Theorem (2.2) (Convolution Theorem)

If

$$\ell[\mathbf{f}(\mathbf{x})] = \mathbf{F}(\mathbf{s}), \quad \ell[\mathbf{g}(\mathbf{x})] = \mathbf{G}(\mathbf{s}).$$

then:

$$\ell[\mathbf{f}(\mathbf{x})^\* \mathbf{g}(\mathbf{x})] = \ell[\mathbf{f}(\mathbf{x}) \mathbf{g}(\mathbf{x})] = \mathbf{f}(\mathbf{s})\mathbf{g}(\mathbf{s})$$

or equivalently,

$$\ell^{-1}[\mathbf{F}(\mathbf{s})\mathbf{G}(\mathbf{s})] = \mathbf{f}(\mathbf{x})^\*\mathbf{g}(\mathbf{x})$$

Consider the differential equation,

$$\mathbf{L}\left[\mathbf{y}(\mathbf{x})\right] + \mathbf{R}\left[\mathbf{y}(\mathbf{x})\right] + \mathbf{N}\left[\mathbf{y}(\mathbf{x})\right] + \mathbf{N}^\*\left[\mathbf{y}(\mathbf{x})\right] = \mathbf{0} \tag{1}$$

With the initial conditions

$$\mathbf{y}(0) = \mathbf{h}(\mathbf{x}), \quad \mathbf{y}'(0) = \mathbf{k}(\mathbf{x}) \tag{2}$$

where L is a linear second-order operator, R is a linear first-order operator, N is the nonlinear operator and <sup>N</sup><sup>∗</sup>½ � y xð Þ is the nonlinear convolution term which is defined by:

$$\mathbf{N}^\*\left[\mathbf{y}(\mathbf{x})\right] = f\left(\mathbf{y}, \mathbf{y}', \mathbf{y}'', \dots, \mathbf{y}^{(\mathbf{n})}\right)^\* \mathbf{g}\left(\mathbf{y}, \mathbf{y}', \mathbf{y}'', \dots, \mathbf{y}^{(\mathbf{n})}\right)^\*$$

According to the variational iteration method, we can construct a correction functional as follows:

$$\mathbf{y}\_{n+1}(\mathbf{x}) = \mathbf{y}\_n(\mathbf{x}) + \int\_0^\mathbf{x} \lambda(\xi) \left[ \mathbf{L} \mathbf{y}\_n(\xi) + \mathbf{R} \tilde{\mathbf{y}}\_n(\xi) + N \tilde{\mathbf{y}}\_n(\xi) + \mathbf{N}^\* \tilde{\mathbf{y}}\_n(\xi) \right] d\xi \tag{3}$$

Rynð Þ <sup>ξ</sup> , Ny~nð Þ <sup>ξ</sup> and <sup>N</sup><sup>∗</sup>y~nð Þ <sup>ξ</sup> are considered as restricted variations, that is,

$$
\delta \mathbf{R} \mathbf{\tilde{y}}\_n = 0, \delta \mathbf{N} \mathbf{\tilde{y}}\_n = 0 \text{ and } \delta \mathbf{N}^\* \mathbf{\tilde{y}}\_n = 0, \ \lambda = -1
$$

Then, the variational iteration formula can be obtained as:

$$\mathbf{y}\_{n+1}(\mathbf{x}) = \mathbf{y}\_n(\mathbf{x}) - \int\_0^\mathbf{x} \left[ \mathbf{L} \mathbf{y}\_n(\xi) + \mathbf{R} \mathbf{y}\_n(\xi) + \mathbf{N} \mathbf{y}\_n(\xi) + \mathbf{N}^\* \tilde{\mathbf{y}}\_n(\xi) \right] d\xi \tag{4}$$

<sup>1</sup> � <sup>y</sup>00ð Þ<sup>x</sup> -2 <sup>þ</sup> 2y0<sup>∗</sup>y00-y0<sup>∗</sup> <sup>y</sup><sup>00</sup> � �<sup>2</sup> <sup>¼</sup> <sup>0</sup>, y 0ð Þ¼ <sup>y</sup><sup>0</sup>

<sup>s</sup> <sup>¼</sup> <sup>ℓ</sup> <sup>y</sup>0<sup>∗</sup> <sup>y</sup><sup>00</sup> � �<sup>2</sup> � 2y0<sup>∗</sup>y<sup>00</sup> h i

