If y=y(x) is the solution of the differential equation √(4-x²) dy/dx = ((sin⁻¹(x/2))²-y)sin⁻¹(x/2) ,

Q: Why do we divide by $\sqrt{4-x^2}$?

We divide by $\sqrt{4-x^2}$ to isolate $\frac{dy}{dx}$ and bring the equation into the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$, which allows us to use the Integrating Factor method.

Q: How do we integrate $t^3 e^{t^2/2}$?

We use the substitution $u = t^2/2$, which implies $du = t dt$. This transforms the integral into $\int 2u e^u du$, which is easily solved using integration by parts.

Q: What if the initial condition $y(2)$ was different?

If $y(2)$ were different, the constant $C$ would take a non-zero value, shifting the entire family of curves. The method remains identical; the boundary condition simply selects the specific curve from the family.

Question

If $y=y(x)$ is the solution of the differential equation $\sqrt{4-x^{2}}\frac{dy}{dx}=((\sin^{-1}(\frac{x}{2}))^{2}-y)\sin^{-1}(\frac{x}{2})$ , $-2\le x\le2$, $y(2)=\frac{\pi^{2}-8}{4}$ , then $y^{2}(0)$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given differential equation is:
$$\sqrt{4-x^2} \frac{dy}{dx} = \left( \left( \sin^{-1} \left( \frac{x}{2} \right) \right)^2 - y \right) \sin^{-1} \left( \frac{x}{2} \right)$$

By expanding the right-hand side, we obtain:
$$\sqrt{4-x^2} \frac{dy}{dx} = \left( \sin^{-1} \left( \frac{x}{2} \right) \right)^3 - y \sin^{-1} \left( \frac{x}{2} \right)$$

Rearranging the terms to isolate $y$ on the left side:
$$\sqrt{4-x^2} \frac{dy}{dx} + \sin^{-1} \left( \frac{x}{2} \right) y = \left( \sin^{-1} \left( \frac{x}{2} \right) \right)^3$$

Dividing the entire equation by $\sqrt{4-x^2}$ reveals the standard **linear differential equation** form $\frac{dy}{dx} + P(x)y = Q(x)$:
$$\frac{dy}{dx} + \frac{\sin^{-1} \left( \frac{x}{2} \right)}{\sqrt{4-x^2}} y = \frac{\left( \sin^{-1} \left( \frac{x}{2} \right) \right)^3}{\sqrt{4-x^2}}$$

### The Integrating Factor

We identify $P(x) = \frac{\sin^{-1} \left( \frac{x}{2} \right)}{\sqrt{4-x^2}}$. The Integrating Factor (IF) is defined as $IF = e^{\int P(x) dx}$.

Using the substitution $t = \sin^{-1} \left( \frac{x}{2} \right)$, we have $dt = \frac{1}{\sqrt{4-x^2}} dx$. The integral becomes:
$$\int t \, dt = \frac{t^2}{2}$$

Thus, our Integrating Factor is:
$$IF = e^{\frac{t^2}{2}}$$

### The Integration Process

The general solution is given by $y \cdot IF = \int Q(x) \cdot IF \, dx$. Substituting our expressions:
$$y \cdot e^{\frac{t^2}{2}} = \int t^3 e^{\frac{t^2}{2}} \, dt$$

To solve the integral, let $u = \frac{t^2}{2}$, which implies $du = t \, dt$. The integral transforms into:
$$\int 2u e^u \, du$$

Applying integration by parts, we get $2(u e^u - e^u) + C$. Substituting back $u = \frac{t^2}{2}$, the solution becomes:
$$y \cdot e^{\frac{t^2}{2}} = e^{\frac{t^2}{2}} \left( \frac{t^2}{2} \cdot 2 - 2 \right) + C = e^{\frac{t^2}{2}} (t^2 - 2) + C$$

### Finding the Specific Curve

The general solution is:
$$y(x) = \left( \sin^{-1} \left( \frac{x}{2} \right) \right)^2 - 2 + C e^{-\frac{\left( \sin^{-1} \left( \frac{x}{2} \right) \right)^2}{2}}$$

