In this section, we present the CH-A for solving CL-M; the CH-A is a hybrid method in which it combines between the HAM and the Adomian decomposition method (ADM). The calculations involved in this technique are more easy than the standard HAM especially when the non-linear term in the CL-M is decomposed by using the Adomian polynomials. To start the procedure, let us consider the general form of CL-M of order *α*>0:

$$ C{{D}^{2\alpha }}y+\frac{2\alpha }{{{x}^{\alpha }}}+C{{D}^{\alpha }}y+f(y)=0 $$

(3)

where *x*>0 and 0<*α*≤1.

To solve this problem using CH-A, first, rewrite Eq. (3), as follows:

$$ C{{D}^{\alpha }}[{{x}^{2\alpha }}C{{D}^{\alpha }}y]=-{{x}^{2\alpha }}f(y) $$

(4)

Then, integrate Eq. (4) twice with respect to *x*, so the general fractional solution of Eq. (3) is given by:

$$ y(x)={{c}_{2}}+\int{{{c}_{1}}{{x}^{-2\alpha }}{{d}_{\alpha }}x-[\int{{{x}^{-2\alpha }}[\int{{{x}^{2\alpha }}f(y){{d}_{\alpha }}x]{{d}_{\alpha }}x]}}} $$

(5)

where *c*_{1} and *c*_{2} are constants and *d*_{α}*x*=*x*^{1−α}*d**x*.

In order to solve Eq. (5) by means of HAM, we need to seek the auxiliary linear operator:

$$ \mathcal{L}[\Phi (x,q)]=\Phi (x,q) $$

(6)

We now define the non-linear operator as:

$$ N[\Phi (x,q)]=\Phi (x,q)-{{c}_{2}}-\int{{{c}_{1}}{{x}^{-2\alpha }}{{d}_{\alpha }}x+[\int{{{x}^{-2\alpha }}[\int{{{x}^{2\alpha }}f(y)}}}{{d}_{\alpha }}x]{{d}_{\alpha }}x] $$

(7)

Consequently, the *m*th-order (*m*≥1) deformation equations can be expressed using the Adomian polynomials as:

$$ [{{y}_{m}}-{{\text{ }\!\!\chi\!\!\text{ }}_{m}}{{y}_{m-1}}]=\hbar H{{R}_{m}}({{\overset{\scriptscriptstyle\rightharpoonup}{y}}_{m-1}}) $$

(8)

where

$$ {{R}_{m}}({{\overset{\scriptscriptstyle\rightharpoonup}{y}}_{m-1}})={{y}_{m-1}}-(1-{{\text{ }\!\!\chi\!\!\text{ }}_{m}})({{c}_{2}}+\int{{{c}_{1}}{{x}^{-2\alpha }}{{d}_{\alpha }}x-[\int{{{x}^{-2\alpha }}}[\int{{{x}^{2\alpha }}{{A}_{m}}}}{{d}_{\alpha }}x]{{d}_{\alpha }}x]) $$

(9)

Hence

$$ {{y}_{m}}={{\text{ }\!\!\chi\!\!\text{ }}_{m}}{{y}_{m-1}}+\hbar H{{R}_{m}}({{\overset{\scriptscriptstyle\rightharpoonup}{y}}_{m-1}}) $$

(10)

Starting with an initial approximation:

$$ {{y}_{0}}={{c}_{2}}+\int{{{c}_{1}}{{x}^{-2\alpha }}{{d}_{\alpha }}x} $$

(11)

So, we have:

$$ {{y}_{m+1}}(x)=-[\int{{{x}^{-2\alpha }}}[\int{{{x}^{2\alpha }}{{A}_{m}}}{{d}_{\alpha }}x]{{d}_{\alpha }}x],\,\,\,\,m=0,1,2,...\,\,. $$

(12)

where

$$ {{\chi}_{m}}=\left\{ \begin{array}{ll} 0, & m\le 1 \\ 1, m>1 \\ \end{array} \right.\\ $$

and \(\left \{ {{A}_{m}} \right \}_{m=0}^{+\infty }\) is the set of Adomian polynomials of *f*(*y*) which is defined as follows:

$$ {{A}_{m}}=\frac{1}{m!}\frac{{{d}^{m}}}{d{{\theta }^{m}}}[f(\sum\limits_{i=0}^{+\infty }{{{\theta }^{i}}{{y}_{i}}){{]}_{\theta =0}},\,\,\,\,m=0,1,2,...\,\,.} $$

(13)

Finally, the exact solution of Eq. (3) can be calculated by:

$$ y(x)=\sum\limits_{m=0}^{+\infty }{{{y}_{m}}(x)} $$

(14)

and the *n*th-order approximate solution of Eq. (3) is given by:

$$ {{y}_{n}}(x)=\sum\limits_{m=0}^{n}{{{y}_{m}}(x)} $$

(15)

Now, we examine some known and charming cases that have been resulting from Eq. (3) as follows:

###
**Case 1**

Set *f*(*y*)=*y*^{k}, *k*=0,1,2,.... in Eq.(3), then the following equation for *x*>0 is called:

$$ C{{D}^{2\alpha }}y+\frac{2\alpha }{{{x}^{\alpha }}}C{{D}^{\alpha }}y+{{y}^{k}}=0,\,\,\,0<\alpha \le 1,\,\,\,k=0,1,2,...\,\,\,. $$

(16)

The CL-M of the first kind.

