JEE Main 2025 (Online) 28th January Evening ShiftQuestion 1
If $y=y(x)$ is the solution of the differential equation, $\sqrt{4-x^{2}}\dfrac{dy}{dx}=\left(\left(\sin^{-1}\left(\dfrac{x}{2}\right)\right)^{2}-y\right)\sin^{-1}\left(\dfrac{x}{2}\right)$, $-2\le x\le2$, $y(2)=\dfrac{\pi^{2}-8}{4}$, then $y^{2}(0)$ is equal to $\_\_\_\_\_\_\_\_$ .
Differential Equation
Answer$4$
Solution
Concept. With $u=\sin^{-1}\frac x2$, $\sqrt{4-x^2}\,\frac{dy}{dx}=\frac{dy}{du}$; the equation becomes linear in $u$. Formulas used. (1) $\dfrac{du}{dx}=\dfrac{1}{\sqrt{4-x^2}}$(2) $\int u^3e^{u^2/2}\,du=(u^2-2)\,e^{u^2/2}$Step 1 - Change variable. By (1): $$\frac{dy}{du}=(u^2-y)\,u\ \Longrightarrow\ \frac{dy}{du}+uy=u^3.$$ Step 2 - Solve. IF $=e^{u^2/2}$. By (2), $$y\,e^{u^2/2}=(u^2-2)e^{u^2/2}+C\ \Longrightarrow\ y=u^2-2+Ce^{-u^2/2}.$$ Step 3 - Constant. At $x=2$, $u=\frac\pi2$: $y=\frac{\pi^2}{4}-2=\frac{\pi^2-8}{4}$ already, so $C=0$. Step 4 - At $x=0$. $u=0$, $y=-2$, so $y^2(0)=4$. Answer: 4