (-1, 1)$, we have $x^2 < 1$, so $x^2 - 1 < 0$. Thus, $|x^2-1| = 1-x^2$. $\int P(x) dx = \frac{1}{2} \ln (1-x^2).$ The integrating factor is then $I.F. = e^{\int P(x) dx} = e^{\frac{1}{2} \ln (1-x^2)} = e^{\ln \sqrt{1-x^2}} = \sqrt{1-x^2}.$ The correct handling of the domain $x\in(-1,1)$ is crucial here.

Step 3: Find the General Solution

The general solution is given by $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$. Substituting $Q(x)$ and $I.F.$, we have $y \sqrt{1-x^2} = \int \frac{x^4+2x}{\sqrt{1-x^2}} \cdot \sqrt{1-x^2} dx + C = \int (x^4+2x) dx + C.$ Integrating, we get $y \sqrt{1-x^2} = \frac{x^5}{5} + x^2 + C.$

Step 4: Apply the Initial Condition

Since the solution curve passes through the origin $(0,0)$, we have $f(0) = 0$. Substituting $x=0$ and $y=0$ into the general solution, we get $0 \cdot \sqrt{1-0^2} = \frac{0^5}{5} + 0^2 + C \implies 0 = 0 + 0 + C \implies C=0.$

Step 5: Find the Particular Solution

With $C=0$, the specific solution is $y \sqrt{1-x^2} = \frac{x^5}{5} + x^2.$ Solving for $y$, we have $f(x) = y = \frac{\frac{x^5}{5} + x^2}{\sqrt{1-x^2}} = \frac{x^5+5x^2}{5\sqrt{1-x^2}}.$

Step 6: Evaluate the Definite Integral

We need to evaluate $\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) dx = \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^5+5x^2}{5\sqrt{1-x^2}} dx = \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^5}{5\sqrt{1-x^2}} dx + \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{5x^2}{5\sqrt{1-x^2}} dx.$ The function $\frac{x^5}{5\sqrt{1-x^2}}$ is odd since $\frac{(-x)^5}{5\sqrt{1-(-x)^2}} = \frac{-x^5}{5\sqrt{1-x^2}} = -\frac{x^5}{5\sqrt{1-x^2}}$. Therefore, $\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^5}{5\sqrt{1-x^2}} dx = 0$. The function $\frac{x^2}{\sqrt{1-x^2}}$ is even since $\frac{(-x)^2}{\sqrt{1-(-x)^2}} = \frac{x^2}{\sqrt{1-x^2}}$. Therefore, $\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} dx = 2 \int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} dx.$ Let $x = \sin \theta$, then $dx = \cos \theta d\theta$. When $x=0$, $\theta = 0$. When $x = \frac{\sqrt{3}}{2}$, $\theta = \frac{\pi}{3}$. $2 \int_{0}^{\frac{\pi}{3}} \frac{\sin^2 \theta}{\sqrt{1-\sin^2 \theta}} \cos \theta d\theta = 2 \int_{0}^{\frac{\pi}{3}} \frac{\sin^2 \theta}{\cos \theta} \cos \theta d\theta = 2 \int_{0}^{\frac{\pi}{3}} \sin^2 \theta d\theta.$ Using the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$, we have $2 \int_{0}^{\frac{\pi}{3}} \frac{1 - \cos 2\theta}{2} d\theta = \int_{0}^{\frac{\pi}{3}} (1 - \cos 2\theta) d\theta = \left[ \theta - \frac{1}{2} \sin 2\theta \right]_{0}^{\frac{\pi}{3}} = \left( \frac{\pi}{3} - \frac{1}{2} \sin \frac{2\pi}{3} \right) - \left( 0 - \frac{1}{2} \sin 0 \right).$ $= \frac{\pi}{3} - \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\pi}{3} - \frac{\sqrt{3}}{4}.$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs, especially when dealing with absolute values and substitutions.
• Domain Awareness: Always consider the given domain and how it affects absolute values and other functions. In this problem, $x\in (-1,1)$ was crucial.
• Odd/Even Functions: Recognizing odd and even functions can greatly simplify definite integrals over symmetric intervals.

Summary

We solved the first-order linear differential equation by finding the integrating factor, using the initial condition to find the constant of integration, and then evaluating the definite integral of the solution. Recognizing the odd/even properties of the integrand simplified the evaluation process. The final result of the definite integral is $\frac{\pi}{3} - \frac{\sqrt{3}}{4}$.

Final Answer

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