Let f(x) be differentiable on the interval (0 | vedclass

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(A) Given the limit $\lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$. Applying $L$'$H$ôpital's rule with respect to $t$: $\lim _{t \rightarro

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$\frac{1}{3x} + \frac{2x^2}{3}$

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(A) Given the limit $\lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$. Applying $L$'$H$ôpital's rule with respect to $t$: $\lim _{t \rightarrow x} \frac{2t f(x) - x^2 f'(t)}{1} = 1$. Substituting $t=x$,we get $2x f(x) - x^2 f'(x) = 1$. Rearranging the terms: $f'(x) - \frac{2}{x} f(x) = -\frac{1}{x^2}$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{2}{x}$ and $Q(x) = -\frac{1}{x^2}$. The integrating factor $IF = e^{\int P(x) dx} = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = \frac{1}{x^2}$. The solution is $f(x) \cdot \frac{1}{x^2} = \int (-\frac{1}{x^2}) \cdot \frac{1}{x^2} dx = \int -x^{-4} dx = \frac{x^{-3}}{3} + C = \frac{1}{3x^3} + C$. Thus,$f(x) = \frac{1}{3x} + Cx^2$. Given $f(1) = 1$,we have $1 = \frac{1}{3} + C$,which implies $C = \frac{2}{3}$. Therefore,$f(x) = \frac{1}{3x} + \frac{2x^2}{3}$.

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

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