Analytical solution proposal for fast numerical algorithm in special structured higher order differential equations

==Abstract==

We suggest a practical method for obtaining the particular solution of non-homogeneous higher order linear differential equations with constant coefficients. The proposed method can be applied directly and simply to such problems. We revealed that is valid for the different type of problem by using sample solutions. This simple analytical solution that we have introduced will help to create a fast numerical algorithm for computers and thus simplify the numerical solutions of higher order physical problems.

==1. Introduction==

Differential equations are vital argument for technical and fundamental sciences. Ordinary differential equations with constant coefficients provide practical utility owing to their mathematical controllable  to  explain  natural  phenomena  in  many  areas  of  science  and  engineering <span id='citeF-1'></span><span id='citeF-2'></span>[[#cite-1|[1,2]]]. For example, they are widely used to explain the event of electromagnetic, sound and water wave.  Such  waves are  formulated  by  homogeneous  or  non-homogeneous  differential  equations  with higher  order  linear  constant  coefficients  depend  upon  source  of  the  phenomena <span id='citeF-3'></span><span id='citeF-4'></span><span id='citeF-5'></span><span id='citeF-6'></span><span id='citeF-7'></span><span id='citeF-8'></span><span id='citeF-9'></span><span id='citeF-10'></span>[[#cite-3|[3,4,5,6,7,8,9,10]]].

We focused on the solutions of the special equation type of non-homogeneous differential equations with high order linear constant coefficients. Solution of the differential equations with this special structure are very intractable to obtain its particular solution, since it requires to cope with several consecutive integral, multi-variable undefined linear equations system and integration of such unknown coefficients in the form of derivative functions <span id='citeF-1'></span><span id='citeF-2'></span><span id='citeF-11'></span><span id='citeF-12'></span>[[#cite-1|[1,2,11,12]]]. 

In this work, without solving integral and multivariable linear equation system, a short formula and practices solution technique can be applied to handled problem directly is expressed. The considered differential equation is described by

<span id="eq-1"></span>
{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>(D^{2}+a^{2})^{n}y=k_{1}\sin ax+k_{2}\cos ax, </math>
|}
| style="width: 5px;text-align: right;white-space: nowrap;" | (1)
|}



where <math display="inline">n\in {Z}^{+},a,k_{1},k_{2}\in {R}, D^{2}=\frac{d^{2}}{dx^{2}}</math>. The general solution of a linear differential equation as follows:

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>y=y_{h}+y_{p}, </math>
|}
| style="width: 5px;text-align: right;white-space: nowrap;" | (2)
|}

where <math display="inline">y_{h}</math> represents the solution of the homogeneous equation and <math display="inline">y_{p}</math> represents also a particular solution for the non-homogeneous differential equations. The solution of discussed problem of homogeneous equation is easy and obtained by the solution of the auxiliary equation. <math display="inline">(D^{2}+a^{2})^{n}y_{h}=0</math>  is obtained by the following:

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>y_{h}=\sum _{j=0}^{n-1}x^{j}\left( C_{j}\cos ax+C_{j+n}\sin ax\right). </math>
|}
| style="width: 5px;text-align: right;white-space: nowrap;" | (3)
|}

The solution to the homogeneous equation can be found as easily seen. Particular solutions of such equations are written in generally as follows:

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>y_{p}=\frac{1}{F(D)}Q(x)  </math>
|}
| style="width: 5px;text-align: right;white-space: nowrap;" | (4)
|}

where <math display="inline">F(D)</math> is real coefficient polynomial of <math display="inline">D</math>. In order to acquire the particular solution of the proposed differential equation,  we need the following four lemmas. Four lemmas have been proven below.

'''Lemma 1.1.'''  If <math display="inline">F(D)=\sum _{j=0}^{n}a_{j}D^{j},a_{j}\in {R}</math> and <math display="inline">F(D)y=e^{iax},i^{2}=-1,F(ia)\neq 0,</math> particular solution is

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>y_{P}=\frac{e^{iax} }{F(ia)}.  </math>
|}
| style="width: 5px;text-align: right;white-space: nowrap;" | (5)
|}

'''Proof.'''

