Question

A function f(x) is given by the equation x²f'(x) + 2x f(x) – x + 1 = 0, where x ≠ 0. If f(1) = 0, then find the interval of the monotonocity of the function f.

Answer 1

We have x²f'(x) + 2xf(x) – x + 1 = 0

Put y = f(x)

Thus, x²dy/dx + 2xy = x – 1

⇒ d/dx(x²y) = x – 1

On integration, we get,

x²y = x²/2 – x + c

when x = 1, y = 0 , then c = 1/2

Thus, x²y = x²/2 – x + 1/2

⇒ y = 1/2 – 1/x + 1/2x²

⇒ dy/dx = 1/x² – 1/x³ = (x – 1)/x³

So, by the sign scheme, f(x) is increasing in x ∈(–∞, 0) ∪ (1, ∞) and decreasing in x ∈ (0, 1)