Differential Equations
Substitution — $v=x+y+2$
nta_pyq_2024_apr
Grade 12
Question:
Let $y=y(x)$ be the solution of $(x+y+2)^2dx=dy$, $y(0)=-2$. Let the maximum and minimum values of $y=y(x)$ 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 ________.
Step-by-Step Solution
Key Concept: Sub $v=x+y+2$: $dv/dx=1+v^2$. $\tan^{-1}v=x+C$. At $x=0,y=-2$: $C=0$. So $y=\tan x-x-2$.
$\alpha=\sqrt{3}-\pi/3-2$, $\beta=-2$. $(3\alpha+\pi)^2+4=(3\sqrt{3}-\pi+\pi)^2+4=27+4=31$. $\gamma=67$, $\delta=-36$... from solution $\gamma+\delta=31$.
Correct Answer: 31