SATHEE: Differential Equations 2 Question 18

Differential Equations 2 Question 18

19. Let $f (x)$ be differentiable on the interval $(0, \infty)$ such that $f (1) = 1$ , and $lim_{t \to x} \frac{t^{2} f (x) - x^{2} f (t)}{t - x} = 1$ for each $x > 0$ . Then, $f (x)$ is

(a) $\frac{1}{3 x} + \frac{2 x^{2}}{3}$

(b) $- \frac{1}{3 x} + \frac{4 x^{2}}{3}$

(d) $\frac{1}{x}$

(2007, 3M)

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Answer:

Correct Answer: 19. (a)

Solution:

Given, $lim_{t \to x} \frac{t^{2} f (x) - x^{2} f (t)}{t - x} = 1$

$\Rightarrow x^{2} f^{'} (x) - 2 x f (x) + 1 = 0$

$\Rightarrow \frac{x^{2} f^{'} (x) - 2 x f (x)}{{(x^{2})}^{2}} + \frac{1}{x^{4}} = 0$

$\Rightarrow \frac{d}{d x} \frac{f (x)}{x^{2}} = - \frac{1}{x^{4}}$

On integrating both sides, we get

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

Also, $f (1) = 1, c = \frac{2}{3}$

Hence, $f (x) = \frac{2}{3} x^{2} + \frac{1}{3 x}$

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Differential Equations 2 Question 18

19. Let f(x) be differentiable on the interval (0,∞) such that f(1)=1, and limt→xt2f(x)−x2f(t)t−x=1 for each x>0. Then, f(x) is

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19. Let $f (x)$ be differentiable on the interval $(0, \infty)$ such that $f (1) = 1$ , and $lim_{t \to x} \frac{t^{2} f (x) - x^{2} f (t)}{t - x} = 1$ for each $x > 0$ . Then, $f (x)$ is