Solution of Nonlinear Partial Differential Equations by New Laplace Variational Iteration Method

<sup>s</sup><sup>2</sup> <sup>ℓ</sup> <sup>y</sup>0<sup>∗</sup> <sup>y</sup><sup>00</sup> � �<sup>2</sup> � 2y0<sup>∗</sup>y<sup>00</sup> � � h i

<sup>s</sup><sup>2</sup> <sup>ℓ</sup> <sup>y</sup>0<sup>∗</sup> <sup>y</sup><sup>00</sup> � �<sup>2</sup> � 2y0<sup>∗</sup>y<sup>00</sup> � � h i

<sup>ℓ</sup> <sup>y</sup><sup>00</sup> � �<sup>2</sup> � <sup>2</sup><sup>ℓ</sup> <sup>y</sup><sup>0</sup> � �<sup>∗</sup>

, :………, <sup>y</sup>nð Þ¼ <sup>x</sup> <sup>x</sup><sup>2</sup>

<sup>2</sup> � <sup>y</sup><sup>0</sup> � <sup>y</sup><sup>0</sup> � �<sup>2</sup> � 2x <sup>þ</sup> <sup>y</sup>0<sup>∗</sup> <sup>y</sup><sup>00</sup> � �<sup>2</sup> <sup>¼</sup> <sup>0</sup>, y 0ð Þ¼ <sup>1</sup> (12)

<sup>ℓ</sup> <sup>y</sup><sup>00</sup> � � � � h i (11)

#)

<sup>¼</sup> <sup>x</sup><sup>2</sup>

Take Laplace transform of Eq. (10), and making use of initial conditions, we have:

2

s <sup>2</sup>ℓyð Þ� <sup>x</sup>

The inverse Laplace transform of the above equation gives that:

<sup>y</sup>ð Þ¼ <sup>x</sup> <sup>x</sup><sup>2</sup> <sup>þ</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

ynþ<sup>1</sup>ð Þ¼ <sup>x</sup> ynð Þþ <sup>x</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

ynþ<sup>1</sup>ð Þ¼ <sup>x</sup> ynð Þþ <sup>x</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

<sup>y</sup>1ð Þ¼ <sup>x</sup> <sup>x</sup><sup>2</sup> <sup>þ</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

<sup>y</sup>2ð Þ¼ <sup>x</sup> x2

Then, the exact solution of Eq. (10) is <sup>y</sup>ð Þ¼ <sup>x</sup> <sup>x</sup>2.

Take the inverse Laplace transform to obtain

or

Then, we have:

This means that:

By using the new modified (Eq. (9)), we have the new correction functional,

<sup>s</sup><sup>2</sup> <sup>ℓ</sup> <sup>y</sup><sup>0</sup> � �<sup>∗</sup>

<sup>y</sup>0ð Þ¼ <sup>x</sup> <sup>x</sup><sup>2</sup>

<sup>y</sup>0ð Þ¼ <sup>x</sup> <sup>y</sup>1ð Þ¼ <sup>x</sup> y2ð Þ¼ <sup>x</sup> :…… <sup>¼</sup> <sup>y</sup>nð Þ¼ <sup>x</sup> <sup>x</sup><sup>2</sup>

( "

, y3ð Þ¼ <sup>x</sup> <sup>x</sup><sup>2</sup>

Take Laplace transform of Eq. (12), and using the initial condition, we obtain:

<sup>s</sup>ℓ<sup>y</sup> � <sup>1</sup> � <sup>2</sup>

<sup>s</sup><sup>2</sup> <sup>ℓ</sup>ð Þ<sup>4</sup> <sup>ℓ</sup>ð Þ� 2x <sup>2</sup>ℓð Þ 2x <sup>ℓ</sup>ð Þ<sup>2</sup>

<sup>s</sup><sup>2</sup> <sup>¼</sup> <sup>ℓ</sup> <sup>y</sup><sup>0</sup> � �<sup>2</sup> � <sup>y</sup>0<sup>∗</sup> <sup>y</sup><sup>00</sup> � �<sup>2</sup> h i

ð Þ¼ 0 0 (10)

157

http://dx.doi.org/10.5772/intechopen.73291

Eq. (4) can be solved iteratively using y0ð Þx as the initial approximation. Then, the solution is <sup>y</sup>ð Þ¼ <sup>x</sup> lim<sup>n</sup>!<sup>∞</sup> ynð Þx .

