Let f be a differentiable function in the interval [(0, )] such that f(1)=1 and [ _t x (t^…

Question

Let $f$ be a differentiable function in the interval $$(0, \infty)$$ such that $f(1)=1$ and $$\lim _\limits{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$$ for each $x>0$. Then $2 f(2)+3 f(3)$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 24

Accepted Answer

$\underset{t	o x}{lim}\frac{{t}^{2}f(x)-{x}^{2}f(t)}{(t-x)}=1\, (\frac{0}{0})$

$\begin{gathered} \underset{t	o x}{lim}\frac{2tf(x)-{x}^{2}{f}^{'}(t)}{1}=1 \ \Rightarrow 2xf(x)-{x}^{2}{f}^{'}(x)=1 \ \frac{dy}{dx}-\frac{2xy}{{x}^{2}}=\frac{-1}{{x}^{2}} \ \Rightarrow \frac{dy}{dx}-(\frac{2}{x})y=\frac{-1}{{x}^{2}} \ \Rightarrow ={e}^{\int \frac{-2}{x}dx}={e}^{-2\ln x}={x}^{-2}=\frac{1}{{x}^{2}} \end{gathered}$

$\begin{gathered} \Rightarrow y(\frac{1}{{x}^{2}})=\int (\frac{-1}{{x}^{2}})(\frac{1}{{x}^{2}})dx+C \ \frac{y}{{x}^{2}}=\frac{1}{3{x}^{3}}+Cx=1,y=1 \ \Rightarrow 1=\frac{1}{3}+C\Rightarrow C=\frac{2}{3} \ \Rightarrow y=\frac{1}{3x}+\frac{2}{3}{x}^{2}=f(x) \ \Rightarrow 2f(2)+3f(3)=24 \end{gathered}$

Solutions · 1