Key Concepts and Formulas

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

Step-by-Step Solution

Step 1: Transform the Differential Equation into Standard Linear Form

The given differential equation is: $\sqrt{4-x^2} \frac{dy}{dx} = \left( \left( \sin^{-1}\left(\frac{x}{2}\right) \right)^2 - y \right) \sin^{-1}\left(\frac{x}{2}\right)$ We want to rewrite it in the form $\frac{dy}{dx} + P(x)y = Q(x)$. Let $t = \sin^{-1}\left(\frac{x}{2}\right)$. Then, the equation becomes: $\sqrt{4-x^2} \frac{dy}{dx} = (t^2 - y)t = t^3 - yt$ Divide both sides by $\sqrt{4-x^2}$: $\frac{dy}{dx} = \frac{t^3}{\sqrt{4-x^2}} - \frac{yt}{\sqrt{4-x^2}}$ Rearrange the terms: $\frac{dy}{dx} + \frac{t}{\sqrt{4-x^2}}y = \frac{t^3}{\sqrt{4-x^2}}$ Substitute back $t = \sin^{-1}\left(\frac{x}{2}\right)$: $\frac{dy}{dx} + \frac{\sin^{-1}\left(\frac{x}{2}\right)}{\sqrt{4-x^2}}y = \frac{\left(\sin^{-1}\left(\frac{x}{2}\right)\right)^3}{\sqrt{4-x^2}}$ Now we have the standard form with: $P(x) = \frac{\sin^{-1}\left(\frac{x}{2}\right)}{\sqrt{4-x^2}}$ $Q(x) = \frac{\left(\sin^{-1}\left(\frac{x}{2}\right)\right)^3}{\sqrt{4-x^2}}$

Step 2: Calculate the Integrating Factor (IF)

The integrating factor is $IF = e^{\int P(x) \, dx}$. We need to calculate $\int P(x) \, dx$: $\int \frac{\sin^{-1}\left(\frac{x}{2}\right)}{\sqrt{4-x^2}} \, dx$ Let $t = \sin^{-1}\left(\frac{x}{2}\right)$. Then, $\frac{dt}{dx} = \frac{1}{\sqrt{1 - \left(\frac{x}{2}\right)^2}} \cdot \frac{1}{2} = \frac{1}{\sqrt{4-x^2}/2} \cdot \frac{1}{2} = \frac{1}{\sqrt{4-x^2}}$. Therefore, $dt = \frac{dx}{\sqrt{4-x^2}}$. The integral becomes: $\int t \, dt = \frac{t^2}{2} = \frac{1}{2} \left(\sin^{-1}\left(\frac{x}{2}\right)\right)^2$ The Integrating Factor is: $IF = e^{\frac{1}{2}\left(\sin^{-1}\left(\frac{x}{2}\right)\right)^2}$

