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

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.