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).
The variance for continuous probability function \(f(x) = x^2 e^{-x}\) when \(x \ge 0\) is
Consider the loop transfer function \(\frac {K(s+6)}{(s+3)(s+5)}\). In the root locus diagram the centroid will be located at:
When nuclear radiations pass through, gas ionization is produced. This is the principle of which of the following detectors?
If \(f = \text{Tan}^{-1}(xy)\) then \((\frac{\partial f}{\partial x})_{(1,2)}\) = _____ .