Now, we assume that <sup>L</sup> <sup>¼</sup> <sup>d</sup><sup>2</sup> dx<sup>2</sup> in Eq. (1).

Take Laplace transform (ℓ) of both sides of Eq. (1) to find:

$$\ell\left[\mathbf{L}\mathbf{y}(\mathbf{x})\right] + \ell\left[\mathbf{R}\mathbf{y}(\mathbf{x})\right] + \ell\left[\mathbf{N}\mathbf{y}(\mathbf{x})\right] + \ell\left[\mathbf{N}^\*\mathbf{y}(\mathbf{x})\right] = \mathbf{0} \tag{5}$$

$$\mathbf{s}^2 \ell \mathbf{y} - \mathbf{s} \mathbf{y}(0) - \mathbf{y}'(0) = -\ell \{ \mathbf{R} \mathbf{y}(\mathbf{x}) + \mathbf{N} \mathbf{y}(\mathbf{x}) + \mathbf{N}^\* \mathbf{y}(\mathbf{x}) \} = \mathbf{0} \tag{6}$$

By using the initial conditions and taking the inverse Laplace transform, we have:

$$\mathbf{y}(\mathbf{x}) = \mathbf{p}(\mathbf{x}) - \ell^{-1} \left[ \frac{1}{s^2} \mathbf{R} \mathbf{y}(\mathbf{x}) + \mathbf{N} \mathbf{y}(\mathbf{x}) + \mathbf{N}^\* \mathbf{y}(\mathbf{x}) \right] = \mathbf{0} \tag{7}$$

where p xð Þ represents the terms arising from the source term and the prescribed initial conditions. Now, the first derivative of Eq. (7) is given by:

$$\frac{d\mathbf{y}(\mathbf{x})}{d\mathbf{x}} = \frac{d\mathbf{p}(\mathbf{x})}{d\mathbf{x}} - \frac{d}{d\mathbf{x}}\ell^{-1} \left[\frac{1}{s^2}\ell\{\mathbf{R}\mathbf{y}(\mathbf{x}) + \mathbf{N}\mathbf{y}(\mathbf{x}) + \mathbf{N}^\*\mathbf{y}(\mathbf{x})\}\right] = \mathbf{0} \tag{8}$$

By the correction functional of the irrational method, we have:

$$\mathbf{y}\_{n+1}(\mathbf{x}) = \mathbf{y}\_n(\mathbf{x}) - \int\_0^\mathbf{x} \left\{ \left( \mathbf{y}\_n(\xi) \right)\_\xi - \frac{d}{d\xi} p(\xi) - \frac{d}{d\xi} \ell^{-1} \left[ \frac{1}{\mathbf{s}^2} \ell \{ \mathbf{R} \mathbf{y}(\xi) + \mathbf{N} \mathbf{y}(\xi) + \mathbf{N}^\* \mathbf{y}(\xi) \} \right] \right\} d\xi$$

Then, the new correction functional (new modified VIM) is given by:

$$\mathbf{y}\_{n+1}(\mathbf{x}) = \mathbf{y}\_n(\mathbf{x}) + \ell^{-1} \left[ \frac{1}{s^2} \ell \left\{ \mathbf{R} \mathbf{y}\_n(\mathbf{x}) + \mathbf{N} \mathbf{y}\_n(\mathbf{x}) + \mathbf{N}^\* \mathbf{y}\_n(\mathbf{x}) \right\} \right], \text{ n} \ge 0 \tag{9}$$

Finally, we find the answer in the strain; if inverse Laplace transforms exist, Laplace transforms exist.

