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.

A ladder of fixed length \( h \) is to be placed along the wall such that it is free to move along the height of the wall.
Based upon the above information, answer the following questions:
(iii) (b) If the foot of the ladder, whose length is 5 m, is being pulled towards the wall such that the rate of decrease of distance \( y \) is \( 2 \, \text{m/s} \), then at what rate is the height on the wall \( x \) increasing when the foot of the ladder is 3 m away from the wall?