We evaluate the integral for P(x). By substituting u = x²-1, the integral becomes (1)/(2) ln|x²-1|. Since -1 < x < 1, x²-1 is negative, so |x²-1| = -(x²-1) = 1-x². Thus, the integrating factor is e^(ln√(1-x²)), which simplifies to √(1-x²).

We evaluate the integral for P(x). By substituting u = x²-1, the integral becomes (1)/(2) ln|x²-1|. Since -1 < x < 1, x²-1 is negative, so |x²-1| = -(x²-1) = 1-x². Thus, the integrating factor is e^(ln√(1-x²)), which simplifies to √(1-x²).

Now we use the formula for the solution of a linear differential equation: y · I.F. = ∫ Q(x) · I.F. dx. We substitute Q(x) and the integrating factor into the equation. The √(1-x²) terms cancel out, simplifying the integral. We then integrate x^4+4x with respect to x to find the general solution.

Now we use the formula for the solution of a linear differential equation: y · I.F. = ∫ Q(x) · I.F. dx. We substitute Q(x) and the integrating factor into the equation. The √(1-x²) terms cancel out, simplifying the integral. We then integrate x^4+4x with respect to x to find the general solution.

With C=0, the solution to the differential equation is y = f(x) = x^5+10x²5√(1-x²). We need to evaluate the definite integral 6 ∫_-1/2^(1/2) f(x) dx. We can split the integral into two parts: one for the odd function x^5 and one for the even function x².

With C=0, the solution to the differential equation is y = f(x) = x^5+10x²5√(1-x²). We need to evaluate the definite integral 6 ∫_-1/2^(1/2) f(x) dx. We can split the integral into two parts: one for the odd function x^5 and one for the even function x².

The term (x^5)/(5√(1-x²)) is an odd function, so its integral over a symmetric interval [-a, a] is 0. The term 10x²5√(1-x²) = 2x²√(1-x²) is an even function. For an even function, ∫_-a^(a) g(x) dx = 2 ∫_0^(a) g(x) dx. This simplifies the integral to 24 ∫_0^(1/2) x²√(1-x²) dx.

The term (x^5)/(5√(1-x²)) is an odd function, so its integral over a symmetric interval [-a, a] is 0. The term 10x²5√(1-x²) = 2x²√(1-x²) is an even function. For an even function, ∫_-a^(a) g(x) dx = 2 ∫_0^(a) g(x) dx. This simplifies the integral to 24 ∫_0^(1/2) x²√(1-x²) dx.

To evaluate ∫ x²√(1-x²) dx, we use the trigonometric substitution x = sinθ. This transforms the integral into ∫ sin²θ dθ, which can be solved using the identity sin²θ = (1-cos(2θ))/(2). After integrating and substituting back x, we evaluate the definite integral from 0 to 1/2. The result is 2π - 3√3. Comparing this to 2π - α, we find α = 3√3. Finally, we calculate α².

To evaluate ∫ x²√(1-x²) dx, we use the trigonometric substitution x = sinθ. This transforms the integral into ∫ sin²θ dθ, which can be solved using the identity sin²θ = (1-cos(2θ))/(2). After integrating and substituting back x, we evaluate the definite integral from 0 to 1/2. The result is 2π - 3√3. Comparing this to 2π - α, we find α = 3√3. Finally, we calculate α².

Let $y=f(x)$ be the solution of the differential equation $\frac{dy}{dx} + \frac{xy}{x^2-1} = \frac{x^4 + 4x}{\sqrt{1-x^2}}, -1 < x < 1$ such that $f(0)=0$ . If $6 \int_{-1/2}^{1/2} f(x) dx = 2\pi - \alpha$ then $\alpha^2$ is equal to:

Answer: 27

Step-by-step solution

Step 1: Identify the form of the differential equation and find the integrating factor

The given differential equation is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$ . We first identify $P(x)$ and then calculate the integrating factor (I.F.) using the formula $e^{\int P(x) dx}$ . The integral of $\frac{x}{x^2-1}$ can be solved by substitution, letting $u = x^2-1$ , so $du = 2x dx$ .

Step 2: Calculate the integrating factor

We evaluate the integral for $P(x)$ . By substituting $u = x^2-1$ , the integral becomes $\frac{1}{2} \ln|x^2-1|$ . Since $-1 < x < 1$ , $x^2-1$ is negative, so $|x^2-1| = -(x^2-1) = 1-x^2$ . Thus, the integrating factor is $e^{\ln\sqrt{1-x^2}}$ , which simplifies to $\sqrt{1-x^2}$ .

Step 3: Solve the differential equation

Now we use the formula for the solution of a linear differential equation: $y \cdot I.F. = \int Q(x) \cdot I.F. dx$ . We substitute $Q(x)$ and the integrating factor into the equation. The $\sqrt{1-x^2}$ terms cancel out, simplifying the integral. We then integrate $x^4+4x$ with respect to $x$ to find the general solution.

Step 4: Apply the initial condition to find C

We are given the initial condition $f(0)=0$ . We substitute $x=0$ and $y=0$ into the general solution to find the value of the constant $C$ . This substitution directly yields $C=0$ .

Step 5: Determine $f(x)$ and evaluate the integral

With $C=0$ , the solution to the differential equation is $y = f(x) = \frac{x^5+10x^2}{5\sqrt{1-x^2}}$ . We need to evaluate the definite integral $6 \int_{-1/2}^{1/2} f(x) dx$ . We can split the integral into two parts: one for the odd function $x^5$ and one for the even function $x^2$ .

Step 6: Simplify the integral using properties of odd and even functions

The term $\frac{x^5}{5\sqrt{1-x^2}}$ is an odd function, so its integral over a symmetric interval $[-a, a]$ is $0$ . The term $\frac{10x^2}{5\sqrt{1-x^2}} = \frac{2x^2}{\sqrt{1-x^2}}$ is an even function. For an even function, $\int_{-a}^{a} g(x) dx = 2 \int_{0}^{a} g(x) dx$ . This simplifies the integral to $24 \int_{0}^{1/2} \frac{x^2}{\sqrt{1-x^2}} dx$ .

Step 7: Evaluate the remaining integral and find $\alpha^2$

To evaluate $\int \frac{x^2}{\sqrt{1-x^2}} dx$ , we use the trigonometric substitution $x = \sin\theta$ . This transforms the integral into $\int \sin^2\theta d\theta$ , which can be solved using the identity $\sin^2\theta = \frac{1-\cos(2\theta)}{2}$ . After integrating and substituting back $x$ , we evaluate the definite integral from $0$ to $1/2$ . The result is $2\pi - 3\sqrt{3}$ . Comparing this to $2\pi - \alpha$ , we find $\alpha = 3\sqrt{3}$ . Finally, we calculate $\alpha^2$ .

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