In particular, consider the nonlinear ordinary differential equations with convolution terms,

Solution of Nonlinear Partial Differential Equations by New Laplace Variational Iteration Method http://dx.doi.org/10.5772/intechopen.73291 157

$$\left(1 - \mathbf{y}''(\mathbf{x})\text{-}2 + 2\mathbf{y}'^\ast\mathbf{y}''\cdot\mathbf{y}'^\ast\left(\mathbf{y}''\right)^2 = 0, \mathbf{y}(0) = \mathbf{y}'(0) = 0\tag{10}$$

Take Laplace transform of Eq. (10), and making use of initial conditions, we have:

$$s^2 \ell y(\mathbf{x}) - \frac{2}{s} = \ell \left[ \mathbf{y}^{\prime \*} (\mathbf{y}^{\prime \*})^2 - 2 \mathbf{y}^{\prime \*} \mathbf{y}^{\prime \*} \right]$$

The inverse Laplace transform of the above equation gives that:

$$y(\mathbf{x}) = \mathbf{x}^2 + \ell^{-1} \left\{ \frac{1}{s^2} \ell \left[ \mathbf{y}'^\* \left( \mathbf{y}'' \right)^2 - 2 \mathbf{y}'^\* \mathbf{y}'' \right] \right\}^2$$

By using the new modified (Eq. (9)), we have the new correction functional,

$$y\_{n+1}(\mathbf{x}) = y\_n(\mathbf{x}) + \ell^{-1} \left\{ \frac{1}{s^2} \ell \left[ \mathbf{y}'^\* \left(\mathbf{y}''\right)^2 - 2\mathbf{y}'^\* \mathbf{y}'' \right] \right\},$$

or

<sup>δ</sup>Ry~<sup>n</sup> <sup>¼</sup> <sup>0</sup>, <sup>δ</sup>Ny~<sup>n</sup> <sup>¼</sup> 0 and <sup>δ</sup>N<sup>∗</sup>

ð x

0

Eq. (4) can be solved iteratively using y0ð Þx as the initial approximation.

<sup>ℓ</sup> Ly xð Þ � � <sup>þ</sup> <sup>ℓ</sup> Ry xð Þ � � <sup>þ</sup> <sup>ℓ</sup> Ny xð Þ � � <sup>þ</sup> <sup>ℓ</sup> <sup>N</sup><sup>∗</sup>

By using the initial conditions and taking the inverse Laplace transform, we have:

ð Þ¼� <sup>0</sup> <sup>ℓ</sup> Ry xð Þþ Ny xð Þþ <sup>N</sup><sup>∗</sup>

<sup>s</sup><sup>2</sup> Ry xð Þþ Ny xð Þþ <sup>N</sup><sup>∗</sup>

<sup>d</sup><sup>ξ</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

where p xð Þ represents the terms arising from the source term and the prescribed initial conditions.

� �

<sup>s</sup><sup>2</sup> <sup>ℓ</sup> Ry xð Þþ Ny xð Þþ <sup>N</sup><sup>∗</sup>y xð Þ � � � �

<sup>s</sup><sup>2</sup> <sup>ℓ</sup> Ryð Þþ <sup>ξ</sup> Nyð Þþ <sup>ξ</sup> <sup>N</sup><sup>∗</sup>yð Þ <sup>ξ</sup> � � � � � �

<sup>s</sup><sup>2</sup> <sup>ℓ</sup> Rynð Þþ <sup>x</sup> Nynð Þþ <sup>x</sup> <sup>N</sup><sup>∗</sup>ynð Þ<sup>x</sup>

Finally, we find the answer in the strain; if inverse Laplace transforms exist, Laplace transforms

In particular, consider the nonlinear ordinary differential equations with convolution terms,

ynð Þx .

dx<sup>2</sup> in Eq. (1).

<sup>y</sup>ð Þ¼ <sup>x</sup> p xð Þ� <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

dx � <sup>d</sup>

By the correction functional of the irrational method, we have:

<sup>y</sup>nð Þ <sup>ξ</sup> � �

ynþ<sup>1</sup>ð Þ¼ <sup>x</sup> <sup>y</sup>nð Þþ <sup>x</sup> <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

dx <sup>ℓ</sup>�<sup>1</sup> <sup>1</sup>

<sup>ξ</sup> � <sup>d</sup>

Then, the new correction functional (new modified VIM) is given by:

<sup>d</sup><sup>ξ</sup> <sup>p</sup>ð Þ� <sup>ξ</sup> <sup>d</sup>

� �

Take Laplace transform (ℓ) of both sides of Eq. (1) to find:

