Differential Equations
Differential Equations
nta_pyq_2025_apr
Grade 12
Question:
If $y = y(x)$ is the solution of the differential equation $\sqrt{4-x^2}\,\dfrac{dy}{dx} = \left(\!\left(\sin^{-1}\frac{x}{2}\right)^2-y\right)\sin^{-1}\frac{x}{2}$, $-2\leq x\leq 2$, $y(2) = \dfrac{\pi^2}{4}-8$, then $y^2(0)$ is equal to ____.
Step-by-Step Solution
Key Concept: Let $u = \sin^{-1}(x/2)$, $du = dx/\sqrt{4-x^2}$; the ODE becomes $dy/du + y = u^2$, a standard linear ODE with I.F. $= e^u$.
Let $u = \sin^{-1}(x/2)$. ODE becomes $\dfrac{dy}{du}+y = u^2$. I.F. $= e^u$.
$ye^u = \int u^2 e^u\,du = e^u(u^2-2u+2)+C$.
$y = u^2-2u+2+Ce^{-u} = \left(\sin^{-1}\frac{x}{2}\right)^2-2+Ce^{-\sin^{-1}(x/2)}$.
At $x=2$, $u=\pi/2$: $\dfrac{\pi^2}{4}-8 = \dfrac{\pi^2}{4}-2+Ce^{-\pi/2} \Rightarrow C=0$.
$y(0) = 0-2 = -2$. Hence $y^2(0) = (-2)^2 = 4$.
Correct Answer: 4