Let $y=y(x)$ be the solution of the differential equation $(x+y+2)^2 d x=d y, y(0)=-2$. Let the maximum and minimum values of the function $y=y(x)$ in $\left[0, \frac{\pi}{3}\right]$ be $\alpha$ and $\beta$, respectively. If $(3 \alpha+\pi)^2+\beta^2=\gamma+\delta \sqrt{3}, \gamma, \delta \in…

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Let $y=y(x)$ be the solution of the differential equation $(x+y+2)^2 d x=d y, y(0)=-2$. Let the maximum and minimum values of the function $y=y(x)$ in $\left[0, \frac{\pi}{3}\right]$ 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 ______

Aicharya · Accepted Answer

$\frac{d y}{d x}=(x+y+2)^2$ ...(1) $y(0)=-2$
Let $\mathrm{x}+\mathrm{y}+2=\mathrm{v}$ $1+\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dx}}$ from (1) $\frac{d v}{d x}=1+v^2$ $\begin{aligned} & \int \frac{d v}{1+v^2}=\int d x \\ & \tan ^{-1}(v)=x+C \\ & \tan ^{-1}(x+y+2)=x+C \\ & \text { at } x=0 y=-2 \Rightarrow C=0 \end{aligned}$ $\begin{aligned} & \Rightarrow \tan ^{-1}(\mathrm{x}+\mathrm{y}+2)=\mathrm{x} \\ & \mathrm{y}=\tan \mathrm{x}-\mathrm{x}-2 \\ & \mathrm{f}(\mathrm{x})=\tan \mathrm{x}-\mathrm{x}-2, \mathrm{x} \in\left[0, \frac{\pi}{3}\right] \end{aligned}$ $\begin{aligned} & f^{\prime}(x)=\sec ^2 x-1>0 \Rightarrow f(x) \uparrow \\ & f_{\min }=f(0)=-2=\beta \\ & f_{\max }=f\left(\frac{\pi}{3}\right)=\sqrt{3}-\frac{\pi}{3}-2=\alpha \end{aligned}$ now $(3 \alpha+\pi)^2+\beta^2=\gamma+\delta \sqrt{3}$ $\begin{aligned} & \Rightarrow(3 \alpha+\pi)^2+\beta^2=(3 \sqrt{3}-6)^2+4 \\ & \gamma+\delta \sqrt{3}=67-36 \sqrt{3} \\ & \Rightarrow \gamma=67 \text { and } \delta=-36 \Rightarrow \gamma+\delta=31 \end{aligned}$

Let $y=y(x)$ be the solution of the differential equation $(x+y+2)^2 d x=d y, y(0)=-2$. Let the maximum and…

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