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 (IF): For a first-order linear differential equation, the integrating factor is given by $IF = e^{\int P(x) dx}$. The solution is then $y \cdot IF = \int Q(x) \cdot IF \, dx + C$.
• Even and Odd Functions: A function $f(x)$ is even if $f(-x) = f(x)$ and odd if $f(-x) = -f(x)$. The integral of an odd function over a symmetric interval $[-a, a]$ is zero. The integral of an even function over a symmetric interval $[-a, a]$ is $2\int_0^a f(x) dx$.

Step-by-Step Solution

Step 1: Rewrite the given differential equation in standard form.

We are given $(1 - x^2)dy = (xy + (x^3 + 2)\sqrt{1 - x^2})dx$. We want to rewrite this in the form $\frac{dy}{dx} + P(x)y = Q(x)$. Divide both sides by $(1 - x^2)dx$:

$\frac{dy}{dx} = \frac{xy}{1 - x^2} + \frac{(x^3 + 2)\sqrt{1 - x^2}}{1 - x^2}$ $\frac{dy}{dx} - \frac{x}{1 - x^2}y = \frac{x^3 + 2}{\sqrt{1 - x^2}}$

Here, $P(x) = -\frac{x}{1 - x^2}$ and $Q(x) = \frac{x^3 + 2}{\sqrt{1 - x^2}}$.

Step 2: Calculate the Integrating Factor (IF).

The integrating factor is given by $IF = e^{\int P(x) dx}$. So, we need to find $\int P(x) dx = \int -\frac{x}{1 - x^2} dx$. Let $u = 1 - x^2$, then $du = -2x dx$, so $xdx = -\frac{1}{2}du$. Then,

$\int -\frac{x}{1 - x^2} dx = \int \frac{1}{2u} du = \frac{1}{2} \ln |u| + C = \frac{1}{2} \ln |1 - x^2| + C$ Since $-1 < x < 1$, we have $1 - x^2 > 0$, so $|1 - x^2| = 1 - x^2$. $IF = e^{\frac{1}{2} \ln (1 - x^2)} = e^{\ln \sqrt{1 - x^2}} = \sqrt{1 - x^2}$

Step 3: Find the general solution.

The general solution is given by $y \cdot IF = \int Q(x) \cdot IF \, dx + C$. Substituting $IF = \sqrt{1 - x^2}$ and $Q(x) = \frac{x^3 + 2}{\sqrt{1 - x^2}}$, we get:

$y\sqrt{1 - x^2} = \int \frac{x^3 + 2}{\sqrt{1 - x^2}} \cdot \sqrt{1 - x^2} \, dx + C = \int (x^3 + 2) dx + C$ $y\sqrt{1 - x^2} = \frac{x^4}{4} + 2x + C$ $y = \frac{1}{\sqrt{1 - x^2}} \left( \frac{x^4}{4} + 2x + C \right)$

Step 4: Apply the initial condition to find the particular solution.

We are given $y(0) = 0$. Substituting $x = 0$ and $y = 0$ into the general solution:

$0 = \frac{1}{\sqrt{1 - 0^2}} \left( \frac{0^4}{4} + 2(0) + C \right)$ $0 = \frac{1}{1}(0 + 0 + C)$ $C = 0$

Thus, the particular solution is: $y = \frac{1}{\sqrt{1 - x^2}} \left( \frac{x^4}{4} + 2x \right) = \frac{x^4/4 + 2x}{\sqrt{1 - x^2}}$

Step 5: Evaluate the definite integral.

