**Example 2**

Secondly, the non-linear, homogeneous Lane-Emden equation, with variable coefficients is considered:

$$\mathbf{y}'' + \frac{2}{x}\mathbf{y}' + 4(2\mathbf{e}^y + \mathbf{e}^{y/2}) = \mathbf{0}, \text{ x} \ge \mathbf{0}, \quad \text{subject to} \quad \mathbf{y}(0) = \mathbf{0} \quad \text{and} \quad \mathbf{y}'(0) = \mathbf{0} \tag{31}$$

which has the exact solution:

$$y(\mathbf{x}) = -2\ln(1+\mathbf{x}^2).$$

Eq. (31) has been solved by [29, 30, 31] using the lie symmetry, the homotopy-pertubation method and the variational iteration method.

#### **Example 3**

In this example, we consider the non-linear, variable coefficients, homogeneous Lane-Emden equation:

$$\text{y}'' + \frac{2}{\text{x}} \text{y}' - 6\text{y} = 4\text{y} \ln(\text{y}), \text{ x} \ge 0, \quad \text{subject to} \quad \text{y}(0) = \text{l} \quad \text{and} \quad \text{y}'(0) = 0 \tag{32}$$

which has the exact solution:

$$\mathcal{Y}(\mathbf{x}) = e^{x^2}.$$

Ramos [28] solved Eq. (32) using the variation iteration method, while Yildirim [16] solved the same equation using the linearisation method.

#### **Example 4**

We consider the non-linear, homogeneous Lane-Emden equation:

$$\text{y}^{\prime\prime} + \frac{6}{\text{x}} \text{y}^{\prime} + 2\text{y}(7 + \ln(\text{y}^2)) = 0, \text{ x} \ge 0, \quad \text{subject to} \quad \text{y}(0) = \text{l} \quad \text{and} \quad \text{y}^{\prime}(0) = 0 \tag{33}$$

which has the exact solution [32]:

$$\mathcal{Y}(\mathbf{x}) = e^{-x^2} \dots$$

This example was solved by Wazwaz [32] using the Adomian decomposition method.

#### **Example 5**

We consider the non-linear, homogenous Lane-Emden equation, which represent an infinite circular cylinder in astrophysics:

$$\mathbf{y}'' + \frac{2}{\mathbf{x}} \mathbf{y}' + \mathbf{y}^3 - 6 - \mathbf{x}^6 = \mathbf{0}, \text{ x} \ge \mathbf{0}, \quad \text{subject to} \quad \mathbf{y}(0) = \mathbf{0} \quad \text{and} \quad \mathbf{y}'(0) = \mathbf{0}, \tag{34}$$

which has the exact solution [33]:

$$y(x) = x^2.$$

This example was solved by Yin et al. [33] using the modified Laplace decomposition method.

#### **Example 6**

<sup>2</sup> *yx x* ( ) = 2ln(1 ). - +

Eq. (31) has been solved by [29, 30, 31] using the lie symmetry, the homotopy-pertubation

In this example, we consider the non-linear, variable coefficients, homogeneous Lane-Emden

*y y y y yx* 6 = 4 ln( ), > 0, subject to (0) = 1 and (0) = 0 *y y <sup>x</sup>* ¢¢ ¢ + - ¢ (32)

2 ( )= . *<sup>x</sup> yx e*

<sup>6</sup> <sup>2</sup> *y yy y x* 2 (7 ln( )) = 0, > 0, subject to (0) = 1 and (0) = 0 *y y <sup>x</sup>* ¢¢ ¢ ++ + ¢ (33)

2 ( )= . *<sup>x</sup> yx e*-

This example was solved by Wazwaz [32] using the Adomian decomposition method.

We consider the non-linear, homogenous Lane-Emden equation, which represent an infinite

<sup>2</sup> 3 6 *y yy x x* 6 = 0, > 0, subject to (0) = 0 and (0) = 0, *y y <sup>x</sup>* ¢¢ ¢ + + -- ¢ (34)

Ramos [28] solved Eq. (32) using the variation iteration method, while Yildirim [16] solved the

method and the variational iteration method.

150 Numerical Simulation - From Brain Imaging to Turbulent Flows

same equation using the linearisation method.

We consider the non-linear, homogeneous Lane-Emden equation:

2

which has the exact solution:

which has the exact solution [32]:

circular cylinder in astrophysics:

**Example 3**

equation:

**Example 4**

**Example 5**

Lastly, we consider the Lane-Emden equation with the general form:

$$\text{y}'' + \frac{2}{x}\text{y}' + \text{y}''' = 0, \text{ x} \ge 0, \quad \text{subject to} \quad \text{y}(0) = \text{l} \quad \text{and} \quad \text{y}'(0) = 0. \tag{35}$$

Eq. (35) is the standard Lane-Emden equation that models the thermal behaviour of a spherical cloud of glass acting under the mutual attraction of its molecules and subject to classical laws of thermodynamics [34]. The values of *m* in the interval 0,5 are most physically interesting to study. The equation is linear when *m* = 0 and *m* = 1 and non-linear for values of *m*>1. Wazwaz [35] gave the general solution of Eq. (35) in series form as

$$\begin{split} y(\mathbf{x}) &= 1 - \frac{1}{6}\mathbf{x}^2 + \frac{m}{120}\mathbf{x}^4 - \frac{m(8m-5)}{3.7!}\mathbf{x}^6 + \frac{m(70 - 183m + 122m^2)}{9.9!}\mathbf{x}^8 \\ &+ \frac{m(3150 - 1080m + 12642m^2 - 5032m^3)}{45.11!}\mathbf{x}^{10} + \dots \end{split} \tag{36}$$

Analytical solutions for *m*=0,1 and *m*=5 are given as [35]

$$\text{y}(\text{x}) = 1 - \frac{1}{3!} \text{x}^2, \qquad \text{y}(\text{x}) = \frac{\sin \text{x}}{\text{x}}, \qquad \text{and} \qquad \text{y}(\text{x}) = \left(1 + \frac{\text{x}^2}{3}\right)^{\frac{1}{2}}, \tag{37}$$

respectively. In this example, we consider the Lane-Emden Eq. (35) for *m* = 5. We therefore consider the equation:

$$\mathbf{y}'' + \frac{2}{\mathbf{x}} \mathbf{y}' + \mathbf{y}^\circ = \mathbf{0}, \ \mathbf{x} > \mathbf{0}, \quad \text{subject to} \quad \mathbf{y}(\mathbf{0}) = \mathbf{I} \quad \text{and} \quad \mathbf{y}'(\mathbf{0}) = \mathbf{0} \tag{38}$$

with the exact solution:

$$\mathbf{y}(\mathbf{x}) = \left(\mathbf{l} + \frac{\mathbf{x}^2}{3}\right)^{-1/2}.\tag{39}$$
