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

Correct Answer: 31

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