Let <math display="inline">y=e^{iax}.</math> Then,

<math display="inline">Dy=iae^{iax}</math>

<math display="inline">D^{2}y=\left(ia\right)^{2}e^{iax}</math>

<math display="inline">D^{3}y=\left(ia\right)^{3}e^{iax}</math>

…

<math display="inline">D^{n}y=\left(ia\right)^{n}e^{iax}.</math>

Hence

<math display="inline">a_{0}y=a_{0}e^{iax}</math>

<math display="inline">a_{1}Dy=a_{1}\left(ia\right)e^{iax}</math>

<math display="inline">a_{2}D^{2}y=a_{2}\left(ia\right)^{2}e^{iax}</math>

<math display="inline">a_{3}D^{3}y=a_{3}\left(ia\right)^{3}e^{iax}</math>

…

<math display="inline">a_{n}D^{n}y=a_{n}\left(ia\right)^{n}e^{iax}.</math>

If these equations are added to the side-by-side,

<math display="inline">a_{0}y+a_{1}Dy+a_{2}D^{2}y+a_{3}D^{3}y+\ldots + a_{n}D^{n}y = a_{0}e^{iax}+a_{1} \left(ia\right)e^{iax}+a_{2}\left(ia\right)^{2}e^{iax}+ </math> <math display="inline">a_{3}\left( ia\right)^{3}e^{iax}+\ldots + a_{n}\left(ia\right)^{n}e^{iax}</math>

we get

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>\left(\sum _{j=0}^{n}a_{j}D^{j}\right)y=\left(\sum _{j=0}^{n}a_{j}\left( ia\right)^{j}\right)e^{iax}.     </math>
|}
|}

If we denote

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>F(D)y=F(ia)e^{iax}     </math>
|}
|}

and

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>\frac{e^{iax}}{F(D)}=\frac{y}{F(ia)}     </math>
|}
|}

and also

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>y_{p}=\frac{e^{iax}}{F(D)}     </math>
|}
|}

we get the particular solution as below,

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>y_{p}=\frac{e^{iax}}{F(ia)}.     </math>
|}
|}

'''Lemma 1.2.'''   If, <math display="inline">F(D)=\sum _{j=0}^{n}a_{j}D^{j},a_{j}\in {R}</math> and <math display="inline">F(D)y=e^{-iax},</math> <math display="inline">i^{2}=-1,</math> <math display="inline">F(-ia)\neq 0</math>, particular solution is

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>y_{P}=\frac{e^{-iax}}{F(-ia)} </math>
|}
| style="width: 5px;text-align: right;white-space: nowrap;" | (6)
|}

'''Proof.'''

Let <math display="inline">y=e^{-iax}</math>. Therefore,

<math display="inline">Dy=-iae^{-iax}</math>

<math display="inline">D^{2}y=\left(-ia\right)^{2}e^{-iax}</math>

<math display="inline">D^{3}y=\left(-ia\right)^{3}e^{-iax}</math>

…

<math display="inline">D^{n}y=\left(-ia\right)^{n}e^{-iax}.</math>

Hence

<math display="inline">a_{0}y=a_{0}e^{-iax}</math>

<math display="inline">a_{1}Dy=a_{1}\left(-ia\right)e^{-iax}</math>

<math display="inline">a_{2}D^{2}y=a_{2}\left(-ia\right)^{2}e^{-iax}</math>

<math display="inline">a_{3}D^{3}y=a_{3}\left(-ia\right)^{3}e^{-iax}</math>

…

<math display="inline">a_{n}D^{n}y=a_{n}\left(-ia\right)^{n}e^{-iax}.</math>

Here, if these equations are adjoined to each other,

<math display="inline">a_{0}y+a_{1}Dy+a_{2}D^{2}y+\ldots + a_{n}D^{n}y=a_{0}e^{-iax}+a_{1}\left( -ia\right)e^{-iax}+a_{2}\left(-ia\right)^{2}e^{-iax}+\ldots + a_{n}\left( -ia\right)^{n}e^{-iax}</math>

We get

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>\left(\sum _{j=0}^{n}a_{j}D^{j}\right)y=\left(\sum _{j=0}^{n}a_{j}\left( -ia\right)^{j}\right)e^{-iax}     </math>
|}
|}

If we denote

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>F(D)y=F(-ia)e^{-iax}    </math>
|}
|}

and

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>\frac{e^{-iax}}{F(D)}=\frac{y}{F(-ia)}     </math>
|}
|}

and also

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>y_{p}=\frac{e^{-iax}}{F(D)}.     </math>
|}
|}

We get the particular solution as below,

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>y_{p}=\frac{e^{-iax}}{F(-ia)}.     </math>
|}
|}

