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

Question

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 $$\beta$$

Accepted Answer

$\begin{gathered} \frac{dy}{dx}=(x+y+z{)}^{2} \ x+y+z=t \ \Rightarrow 1+\frac{dy}{dx}=\frac{dt}{dx} \ \Rightarrow \frac{dt}{dx}-1={t}^{2} \ \Rightarrow \frac{dt}{1+{t}^{2}}=dx\Rightarrow {	an }^{-1}t=x+c \ \Rightarrow x+y+z=	an (x+c) \ \Rightarrow y(x)=	an (x+c)-x-2 \ \because y(0)=-2\Rightarrow -2=	an c-0-2 \ \, \Rightarrow c=0 \ \Rightarrow y(x)=	an x-x-2 \ \frac{dy}{dx}={\sec }^{2}x-1\ge 0 \end{gathered}$

$\Rightarrow y(x)$ is increasing if $x\in (0,\frac{\pi }{3})$

$\begin{gathered} \Rightarrow \alpha =y(\frac{\pi }{3}),\beta =y(0) \ \Rightarrow \alpha =-\frac{\pi }{3}-2+\sqrt{3}\beta =-2 \end{gathered}$

Now, $(3\alpha +\pi {)}^{2}+{\beta }^{2}=(6+3\sqrt{3}{)}^{2}+(-2{)}^{2}$

$=67-36\sqrt{3}=y+\delta \sqrt{3}$.

$\Rightarrow \gamma +\delta =31$

Solutions · 1