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Q.
A function $f( x )$ satisfying $\int\limits_0^1 f ( tx ) dt = n f( x )$, where $x >0$, is
Differential Equations
Solution:
$ \int\limits_0^1 f(t x) d t=n \cdot f(x)$
put $t x= y \Rightarrow dt =\frac{1}{ x } dy$
$\therefore \frac{1}{ x } \int\limits_0^{ x } f ( y ) dy = nf ( x )$
$\therefore \int\limits_0^{ x } f ( y ) dy = x \cdot n \cdot f ( x )$
Differentiating
$f(x)=n\left[f(x)+x f^{\prime}(x)\right] $
$f(x)(1-n)=n x f^{\prime}(x)$
$\therefore \frac{ f ^{\prime}( x )}{ f ( x )}=\frac{1- n }{ nx } $
$\text { Integrating } \ln f ( x )=\left(\frac{1- n }{ n }\right) \ln c x=\ln ( cx )^{\frac{1- n }{ n }} $
$\therefore f ( x )= c \cdot x ^{\frac{1- n }{ n }}$