IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

doucmecomin

let f(x) be diffn on the interval ]0.

[ such that f(1) = 1 , and _{[ t]}

_{[ x]} t²f(x) - x²f(t)/t-x = 1 for each x>0 . then f(x) is?

raulrag009

Here is my solution

$\lim_{t-x}{\frac{t^{2}f(x)-x^{2}f(t)}{t-x}}=1\\\\ Apply\;L-Hospital \;rule\\\\ u\;get\\\\ \lim_{t-x}{2tf(x)-x^{2}f(t)}=1\\\\ 2xf(x)-x^{2}f^{1}(x)=1\\\\ Let\;f(x)=y\quad\;f^{1}{x}=\frac{dy}{dx}\\\\ \frac{dy}{dx}-2\frac{y}{x}=-\frac{1}{x^{2}}\\\\ Integrating\;factor\\\\ =x^{-2}\\\\ \frac{dy}{x^{2}dx}-\frac{2y}{x^{3}}=-\frac{1}{x^{4}}\\\\ d[y.(\frac{1}{x^{2}})]=\frac{1}{3x^{3}}+C\\\\ \frac{y}{x^{2}}=\frac{1}{3x^{3}}+C\\\\ At\;x=1\;y=1\\\\ C=\frac{2}{3}\\\\ hence\\\\ y=\frac{1}{3x}+\frac{2}{3x^{2}}\\\\$

pls forgive for calculation mistakes

karthik2007

His approach is correct, don't know about the final answer

doucmecomin

its correct

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