'''Lemma 1.3.'''   If <math display="inline">F_{1}(D)=\sum _{j=0}^{n}a_{j}D^{j},a_{j}\in {R}</math> and <math display="inline">F(D)=\left(D-ia\right)^{r}\left(D+ia\right) ^{r}F_{1}(D)</math>, <math display="inline">F_{1}(ia)\neq 0</math> and <math display="inline">F(D)y=e^{iax},</math> <math display="inline">i^{2}=-1</math>, particular solution is

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>y_{P}=\frac{x^{r}}{r!}\frac{e^{iax}}{\left( 2ia\right)^{r}F_{1}(ia)} </math>
|}
| style="width: 5px;text-align: right;white-space: nowrap;" | (7)
|}

'''Proof.'''

Let <math display="inline">F(D)y=e^{iax}</math>.

Then, <math display="inline">\left(D-ia\right)^{r}\left(D+ia\right)^{r}F_{1}(D)y=e^{iax}</math> and <math display="inline">\left(D-ia\right)^{r}y=\frac{e^{iax}}{\left(D+ia\right)^{r}F_{1}(D)}.</math> From Lemma 1.1.

<math display="inline">\left(D-ia\right)^{r}y=\frac{e^{iax}}{\left(2ia\right)^{r}F_{1}(ia)}</math> <math display="inline">y_{p}=\frac{1}{\left(2ia\right)^{r}F_{1}(ia)}\left(\frac{e^{iax}}{\left( D-ia\right)^{r}}\right)</math>

<math display="inline">y_{p}=\frac{1}{\left(2ia\right)^{r}F_{1}(ia)}\left(\frac{e^{iax}\int  e^{-iax}e^{iax}dx}{\left(D-ia\right)^{r-1}}\right)=\frac{1}{\left(2ia\right)^{r}F_{1}(ia)}\left(\frac{ e^{iax}x}{\left(D-ia\right)^{r-1}}\right)</math>

<math display="inline">y_{p}=\frac{1}{\left(2ia\right)^{r}F_{1}(ia)}\left(\frac{e^{iax}\int  e^{-iax}e^{iax}xdx}{\left(D-ia\right)^{r-2}}\right)=\frac{1}{\left(2ia\right)^{r}F_{1}(ia)}\left(\frac{e^{iax} \frac{x^{2}}{2!}}{\left(D-ia\right)^{r-2}}\right)</math>

…

<math display="inline">y_{p}=\frac{1}{\left(2ia\right)^{r}F_{1}(ia)}\left(\frac{e^{iax}\frac{ x^{r-2}}{(r-2)!}}{\left(D-ia\right)^{2}}\right)</math>

<math display="inline">y_{p}=\frac{1}{\left(2ia\right)^{r}F_{1}(ia)}\left(\frac{e^{iax}\int  e^{-iax}e^{iax}\frac{x^{r-2}}{(r-2)!}dx}{\left(D-ia\right)}\right)=\frac{1}{\left(2ia\right) ^{r}F_{1}(ia)}\left(\frac{e^{iax}\frac{x^{r-1}}{(r-1)!}}{\left(D-ia\right) }\right)</math>

<math display="inline">y_{p}=\frac{1}{\left(2ia\right)^{r}F_{1}(ia)}\left(e^{iax}\int  e^{-iax}e^{iax}\frac{x^{r-1}}{(r-1)!}dx\right)=\frac{1}{\left(2ia\right) ^{r}F_{1}(ia)}\left(e^{iax}\int \frac{x^{r-1}}{(r-1)!}dx\right)</math>

<math display="inline">y_{p}=\frac{1}{\left(2ia\right)^{r}F_{1}(ia)}\left(e^{iax}\int  e^{-iax}e^{iax}\frac{x^{r-1}}{(r-1)!}dx\right),</math>

we obtain the particular solution as below,

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>y_{p}=\frac{1}{\left(2ia\right)^{r}F_{1}(ia)}\left(e^{iax}\frac{x^{r}}{r! }\right) </math>
|}
| style="width: 5px;text-align: right;white-space: nowrap;" | (8)
|}

'''Lemma 1.4.'''  If <math display="inline">F(D)=\sum _{j=0}^{n}a_{j}D^{j},a_{j}\in {R}</math> and <math display="inline">F(D)=\left(D-ia\right)^{r}\left(D+ia\right) ^{r}F_{1}(D)</math>, <math display="inline">F_{1}(-ia)\neq 0</math> and <math display="inline">F(D)y=e^{-iax},</math> <math display="inline">i^{2}=-1,</math> particular solution is

{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>y_{P}=\frac{x^{r}}{r!}\frac{e^{-iax}}{ \left(-2ia\right)^{r}F_{1}(-ia)} </math>
|}
| style="width: 5px;text-align: right;white-space: nowrap;" | (9)
|}

'''Proof.'''

