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

We are given the equation:
\[ a|x| = |y| e^{xy - \beta}, \quad a, b \in \mathbb{N} \]

Step 1: Differentiate both sides
\[ x dy - y dx + xy(x dy + y dx) = 0 \] Dividing throughout by \(x y\):
\[ \frac{dy}{y} - \frac{dx}{x} + (x dy + y dx) = 0 \]

Step 2: Integrate both sides
\[ \ln|y| - \ln|x| + xy = c \]

Step 3: Using the condition \( y(1) = 2 \)
Substitute \(x = 1, y = 2\):
\[ \ln|2| - 0 + 2 = c \] \[ c = 2 + \ln 2 \]

Step 4: Substituting the value of \( c \)
\[ \ln|y| - \ln|x| + xy = 2 + \ln 2 \] \[ \ln|x| = \ln\left|\frac{y}{2}\right| - 2 + xy \]

Step 5: Simplifying the equation
\[ |x| = \left|\frac{y}{2}\right| e^{xy - 2} \] \[ 2|x| = |y| e^{xy - 2} \]

Step 6: Comparing with the given form
\[ a|x| = |y| e^{xy - \beta} \] Thus, comparing both sides:
\[ a = 2, \quad \beta = 2 \]

Step 7: Final result
\[ \alpha + \beta = 2 + 2 = 4 \]

Final Answer:
\[ \boxed{\alpha + \beta = 4} \]

Answer 2

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.