<sup>ℓ</sup><sup>y</sup> � syð Þ� <sup>0</sup> <sup>y</sup><sup>0</sup>

Then, the variational iteration formula can be obtained as:

ynþ<sup>1</sup>ð Þ¼ <sup>x</sup> <sup>y</sup>nð Þ� <sup>x</sup>

Then, the solution is <sup>y</sup>ð Þ¼ <sup>x</sup> lim<sup>n</sup>!<sup>∞</sup>

156 Differential Equations - Theory and Current Research

s 2

Now, the first derivative of Eq. (7) is given by:

ð x

0

dx <sup>¼</sup> dpð Þ<sup>x</sup>

dyð Þx

ynþ<sup>1</sup>ð Þ¼ <sup>x</sup> <sup>y</sup>nð Þ� <sup>x</sup>

exist.

Now, we assume that <sup>L</sup> <sup>¼</sup> <sup>d</sup><sup>2</sup>

y~<sup>n</sup> ¼ 0, λ ¼ �1

Lynð Þþ <sup>ξ</sup> <sup>R</sup>ynð Þþ <sup>ξ</sup> Nynð Þþ <sup>ξ</sup> <sup>N</sup><sup>∗</sup>y~nð Þ <sup>ξ</sup> � �d<sup>ξ</sup> (4)

y xð Þ � � <sup>¼</sup> <sup>0</sup> (5)

¼ 0 (7)

¼ 0 (8)

dξ

(9)

y xð Þ � � <sup>¼</sup> <sup>0</sup> (6)

�

, n ≥ 0

y xð Þ

$$y\_{n+1}(\mathbf{x}) = y\_n(\mathbf{x}) + \ell^{-1} \left\{ \frac{1}{s^2} \left[ \ell (\mathbf{y}')^\* \ell (\mathbf{y}'')^2 - 2\ell (\mathbf{y}')^\* \ell (\mathbf{y}'') \right] \right\} \tag{11}$$

Then, we have:

$$y\_0(\mathbf{x}) = \mathbf{x}^2$$

$$y\_1(\mathbf{x}) = \mathbf{x}^2 + \boldsymbol{\ell}^{-1} \left\{ \frac{1}{s^2} \left[ \ell(4)\ell(2\mathbf{x}) - 2\ell(2\mathbf{x})\ell(2) \right] \right\} = \mathbf{x}^2$$

$$y\_2(\mathbf{x}) = \mathbf{x}^2, \quad \mathbf{y}\_3(\mathbf{x}) = \mathbf{x}^2, \quad \dots, \dots, \mathbf{y}\_n(\mathbf{x}) = \mathbf{x}^2$$

This means that:

$$y\_0(\mathbf{x}) = y\_1(\mathbf{x}) = \mathbf{y}\_2(\mathbf{x}) = \dots = \mathbf{y}\_n(\mathbf{x}) = \mathbf{x}^2$$

Then, the exact solution of Eq. (10) is <sup>y</sup>ð Þ¼ <sup>x</sup> <sup>x</sup>2.

$$2 - \mathbf{y'} - \left(\mathbf{y'}\right)^2 - 2\mathbf{x} + \mathbf{y'^\*}\left(\mathbf{y''}\right)^2 = 0,\\ \mathbf{y}(0) = 1 \tag{12}$$

Take Laplace transform of Eq. (12), and using the initial condition, we obtain:

$$s\ell y - 1 - \frac{2}{s^2} = \ell \left[ \left( \mathbf{y'} \right)^2 - \mathbf{y'^\*} \left( \mathbf{y''} \right)^2 \right]$$

Take the inverse Laplace transform to obtain

$$y(\mathbf{x}) = 1 + \mathbf{x}^2 + \ell^{-1} \left\{ \frac{1}{s} \ell \left[ \left( \mathbf{y}' \right)^2 - \mathbf{y}'^\* \left( \mathbf{y}'' \right)^2 \right] \right\}^{-1}$$

