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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.

5xum
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Desepted
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  • I advice you to add the restriction $x>0$ to make things less complicated. Doing so also makes it more 'acceptable' that you accepted the answer of Mary. – drhab Apr 25 '14 at 11:26

2 Answers2

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$$\frac{f'(x)}{f(x)}=\frac{1}{2x} \Rightarrow \ln|f(x)|=\frac{1}{2} \ln{x} +c \Rightarrow \ln|f(x)|= \ln{{x}^{\frac{1}{2}}} +c \Rightarrow f(x)= \pm c_1 \sqrt{x} \Rightarrow f(x)=C \sqrt{x}$$ $$f(1)=1 \Rightarrow 1= C $$ So $$f(x)=\sqrt{x}$$

Mary Star
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    No. You should replace $\frac{1}{2} \ln{x}$ with $\frac{1}{2} \ln{x}+c$ and continue from there. This is only correct now because f(1)=1. Imagine f(1)=e and check the second equality... – Thanos Darkadakis Apr 25 '14 at 10:03
  • This is not complete. Note that more generally function $f\left(x\right)=\left|x\right|^{\frac{1}{2}}$ will do so that the restriction $x\ne 0$ (instead of restriction $x>0$) is enough. – drhab Apr 25 '14 at 10:37
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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.

drhab
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