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)
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Answer:
Correct Answer: 19. (a)
Solution:
- 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, $\quad f(x)=\frac{2}{3} x^{2}+\frac{1}{3 x}$