We are given $\int_{-1/2}^{1/2} \sqrt{1 - x^2} y(x) dx = k$. Substituting the expression for $y(x)$:

$k = \int_{-1/2}^{1/2} \sqrt{1 - x^2} \cdot \frac{x^4/4 + 2x}{\sqrt{1 - x^2}} dx = \int_{-1/2}^{1/2} \left( \frac{x^4}{4} + 2x \right) dx$ We can split this into two integrals: $k = \int_{-1/2}^{1/2} \frac{x^4}{4} dx + \int_{-1/2}^{1/2} 2x dx$ Since $2x$ is an odd function, $\int_{-1/2}^{1/2} 2x dx = 0$. Since $\frac{x^4}{4}$ is an even function, $k = \int_{-1/2}^{1/2} \frac{x^4}{4} dx = 2 \int_0^{1/2} \frac{x^4}{4} dx = \frac{1}{2} \int_0^{1/2} x^4 dx = \frac{1}{2} \left[ \frac{x^5}{5} \right]_0^{1/2} = \frac{1}{2} \left( \frac{(1/2)^5}{5} - 0 \right) = \frac{1}{2} \cdot \frac{1}{32 \cdot 5} = \frac{1}{320}$

Step 6: Calculate k - 1.

We have $k = \frac{1}{320}$. Therefore, $k - 1 = \frac{1}{320} - 1 = \frac{1 - 320}{320} = -\frac{319}{320}$.

The problem statement says the correct answer is 1. Let's re-examine the equation $y = \frac{1}{\sqrt{1 - x^2}} \left( \frac{x^4}{4} + 2x + C \right)$. With $C = 0$, we have $y = \frac{x^4/4 + 2x}{\sqrt{1 - x^2}}$. $k = \int_{-1/2}^{1/2} \sqrt{1 - x^2} y(x) dx = \int_{-1/2}^{1/2} \sqrt{1 - x^2} \cdot \frac{x^4/4 + 2x}{\sqrt{1 - x^2}} dx = \int_{-1/2}^{1/2} \left( \frac{x^4}{4} + 2x \right) dx$ Since $2x$ is an odd function, $\int_{-1/2}^{1/2} 2x dx = 0$. Since $\frac{x^4}{4}$ is an even function, $k = \int_{-1/2}^{1/2} \frac{x^4}{4} dx = 2 \int_0^{1/2} \frac{x^4}{4} dx = \frac{1}{2} \int_0^{1/2} x^4 dx = \frac{1}{2} \left[ \frac{x^5}{5} \right]_0^{1/2} = \frac{1}{2} \left( \frac{(1/2)^5}{5} - 0 \right) = \frac{1}{2} \cdot \frac{1}{32 \cdot 5} = \frac{1}{320}$ So $k = \frac{1}{320}$, and $k-1 = \frac{1}{320} - 1 = -\frac{319}{320}$.

There must be an error in the problem statement or provided correct answer. Let's assume there's a typo in the original problem, and the initial condition is something else. Let's solve for when $k=2$, so $k-1 = 1$. $k = \int_{-1/2}^{1/2} \frac{x^4}{4} + 2x dx = \frac{1}{320}$. We need the integral to evaluate to 2. Then we have $y = \frac{x^4/4 + 2x + C}{\sqrt{1-x^2}}$. $k = \int_{-1/2}^{1/2} \frac{x^4}{4} + 2x + C dx = \frac{1}{320} + \int_{-1/2}^{1/2} C dx = \frac{1}{320} + Cx\Big|_{-1/2}^{1/2} = \frac{1}{320} + C(1/2 - (-1/2)) = \frac{1}{320} + C$ So, $\frac{1}{320} + C = 2$, so $C = 2 - \frac{1}{320} = \frac{639}{320}$. This means $y(0) = \frac{C}{1} = C = \frac{639}{320}$. But we still arrive at $k = \frac{1}{320}$ with the current problem statement.

Common Mistakes & Tips

• Remember to include the constant of integration $C$ when evaluating indefinite integrals.
• When evaluating definite integrals over symmetric intervals, remember to check if the function is even or odd to simplify the calculation.
• Be careful with algebraic manipulations, especially when dealing with square roots and fractions.

Summary

We solved the given first-order linear differential equation by finding the integrating factor, obtaining the general solution, and then using the initial condition to find the particular solution. Finally, we evaluated the definite integral and found that $k = \frac{1}{320}$. Therefore, $k - 1 = -\frac{319}{320}$. However, based on the given "Correct Answer," there appears to be an error in the problem statement, the provided answer, or both. Assuming the problem statement is correct as given, the derivation is correct.

Final Answer

The final answer is \boxed{-\frac{319}{320}}.