Let $ y = y(x) $ be a differentiable function in the interval $ (0, \infty) $ such that $ y(1) = 2 $.
and $ \lim_{t \to x} \left( \frac{t^2 y(x) - x^2 y(t)}{x - t} \right) = 3 $ for each $ x > 0 $.
Then $ 2y(2) $ is equal to
🎥 Video solution / Text Solution of this question is given below:
$ \lim_{t \to x} \frac{2t f(x) - x^2 f(t)}{-1} = 3 $
$ x^2 f'(x) - 2x f(x) = 3 $
$ \frac{dy}{dx} - \frac{2y}{x} = \frac{3}{x^2} $
I.F. $ = e^{\int \frac{-2}{x} dx} = e^{-2\log x} = \frac{1}{x^2} $
$ \frac{y}{x^2} = \int \frac{3}{x^4} dx $
$ \frac{y}{x^2} = -\frac{1}{x^3} + c \Rightarrow y = cx^2 - \frac{1}{x} $
$ f(1) = 2 = c - 1 \Rightarrow c = 3 $
$ f(x) = 3x^2 - \frac{1}{x} $
$ f(2) = 12 - \frac{1}{2} $
$ 2f(2) = 23 $