Question

Let $y = y(x)$ be the solution of the differential equation $(x + y + 2)^2 \, dx = dy, \quad y(0) = -2.$ 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 $(3\alpha + \pi)^2 + \beta^2 = \gamma + \delta\sqrt{3}, \quad \gamma, \delta \in \mathbb{Z},$ then $\gamma + \delta$ equals $\dots$.

Answer 1

Given:

$\frac{dy}{dx} = (x + y + 2)^2, \quad y(0) = -2.$

Step 1: Substitute $v = x + y + 2$:

Let $x + y + 2 = v$. Then, $1 + \frac{dy}{dx} = \frac{dv}{dx}.$ From the differential equation, $\frac{dv}{dx} = 1 + v^2.$

Step 2: Separate Variables and Integrate:

$\int \frac{dv}{1 + v^2} = \int dx.$ We have: $\tan^{-1}(v) = x + C.$

Step 3: Apply Initial Condition:

At $x = 0, y = -2$, so $v = 0$, giving $C = 0$. Thus, $\tan^{-1}(x + y + 2) = x,$ or $y = \tan x - x - 2.$

Step 4: Determine $f_{\text{min}}$ and $f_{\text{max}}$ on $[0, \frac{\pi}{3}]$:

$f(x) = \tan x - x - 2, \quad x \in \left[0, \frac{\pi}{3}\right].$ We find $f'(x) = \sec^2 x - 1 > 0$, so $f(x)$ is increasing in the interval. $f_{\text{min}} = f(0) = -2 = \beta,$ $f_{\text{max}} = f\left(\frac{\pi}{3}\right) = \sqrt{3} - \frac{\pi}{3} - 2 = \alpha.$

Step 5: Calculate $(3\alpha + \pi)^2 + \beta^2$:

$(3\alpha + \pi)^2 + \beta^2 = \left(3\left(\sqrt{3} - \frac{\pi}{3} - 2\right) + \pi\right)^2 + (-2)^2.$ Simplifying, we get: $\gamma + \delta\sqrt{3} = 67 - 36\sqrt{3}.$ Therefore, $\gamma = 67$ and $\delta = -36$.

Step 6: Calculate $\gamma + \delta$:

$\gamma + \delta = 67 - 36 = 31.$

Answer: 31