Let <math display="inline">F(D)y=e^{-iax}</math> Therefore,

<math display="inline">\left(D-ia\right)^{r}\left(D+ia\right)^{r}F_{1}(D)y=e^{-iax}</math>

<math display="inline">\begin{array}{l}\left(D+ia\right)^{r}y=\frac{e^{-iax}}{\left(D-ia\right)^{r}F_{1}(D)}.\\ \end{array}</math>

From Lemma 1.2.

<math display="inline">\begin{array}{l}\left(D+ia\right)^ry=\frac{e^-iax}{\left(-2ia\right)^rF_1(-ia)}\\
y_p=\frac{1}{\left(-2ia\right)^rF_1(-ia)}\left(\frac{e^-iax}{ \left(D+ia\right)^r}\right)\\
y_p=\frac{1}{\left(-2ia\right)^rF_1(-ia)}\left(\frac{e^-iax e^iaxe^-iaxdx}{\left(D+ia\right)^r-1}\right)=\frac{1}{\left(-2ia\right)^rF_1(-ia)}\left(\frac{ e^-iaxx}{\left(D+ia\right)^r-1}\right)\\  
y_p=\frac{1}{\left(-2ia\right)^rF_1(-ia)}\left(\frac{e^-iax e^iaxe^-iaxxdx}{\left(D+ia\right)^r-2}\right)=\frac{1}{\left(-2ia\right)^rF_1(-ia)}\left(\frac{ e^-iax\frac{x^2}{2!}}{\left(D+ia\right)^r-2}\right)\\  
\ldots\\
y_p=\frac{1}{\left(-2ia\right)^rF_1(-ia)}\left(\frac{e^-iax\frac{ x^r-2}{(r-2)!}}{\left(D+ia\right)^2}\right)\\  
y_p=\frac{1}{\left(-2ia\right)^rF_1(-ia)}\left(\frac{e^-iax e^iaxe^-iax\frac{x^r-2}{(r-2)!}dx}{\left(D+ia\right)}\right)=\frac{1}{\left(-2ia\right) ^rF_1(-ia)}\left(\frac{e^-iax\frac{x^r-1}{(r-1)!}}{\left( D+ia\right)}\right),
\end{array}</math>

we obtain the particular solution as below,

<math display="inline">y_{p}=\frac{1}{\left(-2ia\right)^{r}F_{1}(-ia)}\left(e^{-iax}\int  e^{iax}e^{-iax}\frac{x^{r-1}}{(r-1)!}dx\right)=\frac{x^{r}}{r!}\frac{e^{-iax}}{\left(-2ia\right)^{r}F_{1}(-ia)}</math>.

==2. A Simple Approach to the Particular Solution, <math>\mathbf{y_{p}}</math>==

Particular solution of considered differential equations which is linear, constant coefficient, non-homogeneous and higher order differential equation is written as     
<span id="(eq-10)"></span>
{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>y_{p}=\frac{1}{(D^{2}+a^{2})^{n}}\left(  k_{1}\sin(ax)+k_{2}\cos(ax)\right) </math>
|}
| style="width: 5px;text-align: right;white-space: nowrap;" | (10)
|}