Step 3: Find the General Solution

The general solution is given by $y \cdot IF = \int Q(x) \cdot IF \, dx + C$. Substitute $Q(x)$ and $IF$: $y \cdot e^{\frac{1}{2}\left(\sin^{-1}\left(\frac{x}{2}\right)\right)^2} = \int \frac{\left(\sin^{-1}\left(\frac{x}{2}\right)\right)^3}{\sqrt{4-x^2}} \cdot e^{\frac{1}{2}\left(\sin^{-1}\left(\frac{x}{2}\right)\right)^2} \, dx + C$ Let $t = \sin^{-1}\left(\frac{x}{2}\right)$. Then $dt = \frac{dx}{\sqrt{4-x^2}}$. The integral on the right-hand side becomes: $\int t^3 e^{\frac{1}{2}t^2} \, dt$ Let $u = t^2$ and $dv = t e^{\frac{1}{2}t^2} dt$. Then $du = 2t dt$. To find $v$, we integrate $dv$. Let $s = \frac{1}{2}t^2$, so $ds = t dt$. Then $\int t e^{\frac{1}{2}t^2} dt = \int e^s ds = e^s = e^{\frac{1}{2}t^2}$. Thus, $v = e^{\frac{1}{2}t^2}$. Using integration by parts $\int u \, dv = uv - \int v \, du$: $\int t^3 e^{\frac{1}{2}t^2} \, dt = t^2 e^{\frac{1}{2}t^2} - \int e^{\frac{1}{2}t^2} (2t \, dt) = t^2 e^{\frac{1}{2}t^2} - 2 \int t e^{\frac{1}{2}t^2} \, dt$ Since $\int t e^{\frac{1}{2}t^2} \, dt = e^{\frac{1}{2}t^2}$, we have: $\int t^3 e^{\frac{1}{2}t^2} \, dt = t^2 e^{\frac{1}{2}t^2} - 2e^{\frac{1}{2}t^2} = (t^2 - 2)e^{\frac{1}{2}t^2}$ Substituting back $t = \sin^{-1}\left(\frac{x}{2}\right)$, we get: $y \cdot e^{\frac{1}{2}\left(\sin^{-1}\left(\frac{x}{2}\right)\right)^2} = \left(\left(\sin^{-1}\left(\frac{x}{2}\right)\right)^2 - 2\right) e^{\frac{1}{2}\left(\sin^{-1}\left(\frac{x}{2}\right)\right)^2} + C$ Divide both sides by $e^{\frac{1}{2}\left(\sin^{-1}\left(\frac{x}{2}\right)\right)^2}$: $y = \left(\sin^{-1}\left(\frac{x}{2}\right)\right)^2 - 2 + C e^{-\frac{1}{2}\left(\sin^{-1}\left(\frac{x}{2}\right)\right)^2}$

Step 4: Apply the Initial Condition to Find C

We are given $y(2) = \frac{\pi^2 - 8}{4}$. When $x=2$, $\sin^{-1}\left(\frac{2}{2}\right) = \sin^{-1}(1) = \frac{\pi}{2}$. So, $y(2) = \left(\frac{\pi}{2}\right)^2 - 2 + C e^{-\frac{1}{2}\left(\frac{\pi}{2}\right)^2} = \frac{\pi^2}{4} - 2 + C e^{-\frac{\pi^2}{8}}$. We are given $y(2) = \frac{\pi^2 - 8}{4} = \frac{\pi^2}{4} - 2$. Therefore, $\frac{\pi^2}{4} - 2 = \frac{\pi^2}{4} - 2 + C e^{-\frac{\pi^2}{8}}$, which implies $C e^{-\frac{\pi^2}{8}} = 0$, so $C=0$. Thus, the particular solution is: $y(x) = \left(\sin^{-1}\left(\frac{x}{2}\right)\right)^2 - 2$

Step 5: Calculate y(0) and y^2(0)

We want to find $y^2(0)$. First, find $y(0)$: $y(0) = \left(\sin^{-1}\left(\frac{0}{2}\right)\right)^2 - 2 = \left(\sin^{-1}(0)\right)^2 - 2 = 0^2 - 2 = -2$ Therefore, $y^2(0) = (-2)^2 = 4$.

Common Mistakes & Tips

• Substitution: Using the substitution $t = \sin^{-1}\left(\frac{x}{2}\right)$ simplifies the integral calculations significantly.
• Integration by Parts: Remember the integration by parts formula: $\int u \, dv = uv - \int v \, du$. Choose $u$ and $dv$ strategically to simplify the integral.
• Initial Condition: Make sure to use the given initial condition to find the value of the constant of integration, $C$.

Summary

We solved the given differential equation by first transforming it into a standard first-order linear form. We then found the integrating factor and used it to obtain the general solution. Using the provided initial condition, we determined the particular solution. Finally, we evaluated the solution at $x=0$ and squared the result to find $y^2(0)$.

The final answer is \boxed{4}.