JEE Main 2024 (Online) 4th April Evening ShiftQuestion 1
Let $y=y(x)$ be the solution of the differential equation $(x+y+2)^{2}dx=dy$, $y(0)=-2$. Let the maximum and minimum values of the function $y=y(x)$ in $\left[0,\dfrac{\pi}{3}\right]$ be $\alpha$ and $\beta$, respectively. If $(3\alpha+\pi)^{2}+\beta^{2}=\gamma+\delta\sqrt{3}$, $\gamma,\delta\in\mathbb{Z}$, then $\gamma+\delta$ equals $\_\_\_\_\_\_\_\_$ .
Differential Equation
Answer$31$
Solution
Concept. With $v=x+y+2$ the equation becomes $v'=1+v^2$. Formulas used. (1) $\int\dfrac{dv}{1+v^2}=\tan^{-1}v$Step 1 - Substitute. $v'=1+y'=1+v^2$, and $v(0)=0-2+2=0$. By (1), $\tan^{-1}v=x$, so $v=\tan x$: $$y=\tan x-x-2.$$ Step 2 - Max and min. $y'=\sec^2x-1=\tan^2x\ge0$, so $y$ increases on $\left[0,\frac\pi3\right]$: $$\alpha=y\left(\tfrac\pi3\right)=\sqrt3-\tfrac\pi3-2,\qquad\beta=y(0)=-2.$$ Step 3 - Compute. $3\alpha+\pi=3\sqrt3-6$, so $$(3\sqrt3-6)^2+4=27-36\sqrt3+36+4=67-36\sqrt3.$$ Step 4 - Value. $\gamma=67$, $\delta=-36$, so $\gamma+\delta=31$. Answer: 31