If $2xf ' (x) - f(x) = 0$ find $f$ So $2xf '(x) - f(x) = 0$ and we know that $f(1) =1$. So I actually need to find the integral of $2xf'(x) - f(x)$.
Thanks.
 A: $$\frac{f'(x)}{f(x)}=\frac{1}{2x} \Rightarrow \ln|f(x)|=\frac{1}{2} \ln{x} +c \Rightarrow \ln|f(x)|= \ln{{x}^{\frac{1}{2}}} +c \Rightarrow f(x)= \pm c_1 \sqrt{x} \Rightarrow f(x)=C \sqrt{x}$$ 
$$f(1)=1 \Rightarrow 1= C $$
So $$f(x)=\sqrt{x}$$
A: Presume that $f$ takes positive values and set $g\left(x\right):=\ln f\left(x\right)$.
Then $g'\left(x\right)=\frac{f'\left(x\right)}{f\left(x\right)}=\frac{1}{2x}$
leading to $g\left(x\right)=\frac{1}{2}\ln\left|x\right|+c$ and consequently
$f\left(x\right)=d\left|x\right|^{\frac{1}{2}}$ for some positive
constant $d$. 
From $f\left(1\right)=1$ it follows that $d=1$ so
we end up with $f\left(x\right)=\left|x\right|^{\frac{1}{2}}$. 
It is too early to take this as a solution allready, since we made presumptions.
We will exploit the result as follows: 
Set $f\left(x\right)=h\left(x\right)\left|x\right|^{\frac{1}{2}}$
so that the case is solved if we manage to find $h\left(x\right)$.
Based on the original equation we reach the conclusion that
$h'\left(x\right)=0$. 
So $h$ is constant on $\left(0,\infty\right)$
and also on $\left(-\infty,0\right)$. Based on $f\left(1\right)=1$
we find that $h\left(1\right)=1$ and consequently $f\left(x\right)=x^{\frac{1}{2}}$
for $x>0$. But we cannot determine wich value is taken by $h$
on $\left(-\infty,0\right)$. 
In general we come to $$f\left(x\right)=\begin{cases}
\left|x\right|^{\frac{1}{2}}& \text{if }x>0\\
r\left|x\right|^{\frac{1}{2}} & \text{if }x<0\end{cases}$$
Here $r$ is an indetermined constant.
