in the following equation
$$ax+bx^{5/3} = c$$
$a, b$, and $c$ are constant. Through numerical result I know that $x$ is near zero and positive. I want to find an analytical solution or approximation. I really appreciate if you could help me through this matter.
On
As comments say, you can write this as $a u^3 + b u^5 = c$, where $u = x^{1/3}$. This has a real, positive root (if $a, b, c$ are positive, that is). If $u$ is near 1, one (or a few) iterations of Newton's method starting at $u_0 = 1$ should give a precise value. I.e.,
$$u_1 = \frac{2 a + 4 b + c}{3 a + 5 b}$$
(Next one gets hairy...)
First of all, let us consider some numerical aspects which will (probably) be useful later.
I shall assume that constants $(a,b,c)$ are all positive.
We are looking for the zero of function $$f(x)=a x+bx^{5/3}-c$$ for which $$f'(x)=a+\frac{5}{3} b x^{ 2/3} \qquad \text{and} \qquad f''(x)=\frac {10b} {9x^{ 1/3}}\qquad \text{and} \qquad f'''(x)=-\frac{10 b}{27 x^{4/3}}$$
Starting at $\color{red}{x_0=0}$, the first iterate of Newton method is $\color{red}{x_1=\frac c a}$ for which $$f(x_1)=b \left(\frac{c}{a}\right)^{5/3} >0 \qquad \text{and} \qquad f''(x_1)=\frac{10 b}{9 \left(\frac{c}{a}\right)^{1/3}} >0$$ By Darboux theorem, this means that, even if we start with this probably very poor estimate, Newton method will converge without any overshoot of the solution.
As an example, using $a=60$, $b=10000$ and $c=2$ the iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.0333333 \\ 1 & 0.0140051 \\ 2 & 0.0072200 \\ 3 & 0.0055641 \\ 4 & 0.0054253 \\ 5 & 0.0054242 \end{array} \right)$$ So, there is no risk to generate a better starting point $$\color{red}{x_2=\frac{c}{a}-\frac{3 b \left(\frac{c}{a}\right)^{5/3}}{20 \left(2500 \left(\frac{c}{a}\right)^{2/3}+9\right)}}$$ at which we can compute the values $f(x_2)$, $f'(x_2)$, $f''(x_2)$, $f'''(x_2)$.
Now, let us use Householder method which will give $$\color{red}{x_3=x_2+\frac{3 f(x_2) \left(f(x_2) f''(x_2)-2 f'(x_2)^2\right)}{f(x_2)^2 f'''(x_2)+6 f'(x_2)^3-6 f(x_2) f'(x_2) f''(x_2)}}$$ For the worked example, this should give $x_3=0.0056389$. If we repeat once this last step, this should give $x_4=0.0054242$ which is the answer.