Differential Equations 2 Question 18

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

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

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

(c) $-\frac{1}{x}+\frac{2}{x^{2}}$

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

(2007, 3M)

Show Answer

Answer:

Correct Answer: 19. (a)

Solution:

  1. Given, $\quad \lim _{t \rightarrow x} \frac{t^{2} f(x)-x^{2} f(t)}{t-x}=1$

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

$\Rightarrow \quad \frac{x^{2} f^{\prime}(x)-2 x f(x)}{\left(x^{2}\right)^{2}}+\frac{1}{x^{4}}=0$

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

On integrating both sides, we get

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

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

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



NCERT Chapter Video Solution

Dual Pane