We are given the differential equation:
\[ xy \frac{dy}{dx} = 1 + x + y + xy \]
Rearranging the terms to separate variables:
\[ xy \frac{dy}{dx} = (1 + x) + y(1 + x) \]
Now, simplifying the equation:
\[ \frac{dy}{dx} = \frac{(1 + x) + y(1 + x)}{xy} \]
We can factor out \( (1 + x) \) from the numerator:
\[ \frac{dy}{dx} = \frac{(1 + x)(1 + y)}{xy} \]
Separating the variables \( x \) and \( y \), we get:
\[ \frac{dy}{1 + y} = \frac{(1 + x)}{x} \cdot \frac{dx}{y} \]
Integrating both sides, we get the solution:
\[ \log(x(1 + y)) = c \]
This matches option (A).
Let \( y = f(x) \) be the solution of the differential equation\[\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^6 + 4x}{\sqrt{1 - x^2}}, \quad -1 < x < 1\] such that \( f(0) = 0 \). If \[6 \int_{-1/2}^{1/2} f(x)dx = 2\pi - \alpha\] then \( \alpha^2 \) is equal to ______.
Let \( y = y(x) \) be the solution of the differential equation \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] such that \( y(0) = \frac{5}{4} \). Then \[ 12 \left( y\left( \frac{\pi}{4} \right) - e^{-2} \right) \] is equal to _____.