Mathematics - Differential Equation Question with Solution | TestHub

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