Question

A differentiable function f satisfies the relation f(xy) = xf(y) + yf(x) for all x, y in R⁺. If f(1) = 1, find f(x).

Answer 1

Let x = 1 = y, then f(1) = 0

Now, f(x(1 + h)) = xf(1 + h) + (1 + h)f(x)

Integrating, we get

v = log |x| + c

y/x = log |x| + c

when x = 1, y = 0, then c = 0

Thus, y/x = log |x|

y = x log |x|

which is the required solution.