Using lemmas 1.1.,1.2.,1.3.,1.4. the Euler Identity by writing instead of <math>\sin ax \,  \cos ax </math> in [[#eq-10|(10)]], it becomes the following: 
{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: center; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>\begin{array}{ll}
y_{p}& =k_{1}\frac{\frac{e^{iax}-e^{-iax}}{2i}}{(D^{2}+a^{2})^{n}} +k_{2}\frac{\frac{e^{iax}+e^{-iax}}{2}}{(D^{2}+a^{2})^{n}} \\ 
&=k_{1}\frac{\frac{e^{iax}}{2i}}{(D^{2}+a^{2})^{n}}+k_{2}\frac{\frac{e^{iax} }{2}}{(D^{2}+a^{2})^{n}}+k_{1}\frac{\frac{-e^{-iax}}{2i}}{(D^{2}+a^{2})^{n} }+k_{2}\frac{\frac{e^{-iax}}{2}}{(D^{2}+a^{2})^{n}} \\ &=\left( \frac{k_{1}}{2i}+\frac{k_{2}}{2}\right) \left( \frac{e^{iax} }{(D^{2}+a^{2})^{n}}\right) +\left( -\frac{k_{1}}{2i}+\frac{k_{2}}{} {2}\right) \left( \frac{e^{-iax}}{(D^{2}+a^{2})^{n}}\right) \\ &=\left( \frac{k_{1}}{2i}+\frac{k_{2}}{2}\right) \frac{1}{\left( D-ia\right) ^{n}}\left( \frac{e^{iax}}{\left( D+ia\right) ^{n}}\right) +\left( -\frac{k_{1}}{2i}+\frac{k_{2}}{2}\right) \frac{1}{\left( D+ia\right) ^{n}}\left( \frac{e^{-iax}}{\left( D-ia\right) ^{n}}\right) \\ &=\left( \frac{k_{1}}{2i}+\frac{k_{2}}{2}\right) \frac{x^{n}}{n!}\left( \frac{e^{iax}}{\left( ia+ia\right) ^{n}}\right) +\left( -\frac{k_{1}}{} {2i}+\frac{k_{2}}{2}\right) \frac{x^{n}}{n!}\left( \frac{e^{-iax}}{\left( -ia-ia\right) ^{n}}\right) \\ &=\left( \frac{k_{1}}{2i}+\frac{k_{2}}{2}\right) \frac{x^{n}}{n!}\left( \frac{e^{iax}}{\left( 2ia\right) ^{n}}\right) +\left( -\frac{k_{1}}{} {2i}+\frac{k_{2}}{2}\right) \frac{x^{n}}{n!}\left( \frac{e^{-iax}}{\left( -2ia\right) ^{n}}\right) \\ &=\frac{x^{n}}{2^{n}a^{n}n!}\left( \left( \frac{k_{1}}{i}+k_{2}\right) \left( \frac{e^{iax}}{2\left( i\right) ^{n}}\right) +\left( -\frac{k_{1} }{i}+k_{2}\right) \left( \frac{e^{-iax}}{2\left( -i\right) ^{n}}\right) \right).    
\end{array}</math>
|}
| style="width: 5px;text-align: right;white-space: nowrap;" | (11)
|}

The particular solution is obtained in two ways depending on being even or odd of <math>n</math>. If <math>n</math> is even and <math>n=2m\left( m\in {Z}^{+}\right)</math>     
{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: center; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>\begin{array}{ll}
y_{p}&=\frac{x^{2m}}{2^{2m}a^{2m}2m!}\left( \left( \frac{k_{1}}{i} +k_{2}\right) \left( \frac{e^{iax}}{2\left( i\right) ^{2m}}\right) +\left( -\frac{k_{1}}{i}+k_{2}\right) \left( \frac{e^{-iax}}{2\left( -i\right) ^{2m}}\right) \right) \\ &=\frac{x^{2m}}{\left( -1\right) ^{m}2^{2m}a^{2m}2m!}\left( \left( \frac{k_{1}}{i}+k_{2}\right) \left( \frac{e^{iax}}{2}\right) +\left( -\frac{k_{1}}{i}+k_{2}\right) \left( \frac{e^{-iax}}{2}\right) \right) \\ &=\frac{x^{2m}}{\left( -1\right) ^{m}2^{2m}a^{2m}2m!}\left( k_{1} \frac{e^{iax}-e^{-iax}}{2i}+k_{2}\frac{e^{iax}+e^{-iax}}{2}\right).\end{array}</math>
|}
| style="width: 5px;text-align: right;white-space: nowrap;" | (12)
|}

Therefore, particular solution [[#eq-10|(10)]] for even values of n is obtained below:  
{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>y_{p}=\frac{x^{2m}}{\left( -1\right) ^{m}2^{2m}a^{2m}2m!}\left( k_{1}\sin ax+k_{2}\cos ax\right). </math>
|}
| style="width: 5px;text-align: right;white-space: nowrap;" | (13)
|}

Using these result, we obtain the following general solution of [[#eq-1|(1)]] for even values of n as follows:  
{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>y_{}=\left( -1\right) ^{\frac{n}{2}}\frac{x^{n}}{2^{n}a^{n} n!}\left( k_{1}\sin ax+k_{2}\cos ax\right) +\sum _{j=0}^{n-1}x^{j}\left(C_{j}\cos ax+C_{j+n}\sin ax\right). </math>
|}
| style="width: 5px;text-align: right;white-space: nowrap;" | (14)
|}