Given the boundary condition $y(2) = \frac{\pi^2-8}{4}$, we note that at $x=2$, $t = \sin^{-1}(1) = \frac{\pi}{2}$. Substituting these values:
$$\frac{\pi^2-8}{4} = \left( \frac{\pi}{2} \right)^2 - 2 + C e^{-\frac{\pi^2}{8}} \Rightarrow C = 0$$

The specific solution is therefore:
$$y(x) = \left( \sin^{-1} \left( \frac{x}{2} \right) \right)^2 - 2$$

### Final Calculation

We are required to find $y^2(0)$. Evaluating at $x=0$:
$$y(0) = (\sin^{-1}(0))^2 - 2 = -2$$

Squaring this result, we obtain:
$$y^2(0) = (-2)^2 = 4$$

The final answer is **4**.

Animated Solution for Mathematics - Differential Equations: If $y = y (x)$ is the solution of the differential equation $4 - x^{2} \frac{d y}{d x} = ((sin^{- 1} (\frac{x}{2}))^{2} - y) sin^{- 1} (\frac{x}{2})$ , $- 2 \leq x \leq 2$ , $y (2) = \frac{π ^{2} - 8}{4}$ , then $y^{2} (0)$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Differential Equation

Rearrange to Linear Form

Identify $P (x)$ and $Q (x)$

Setup Integrating Factor (IF)

Solve for IF

General Solution Setup

Integrate by Substitution

Integration by Parts

General Solution for $y (x)$

Apply Initial Condition

Find $y (0)$ and $y^{2} (0)$

Summary and Takeaway

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Integrating Factor

The Integration Process

Finding the Specific Curve

Final Calculation

Similar Questions

Let $y = y (x)$ be the solution of the differential equation $(x^{2} + 4)^{2} d y + (2 x^{3} y + 8 x y - 2) d x = 0$ . If $y (0) = 0$ , then $y (2)$ is equal to

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + 2 y = sin (2 x), y (0) = \frac{3}{4}$ , then $y (\frac{π}{8})$ is equal to:

Let $y = y (x)$ be the solution of the differential equation $cosec^{2} x d y + 2 d x = (1 + y cos 2 x) cosec^{2} x d x$ , with $y (\frac{π}{4}) = 0$ . Then, the value of $(y (0) + 1)^{2}$ is equal to:

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + (tan x) y = sin x, 0 \leq x \leq \frac{π}{3}$ , with $y (0) = 0$ , then $y (\frac{π}{4})$ equal to :

Let $y = y (x)$ be solution of the following differential equation $e^{y} \frac{d y}{d x} - 2 e^{y} sin x + sin x cos^{2} x = 0, y (\frac{π}{2}) = 0$ . If $y (0) = lo g_{e} (α + β e^{- 2})$ , then $4 (α + β)$ is equal to

Let $y = y (x)$ be the solution of the differential equation $2 cos x \frac{d y}{d x} = sin 2 x - 4 y sin x, x \in (0, \frac{π}{2}) .$ If $y (\frac{π}{3}) = 0$ then $y^{'} (\frac{π}{4}) + y (\frac{π}{4})$ is equal to

Let the solution curve $y = y (x)$ of the differential equation $(4 + x^{2}) d y - 2 x (x^{2} + 3 y + 4) d x = 0$ pass through the origin. Then $y (2)$ is equal to _______.

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} = (tan x - y) sec^{2} x, x \in (- π /2, π /2)$ , such that $y (0) = 0$ , then $y (- π /4)$ is equal to :

A solution of the differential equation $(\frac{d y}{d x})^{2} - x \frac{d y}{d x} + y = 0$ is

Let $y = y (x)$ be the solution of the differential equation $x lo g_{e} x \frac{d y}{d x} + y = x^{2} lo g_{e} x, (x > 1)$ . If $y (2) = 2$ , then $y (e)$ is equal to