How prove $f'(x)$ must bounded 
let $f(x)$ have two derivatives,and $c$ is constant, for all real $x$ such 
  $$f(x)(x^2-1)+xf'(x)=c$$
show that:$f'(x)$ is  bounded function,In other words: there exist  $M>0$,such
  $|f'(x)|\le M,x\in R$

My try:if we take the ODE

$$y(x^2-1)+x\dfrac{dy}{dx}=c$$
  then
  $$\dfrac{dy}{dx}+\dfrac{x^2-1}{x}y=\dfrac{c}{x}$$
  then we have
  $$y=e^{-\int \dfrac{x^2-1}{x}dx}\left(\int \dfrac{c}{x}e^{\int \dfrac{x^2-1}{x}dx}dx+C\right)$$
  then we have
  $$y=\dfrac{x}{e^{\frac{1}{2}x^2}}\left(\int\dfrac{ce^{\frac{1}{2}x^2}}{x^2}dx+C\right)$$

Then I can't,Thank you evryone can help.maybe My try is not good,I think have other idea.Thank you very much!
 A: I agree with where you've gotten so far. Let's make some notes so that I can follow your steps:
1) The solution of $f'(x) = a(x) f(x) + b(x)$ with initial condition $f(x_0)$ is
$$
f(x) = e^{\int_{x_0}^x a(s)ds}\left( f(x_0)   + \int_{x_0}^x b(s) e^{-\int_{x_0}^s a(t)dt} ds \right)
$$
2) In this case $a(x) = \frac{1-x^2}{x} = \frac{1}{x} - x$ and $b(x) = \frac{c}{x}$.
3)  $\int_{x_0}^x a(s)ds = \ln(x) - \frac{1}{2}x^2 + \text{const}$, so
$$
e^{\epsilon \int_{x_0}^x a(s)ds} = x^{\epsilon}e^{-\frac{\epsilon}{2}x^2}
$$
where I'm thinking $\epsilon = \pm 1$
4) Putting this together, you have
$$
f(x) = \kappa_1 xe^{-\frac{1}{2}x^2}\left( f(x_0) + \int_{x_0}^x \frac{c}{s^2}e^{\frac{1}{2}s^2} ds \right)
$$
where $\kappa_1$ is just some constant. Now I believe we're where you left off.
Let's move forward by the reasonable assumption that $0 << |x_0|<|x|$ and sign($x_0$)$=$sign($x$). From here let's get some feel for the integral term
$$
\int_{x_0}^x s^{-2}e^{s^2/2}ds = \int_{x_0}^x s^{-3}(s e^{s^2/2})ds = x^{-3}e^{x^2/2} + 3\int_{x_0}^xs^{-4}e^{s^2/2}ds+\kappa_2
$$
The last equality came from an integration by parts with $u=s^{-3}$ and $dv = se^{s^2/2}$, and as before $\kappa_2$ is just some constant. We will use the following observation
$$
\left| e^{-x^2/2} \int_{x_0}^x h(s) e^{s^2/2} ds \right| \leq e^{-\frac{1}{2}\big(x^2 - \frac{(x-x_0)^2}{2}\big)} \int_{x_0}^{(x-x_0)/2} |h(s)| ds + \int_{(x-x_0)/2}^x |h(s)|ds \\
\leq e^{-x^2/4}\int_{x_0}^x |h(s)|ds + \int_{(x-x_0)/2}^x |h(s)|ds.
$$
The last inequality wasn't necessary, but just made things look nicer. Putting these pieces back together
$$
|f(x)| \leq C_1 |x| e^{-x^2/2} + C_2 |x|^{-2}+C_3 |x| e^{-x^2/4}\int_{x_0}^x s^{-4} ds  + C_4 |x| \int_{(x-x_0)/2}^x s^{-4}ds \\ \leq C_1 |x|e^{-x^2/2} + C_2 |x| e^{-x^2/4} + C_3 |x|^{-2}
$$
(we let our constants $C_i$ change from the first inequality to the second, but they are still constants independent of $x$).
Plugging our expression for $f(x)$ back into the equation $f'(x) = a(x) f(x) + b(x)$ we get 
$$
|f'(x)| \leq  \left( \frac{1}{|x|} + |x| \right) |f(x)| + \frac{|c|}{|x|} 
$$ 
After just a mild bit of more work, you will see that the right hand side is bounded as $|x| \to \infty$. Now, you only need to look at $|x| > |x_0| >> 0$ since by assumption $f''(x)$ exists, implying that $f'(x)$ is continuous and therefore will be bounded on any compact set (in particular one of the type $\{x : |x| \leq R\}$ for large $R$).
