Step 1: Rewrite the equation
The given equation is: \[ x \log x \frac{dy}{dx} + y = 2 \log x. \] Rearranging: \[ \frac{dy}{dx} + \frac{y}{x \log x} = \frac{2}{x \log x}. \]
Step 2: Check the form of the equation
This is a first-order linear differential equation of the form: \[ \frac{dy}{dx} + P(x)y = Q(x), \] where \( P(x) = \frac{1}{x \log x} \) and \( Q(x) = \frac{2}{x \log x} \).
Step 3: Conclude the result
The equation is a first-order linear differential equation.
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.