Question

Solution of the differential equation \(\frac{dy}{dx} = \frac{y(x-y ln⁡y)}{x(x ln x-y)}\) is _____________

(a) \(\frac{x ln⁡x+y ln⁡y}{xy} = c\)

(b) \(\frac{x ln⁡x-y ln⁡y}{xy} = c\)

(c) \(\frac{ln⁡x}{x} + \frac{ln⁡y}{y} = c\)

(d) \(\frac{ln⁡x}{x} – \frac{ln⁡y}{y} = c\)

This question was addressed to me during a job interview.

Question is from Separable and Homogeneous Equations topic in division Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Right choice is (a) \(\frac{x ln⁡x+y ln⁡y}{xy} = c\)

Best explanation: \(\frac{dy}{dx} = \frac{y(x-y ln⁡y)}{x(x ln x-y)}\)

–> x^2 ln⁡x dy-xy dy=xy dx – y^2  ln⁡y dx …….dividing by x^2 y^2 then

\(\frac{ln⁡y}{y^2} \,dy\, – \frac{1}{xy} \,dy\, = \frac{1}{xy} \,dx\, – \frac{ln⁡y}{x^2}  \,dx\)

\((ln x(\frac{1}{-y^2}dy) + \frac{1}{xy} \,dx) + (ln y(\frac{1}{-x^2}dx) + \frac{1}{xy} \,dy) = 0 \)

\(d(\frac{ln⁡x}{y}) + d(\frac{ln⁡y}{x}) = 0\)

on integrating we get

\(\int d(\frac{ln⁡x}{y}) + \int d(\frac{ln⁡y}{x})\)

\( \frac{ln⁡x}{y} + \frac{ln⁡y}{x}\) = c…. where c is a constant of integration.