Using Eq. (9) to find the new correction functional in the form

$$y\_{n+1}(\mathbf{x}) = y\_n(\mathbf{x}) + \ell^{-1} \left\{ \frac{1}{s} \ell \left[ \left( \mathbf{y}' \right)^2 - \mathbf{y}'\_n \left( \mathbf{y}''\_n \right)^2 \right] \right\}$$

or

$$y\_{n+1}(\mathbf{x}) = y\_n(\mathbf{x}) + \ell^{-1} \left\{ \frac{1}{s} \left[ \ell \left[ \left( \mathbf{y'} \right)^2 \right] - \ell \left[ \mathbf{y'}\_n \right] \ell \left[ \left( \mathbf{y''}\_n \right)^2 \right] \right] \right\} \tag{13}$$

<sup>ℓ</sup> <sup>∂</sup>fð Þ <sup>x</sup>; <sup>t</sup> ∂t � �

fð Þ x; t ∂t 2 � �

<sup>ℓ</sup> <sup>∂</sup><sup>2</sup>

where f xð Þ ;s is the Laplace transform of ð Þ x; t .

unþ<sup>1</sup> ¼ un þ

is considered as a restricted variation, that is, δu~<sup>n</sup> ¼ 0.

Eq. (17) is called a correction functional.

ðt

0

In this section, we assume that L is an operator of the first-order <sup>∂</sup>

Applying the inverse Laplace transform on both sides of Eq. (19), we find:

Taking Laplace transform on both sides of Eq. (15), we get:

with the initial condition

tional as follows:

given by:

¼ d

¼ d2

We can easily extend this result to the nth partial derivative by using mathematical induction. To illustrate the basic concept of He's VIM, we consider the following general differential equations,

where L is a linear operator of the first-order, N is a nonlinear operator and g xð Þ ; t is inhomogeneous term. According to variational iteration method, we can construct a correction func-

where λ is a Lagrange multiplier ð Þ λ ¼ �1 , the subscripts n denotes the nth approximation, u~<sup>n</sup>

The successive approximation unþ<sup>1</sup> of the solution u will be readily obtained by using the determined Lagrange multiplier and any selective function u0; consequently, the solution is

<sup>u</sup> <sup>¼</sup> lim<sup>u</sup>!<sup>∞</sup> un

Using the differentiation property of Laplace transform and initial condition (16), we have:

dx ½ � <sup>F</sup>ð Þ <sup>x</sup>; <sup>s</sup> ,

Solution of Nonlinear Partial Differential Equations by New Laplace Variational Iteration Method

dx<sup>2</sup> ½ � <sup>F</sup>ð Þ <sup>x</sup>; <sup>s</sup> :

<sup>ℓ</sup>½ �þ Luð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup>½ �¼ Nuð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup>½ � g xð Þ ; <sup>t</sup> (15)

u xð Þ¼ ; 0 h xð Þ (16)

http://dx.doi.org/10.5772/intechopen.73291

159

λ½ � Lunð Þþ x;s Nu~nð Þ� x; s g xð Þ ; s ds (17)

<sup>∂</sup><sup>t</sup> in Eq. (15).

<sup>ℓ</sup>½ �þ Luð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup>½ �¼ Nuð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup>½ � g xð Þ ; <sup>t</sup> (18)

<sup>s</sup>ℓ½ �� <sup>u</sup>ð Þ <sup>x</sup>; <sup>t</sup> <sup>h</sup>ð Þ¼ <sup>x</sup> <sup>ℓ</sup>½ �� <sup>g</sup>ð Þ <sup>x</sup>; <sup>t</sup> <sup>ℓ</sup>½ � Nuð Þ <sup>x</sup>; <sup>t</sup> (19)

Then, we have:

$$y\_0(\mathbf{x}) = 1 + \mathbf{x}^2$$

$$y\_1(\mathbf{x}) = 1 + \mathbf{x}^2 + \ell^{-1} \frac{1}{s} \left\{ \ell (4\mathbf{x}^2) - \ell (2\mathbf{x})\ell (4) \right\} = 1 + \mathbf{x}^2 + \ell^{-1} \frac{1}{s} \left\{ \frac{8}{s^3} - \left(\frac{2}{s^2}\right) \left(\frac{4}{s}\right) \right\} = 1 + \mathbf{x}^2$$

$$y\_0(\mathbf{x}) = y\_1(\mathbf{x}) = y\_2(\mathbf{x}) = \dots = y\_n(\mathbf{x}) = 1 + \mathbf{x}^2$$

Then, the exact solution of Eq. (12) is:

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