Question

Let \(\alpha |x| = |y| e^{xy - \beta}\), \(\alpha, \beta \in \mathbb{N}\) be the solution of the differential equation \[ xdy - ydx + xy(xdy + ydx) = 0, \quad y(1) = 2. \] Then \(\alpha + \beta\) is equal to _.

Answer 1

Correct Answer: 4

Consider the given differential equation:
\[xdy - ydx + xy(xdy + ydx) = 0.\]
Rearranging terms:
\[(x + xy)dy = (y - xy)dx.\]
Dividing both sides by \(xy\):
\[\frac{dy}{dx} = \frac{y - xy}{x + xy}.\]
Given that \(\alpha|x| = |y|e^{xy - \beta}\), substituting the initial condition \(y(1) = 2\) into the expression:
\[\alpha|1| = |2|e^{1 \cdot 2 - \beta}.\]
Simplifying:
\[\alpha = 2e^{2 - \beta}.\]
Since \(\alpha, \beta \in \mathbb{N}\), assume values for \(\beta\) such that \(\alpha\) is an integer. Let \(\beta = 2\):
\[\alpha = 2e^0 = 2.\]
Calculating \(\alpha + \beta\):
\[\alpha + \beta = 2 + 2 = 4.\]
Answer: 4.