Question

Let \(y = y(x)\) be the solution of the differential equation
\(\left((x + 2)e^{\left(\frac{y + 1}{x + 2}\right)} + (y + 1)\right)dx = (x + 2)\,dy\), \(y(1) = 1\).
If the domain of \(y = y(x)\) is an open interval \((\alpha, \beta)\), then \(|\alpha + \beta|\) is equal to _____________.

Hint:

Put \(t = \frac{y + 1}{x + 2}\) to separate variables, use \(y(1) = 1\), and find where \(e^{-t} \gt 0\) holds; the domain is the interval \(|x + 2| \lt 3e^{e^{-2/3}}\).

Answer 1

Correct Answer: 4

The equation is \(\frac{dy}{dx} = e^{\frac{y+1}{x+2}} + \frac{y+1}{x+2}\). Let \(t = \frac{y + 1}{x + 2}\), so \(y + 1 = t(x + 2)\) and \(\frac{dy}{dx} = t + (x + 2)\frac{dt}{dx}\).
Then \((x + 2)\frac{dt}{dx} = e^t\), so \(e^{-t}dt = \frac{dx}{x + 2}\).
Integrating: \(-e^{-t} = \ln|x + 2| + C\).
At \(x = 1, y = 1\): \(t = \frac23\), so \(-e^{-2/3} = \ln 3 + C\), giving \(C = -\ln 3 - e^{-2/3}\).
Therefore \(e^{-t} = \ln 3 - \ln|x + 2| + e^{-2/3}\).
Since \(e^{-t} \gt 0\), we need \(\ln|x + 2| \lt \ln 3 + e^{-2/3}\), i.e. \(|x + 2| \lt 3e^{e^{-2/3}} = K\).
Taking the domain as \(-K \lt x + 2 \lt K\), i.e. \((\alpha, \beta) = (-2 - K, -2 + K)\), we get \(\alpha + \beta = -4\) and \(|\alpha + \beta| = 4\).
Answer: 4