Calculus – Differential Equations – JEE Main 28 Jan 2025 Shift 2

Solution:

Divide by $\sqrt{4-x^2}$:
$$\frac{dy}{dx} = \frac{(\sin^{-1}\frac{x}{2})^2 \sin^{-1}\frac{x}{2}}{\sqrt{4-x^2}} – \frac{y \sin^{-1}\frac{x}{2}}{\sqrt{4-x^2}}$$
$$\frac{dy}{dx} + \frac{\sin^{-1}\frac{x}{2}}{\sqrt{4-x^2}} y = \frac{(\sin^{-1}\frac{x}{2})^3}{\sqrt{4-x^2}}$$
This is a Linear Differential Equation (LDE) of form $\frac{dy}{dx} + Py = Q$.

Integrating Factor (IF):
Let $\sin^{-1}\frac{x}{2} = t \Rightarrow \frac{1}{\sqrt{1-(x/2)^2}} \cdot \frac{1}{2} dx = dt \Rightarrow \frac{dx}{\sqrt{4-x^2}} = dt$.
$$IF = e^{\int \frac{\sin^{-1}(x/2)}{\sqrt{4-x^2}} dx} = e^{\int t dt} = e^{t^2/2} = e^{\frac{1}{2}(\sin^{-1}\frac{x}{2})^2}$$

Solution is:
$$y \cdot e^{t^2/2} = \int t^3 e^{t^2/2} dt$$
Let $t^2/2 = u \Rightarrow t dt = du$. Then $t^2 = 2u$.
$$\int (2u) e^u du = 2(u e^u – e^u) = 2e^u(u-1)$$
Substituting back:
$$y \cdot e^{t^2/2} = 2e^{t^2/2}(\frac{t^2}{2} – 1) + C$$
$$y = t^2 – 2 + C e^{-t^2/2}$$
$$y = (\sin^{-1}\frac{x}{2})^2 – 2 + C e^{-\frac{1}{2}(\sin^{-1}\frac{x}{2})^2}$$

Given $y(2) = \frac{\pi^2 – 8}{4}$.
At $x=2, \sin^{-1}(1) = \frac{\pi}{2}$. $t = \frac{\pi}{2}$.
$$\frac{\pi^2}{4} – 2 = (\frac{\pi}{2})^2 – 2 + C e^{-\pi^2/8}$$
$$0 = C e^{-\pi^2/8} \Rightarrow C = 0$$
So, $y(x) = (\sin^{-1}\frac{x}{2})^2 – 2$.

Find $y(0)$:
At $x=0, \sin^{-1}(0) = 0$.
$$y(0) = 0 – 2 = -2$$
We need $y^2(0)$:
$$y^2(0) = (-2)^2 = 4$$

Ans. (4)