Particle slightly disturbed radially in stable circular motion under central force 
A particle of mass $m$, moves around a central force whose potential is $$V(r) = kmr^3 \; (k>0),$$ 
  then
  $(c)$ if the particle is slightly disturbed from the circular motion, what is the period of small radial oscillation about $r=a$?

I have the solution but I am not understanding it. It says that the equation of motion to the first order in $x$ will be 
$$m \ddot x + \left( \frac{3V'(a)}{a} + V''(a)\right)x = 0$$
Where $a$ is the radius of stable orbit when the particle will be in circular motion, and $r$ being radial distance, $r = a + x(t)$. Taylor expansion of $$V(a+x) = V(a) + V'(a)x + \frac 1 2 V''(a) x^2 + \dots $$ has been used. I don't understand the logic behind it, I am not even sure if this is correct, kind of looks like hooks law. Please explain if it's correct and provide hints to correct answer if it's wrong.
The final answer is given as $\sqrt{15ka}$, which is angular frequency of the above equation.
 A: We have : 
$\vec r = (a+x) \vec e_r$
$\dot{\vec{ r}} = (a+x) \dot \theta \vec e_\theta + \dot x \vec e_r$
$\ddot{\vec{ r}} = - (a+x) \dot \theta^2 \vec e_r + (a+x) \ddot \theta  \vec e_\theta+  \dot x \dot \theta \vec e_\theta +  \dot x \dot \theta \vec e_\theta +\ddot x\vec e_r $
Finally : 
$$\ddot{\vec{ r}} = ( \ddot x - (a+x) \dot \theta^2 ) \vec e_r + ((a+x) \ddot \theta + 2\dot x \dot \theta) \vec e_\theta \quad\quad\quad\quad (1)$$
The equations of movement are : 
$$m(\ddot x - (a+x) \dot \theta^2 ) = - V'(a) - x V''(a) \quad \quad \quad\quad (2)$$
$$(a+x) \ddot \theta + 2\dot x \dot \theta = 0 \quad \quad \quad \quad \quad  \quad \quad \quad\quad  (3)$$
We know suppose that the angular velocity is slowly varying with $x$, with the law : 
$$\dot \theta = \dot \theta_0(1 +\gamma x) \quad \quad \quad \quad \quad  \quad \quad \quad\quad  (4)$$ 
We now plug this law in equation (3) and keeping only the term in first order in $x, \dot x$, which gives : 
$$a\dot \theta_0 \gamma \dot x + 2 \dot x \dot \theta_0 = 0\quad \quad \quad \quad \quad  \quad \quad \quad\quad  (5)$$
So, finally $\gamma = \frac{-2}{a}$, and :$\dot \theta = \dot \theta_0(1 - 2 \frac{x}{a})$. Getting the square of the last expression, and keeping only the first term is x, we get :  
$$\dot \theta^2 = \dot \theta_0^2 (1-  4 \frac{x}{a})\quad \quad \quad \quad \quad  \quad \quad \quad\quad  (6)$$
Applying equation $(2)$ with $x = 0$, we get : 
$$ma\dot \theta_0^2 = V'(a) \quad \quad \quad \quad \quad  \quad \quad \quad\quad  (7)$$
Finally, from equation $(2),(6)$, we get : 
$$m(\ddot x - (a+x)(1-  4 \frac{x}{a}) \dot \theta_0^2)  = - V'(a) - x V''(a) \quad \quad \quad\quad (8)$$
Keeping only the first order in $x$, we get :
$$m\ddot x - ma \dot \theta_0^2 + 3xm \dot \theta_0^2  = - V'(a) - x V''(a) \quad \quad \quad\quad (9)$$
Finally, using (7), we get : 
$$m\ddot x + (3 \frac{V'(a)}{a} +V''(a))x = 0\quad \quad \quad\quad (10)$$
With $V(a)= kmr^3$, this gives : 
$$\ddot x + (15ka)x = 0\quad \quad \quad\quad (11)$$