To solve this problem, we apply Eq. (13) to compute the set of Adomian polynomials \(\left \{ {{A}_{k}} \right \}_{k=0}^{+\infty }\) of the non-linear function *f*(*y*)=*y*^{k}.

Hence, according to Eq. (14), the general fractional solution of Eq. (16) is given as follows:

$$\ y(x)=\sum\limits_{m=0}^{+\infty }{{{y}_{m}}(x)} $$

where *y*_{0} is given by Eq. (11) and *y*_{m}, *m*=1,2,3,..., is given by Eq. (12).

###
**Problem 1**

Set the initial conditions *y*(0)=1 and *y*^{′}(0)=0 into Eq. (16), then the general fractional solution of Eq. (16) in terms of Adomian polynomials is given by Eqs. (11), (12), and (14), respectively.

###
**Problem 2**

Set *k*=0 into Eq. (16), then we have:

$$ C{{D}^{2\alpha }}y+\frac{2\alpha }{{{x}^{\alpha }}}C{{D}^{\alpha }}y+1=0,\,\,\,0<\alpha \le 1 $$

(17)

According to the initial conditions given in problem 1 and by integrating twice, then, the exact solution of this equation (see Fig. 1) can be obtained by :

$$ y(x)=1-\frac{{{x}^{2\alpha }}}{6{{\alpha }^{2}}} $$

(18)

###
**Problem 3**

Set *k*=1 into Eq. (16) and by using the initial conditions *y*(0)=1 and *y*^{′}(0)=0, then we have:

$$ C{{D}^{2\alpha }}y+\frac{2\alpha }{{{x}^{\alpha }}}C{{D}^{\alpha }}y+y=0,\,\,\,0<\alpha \le 1 $$

(19)

According to Eqs.(11) and (12), with *A*_{0}=*y*_{0}, *A*_{1}=*y*_{1}, *A*_{2}=*y*_{2}, *e**t**c*., then the solution of Eq. (19) is (see Fig. 2):

$$ y(x)=1-\frac{{{x}^{2\alpha }}}{6{{\alpha }^{2}}}+\frac{{{x}^{4\alpha }}}{120{{\alpha }^{4}}}-\frac{{{x}^{6\alpha }}}{5040{{\alpha }^{6}}} $$

(20)

###
**Problem 4**

Set *k*=2 into Eq. (16) and by using the initial conditions *y*(0)=1 and *y*^{′}(0)=0, then we have:

$$ C{{D}^{2\alpha }}y+\frac{2\alpha }{{{x}^{\alpha }}}C{{D}^{\alpha }}y+{{y}^{2}}=0,\,\,\,0<\alpha \le 1 $$

(21)

Also by using Eqs.(11) and (12), with \({{A}_{0}}=y_{0}^{2},\,\,{{A}_{1}}=2{{y}_{0}}{{y}_{1}},\,\,{{A}_{2}}=2{{y}_{0}}{{y}_{2}}+y_{1}^{2},\,etc.\), then the solution of Eq. (21) is (see Fig. 3):

$$ y(x)=1-\frac{{{x}^{2\alpha }}}{6{{\alpha }^{2}}}+\frac{{{x}^{4\alpha }}}{60{{\alpha }^{4}}}-\frac{11{{x}^{6\alpha }}}{7560{{\alpha }^{6}}} $$

(22)

###
**Case 2**

Consider *f*(*y*)=*e*^{y} as in Eq. (3), then the following equation for *x*>0 is called:

$$ C{{D}^{2\alpha }}y+\frac{2\alpha }{{{x}^{\alpha }}}C{{D}^{\alpha }}y+{{e}^{y}}=0,\,\,\,0<\alpha \le 1 $$

(23)

The CL-M of the second kind.

To solve problem (23) using CH-A and according to Eqs.(11) and (12), with \({{A}_{0}}=1,\,\,{{A}_{1}}={{y}_{1}},\,\,{{A}_{2}}={{y}_{2}}+\frac {1}{2}y_{1}^{2},\,\,{{A}_{3}}={{y}_{3}}+{{y}_{1}}{{y}_{2}}+\frac {1}{3!}y_{1}^{3},\,...,\) the solutions will be drawn as:

$$ {{y}_{0}}=0,\,\,{{y}_{k+1}}(x)=-[\int{{{x}^{-2\alpha }}}[\int{{{x}^{2\alpha }}{{A}_{k}}}{{d}_{\alpha }}x]{{d}_{\alpha }}x],\,\,\,\,k=0,1,2,...\,\, $$

(24)

Therefore, the general fractional power series solution is (see Fig. 4) given by:

$$ y(x)=-\frac{{{x}^{2\alpha }}}{6{{\alpha }^{2}}}+\frac{{{x}^{4\alpha }}}{120{{\alpha }^{4}}}-\frac{{{x}^{6\alpha }}}{1890{{\alpha }^{6}}} $$

(25)