Let y = y ( x ) be a differentiable function in the interval ( 0 , ) such that y ( 1 ) = 2 , and lim t x ( t 2 y ( x ) - x 2 y ( t ) x - t ) = 3 for each x 0 . Then 2 y ( 2 ) is equal to [2026]

Question

Let $y = y (x)$ be a differentiable function in the interval $(0, \infty)$ such that $y (1) = 2$ , and $\lim_{t \to x} (\frac{t^{2} y (x) - x^{2} y (t)}{x - t}) = 3$ for each $x > 0$ . Then $2 y (2)$ is equal to [2026]

Smart Achievers · Accepted Answer

(1)

$\lim_{t \to x} \frac{2 t f (x) - x^{2} f' (t)}{- 1} = 3$

$x^{2} f' (x) - 2 x f (x) = 3$

$\frac{d y}{d x} - \frac{2 y}{x} = \frac{3}{x^{2}}$

$I.F. = e^{\int \frac{- 2}{x} d x} = e^{- 2 \log_{e} x} = \frac{1}{x^{2}}$

$y \cdot \frac{1}{x^{2}} = \int \frac{3}{x^{4}} d x$

$\frac{y}{x^{2}} = - \frac{1}{x^{3}} + c \Rightarrow y = c x^{2} - \frac{1}{x} = f (x)$

$f (1) = 2 = c - 1 \Rightarrow c = 3$

$f (x) = 3 x^{2} - \frac{1}{x}$

$f (2) = 12 - \frac{1}{2} \Rightarrow 2 f (2) = 23$

Solution

Q.
Let $y = y (x)$ be a differentiable function in the interval $(0, \infty)$ such that $y (1) = 2$ , and $\lim_{t \to x} (\frac{t^{2} y (x) - x^{2} y (t)}{x - t}) = 3$ for each $x > 0$ . Then $2 y (2)$ is equal to [2026]

1 23

2 27

3 18

4 12

Want Us to Email you About Special Offers & Updates?