A solution of a differential equation of first order in the large-variable limit The differential equation reads:

$  \dfrac{\partial R (t)}{\partial t} =   \dfrac{c_2}{R^2} + \dfrac{c_3}{R^3} + O(R^{-4})$,

Where $c2 > 
 0$ and $c3 > 0$, how to get the solution of the differential equation
for large $t$ and $R$?
 A: As Mann commented, this is a separable first order differential equation which write $$\frac{dt}{dR}=\frac{R^3}{c_2R+c_3}=-\frac{c_3^3}{c_2^3 (c_2 R+c_3)}+\frac{c_3^2}{c_2^3}-\frac{c_3 R}{c_2^2}+\frac{R^2}{c_2}$$ So, integration leads to $$t+C=-\frac{{c_3}^3 \log ({c_2} R+{c_3})}{{c_2}^4}+\frac{{c_3}^2
   R}{{c_2}^3}-\frac{{c_3} R^2}{2 {c_2}^2}+\frac{R^3}{3 {c_2}}$$ from which you cannot extract any analytical expression for $R(t)$.
Now, the asymptotic behavior of $R(t)$ may be given by the highest power of $R(t)$ which would mean that the $c_3$ term is ignored in the original equation, leading to $$R(t)\approx \sqrt[3]{C+3 {c_2} t}$$ but I am confess that I am feeling very uncomfortable with that.
A: Actually， this problem is from a paper (Phys. Rev. B Vol.34, p7845 (1986)) 
In the paper, the solution is given as
   R(t)=\sqrt[3]{3} \sqrt[3]{c_2 t}+\frac{c_3}{2c_2}+\frac{O}{\sqrt[3]{t}}

Following Claude Leibovici'post, I arrived this,
t = -\frac{c_3 R^2}{2 c_2^2}+\frac{c_3^2 R}{c_2^3}+\frac{c_3^3 \log \left(c_2 R+c_3\right)}{c_2^4}+\frac{R^3}{3 \text{c2}}=\frac{R^3-\frac{3 c_3 R^2}{2 c_2}}{3 \text{c2}}+O(R)=\frac{\left(R-\frac{c_3}{2 c_2^2}\right){}^3}{3

\text{c2}}+O(R)
Leaving the O(R), we came across, 
R = \sqrt[3]{3} \sqrt[3]{c_2 t}+\frac{c_3}{2 c_2}
Although this is the same with the paper, I don't know whether this is a reasonable approximation. Furthermore, how can I have an estimation of the deviation, i.e., the term O(t^-1/3) in the paper ?
Thank you for your advice.
