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Mathematics Question on Differential equations

Let f(x)f(x) be differentiable on the interval (0,∞)(0, \infty) such that f(1)=1f(1) = 1 and limt→xt2f(x)−x2f(t)t−x=1lim_{t \rightarrow x} \frac {t^2f(x)-x^2f(t)}{t-x} =1 for each x>0x>0 . Then, f(x)f(x) is

A

13x+2x23\frac{1}{3x} + \frac{2x^2}{3}

B

−13x+4x23-\frac{1}{3x} + \frac{4x^2}{3}

C

−1x+2x2-\frac{1}{x} + \frac{2}{x^2}

D

1x\frac{1}{x}

Answer

13x+2x23\frac{1}{3x} + \frac{2x^2}{3}

Explanation

Solution

Given, lim⁡t→xt2f(x)−x2f(t)t−x=1\displaystyle\lim_{t\to x} \frac{t^{2}f\left(x\right)-x^{2}f\left(t\right)}{t-x}=1
Apply L' Hospital rule lim⁡t→x2t f(x)−x2f′(t)1=1\displaystyle\lim_{t\to x} \frac{2t\,f \left(x\right)-x^{2}f'\left(t\right)}{1}=1
⇒2xf(x)−x2f′(x)=1\Rightarrow 2x f \left(x\right)-x^{2}f'\left(x\right)=1
⇒x2f′(x)−2xf(x)=−1\Rightarrow x^{2}f' \left(x\right)-2x f\left(x\right)=-1
⇒f′(x)−2xf(x)=−1x2\Rightarrow f'\left(x\right)-\frac{2}{x} f \left(x\right)=-\frac{1}{x^{2}}
It is a linear differential equation
∴IF=e∫2xdx\therefore IF =e^{\int \frac{2}{x}dx}
=e−2 log x=e^{-2\,log\,x}
=elog(1x2)=e^{log \left(\frac{1}{x^{2}}\right)}
=1x2=\frac{1}{x^{2}}
∴\therefore Solution is
f(x)×1x2=∫−1x2×1x2dx+Cf \left(x\right)\times\frac{1}{x^{2}}=\int-\frac{1}{x^{2}}\times\frac{1}{x^{2}} dx+C
f(x)×1x2=13x3+Cf \left(x\right)\times\frac{1}{x^{2}}=\frac{1}{3x^{3}}+C
⇒f(x)=13x+Cx2\Rightarrow f\left(x\right)=\frac{1}{3x}+Cx^{2}
At f(1)=1f\left(1\right)=1
⇒1=13+C\Rightarrow 1=\frac{1}{3}+C
⇒C=23\Rightarrow C=\frac{2}{3}
∴f(x)=13x+23x2\therefore f\left(x\right)=\frac{1}{3x}+\frac{2}{3}x^{2}