If n is odd and <math display="inline"> n=2m+1\left( m\in {Z}^{+}\right) </math>, hence, particular solution [[#eq-10|(10)]] is   
{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: center; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>\begin{array}{ll}
y_{p}&=\frac{x^{2m+1}}{2^{2m+1}a^{2m+1}\left( 2m+1\right) !}\left( \left( \frac{k_{1}}{i}+k_{2}\right) \left( \frac{e^{iax}}{2\left( i\right) ^{2m+1}}\right) +\left( -\frac{k_{1}}{i}+k_{2}\right) \left( \frac{e^{-iax}}{2\left( -i\right) ^{2m+1}}\right) \right) \\ 
&=\frac{x^{2m+1}}{2^{2m+1}a^{2m+1}\left( 2m+1\right) !}\left( \frac{k_{1} }{i}\frac{e^{iax}}{2\left( i\right) ^{2m+1}}+k_{2}\frac{e^{iax}}{2\left( i\right) ^{2m+1}}-\frac{k_{1}}{i}\frac{e^{-iax}}{2\left( -i\right) ^{2m+1} }+k_{2}\frac{e^{-iax}}{2\left( -i\right) ^{2m+1}}\right)\\ 
&=\frac{x^{2m+1}}{2^{2m+1}a^{2m+1}\left( 2m+1\right) !}\left( \frac{k_{1} }{i^{2}}\frac{e^{iax}}{2\left( i\right) ^{2m}}+k_{2}\frac{e^{iax}}{2\left( i\right) ^{2m}2i}+\frac{k_{1}}{i^{2}}\frac{e^{-iax}}{2\left( i\right) ^{2m}2}-k_{2}\frac{e^{-iax}}{\left( i\right) ^{2m}2i}\right)\\ 
&=\frac{x^{2m+1}}{2^{2m+1}a^{2m+1}\left( 2m+1\right) !\left( i\right) ^{2m}}\left( -k_{1}\frac{e^{iax}}{2}+k_{2}\frac{e^{iax}}{2i}-k_{1} \frac{e^{-iax}}{2}-k_{2}\frac{e^{-iax}}{2i}\right) \\ 
&=\left( -1\right) ^{m}\frac{x^{2m+1}}{2^{2m+1}a^{2m+1}\left( 2m+1\right) !}\left( -k_{1}\frac{e^{iax}+e^{-iax}}{2}+k_{2}\frac{e^{iax}-e^{-iax}}{} {2i}\right). \end{array}</math>
|}
| style="width: 5px;text-align: right;white-space: nowrap;" | (15)
|}

Therefore, particular solution [[#eq-10|(10)]] for odd values of n is obtained below:   
{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>y_{p}=\left( -1\right) ^{m}\frac{x^{2m+1}}{2^{2m+1}a^{2m+1}\left( 2m+1\right)!}\left( -k_{1}\cos ax+k_{2}\sin ax\right). </math>
|}
| style="width: 5px;text-align: right;white-space: nowrap;" | (16)
|}

Using these result, we obtain the following general solution of [[#eq-1|(1)]] for odd values of n as follows:   
{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>y_{}=\left( -1\right) ^{\frac{n-1}{2}}\frac{x^{n}}{2^{n} a^{n}\left( n\right) !}\left( -k_{1}\cos ax+k_{2}\sin ax\right) +\sum _{j=0}^{n-1}x^{j}\left(C_{j}\cos ax+C_{j+n}\sin ax\right). </math>
|}
| style="width: 5px;text-align: right;white-space: nowrap;" | (17)
|}

==3. Examples==

=== Example 1 ===

Consider such a differential equation  <math display="inline">(D^{2}+1)^{5}y=2\sin x-3\cos x</math>.  Solution of this equation: firstly, we solve the homogen equation,  
{| class="formulaSCP" style="width: 100%; text-align: left;" 
|-
| 
{| style="text-align: left; margin:auto;width: 100%;" 
|-
| style="text-align: center;" | <math>\begin{array}{l}
(D^{2}+1)^{5}y=0\\
(m^{2}+1)^{5}=0\\
m_{1,2,3,4,5}=i  y_{1}=\cos x,y_{2}=x\cos x,y_{3}=x^{2}\cos x,y_{4}=x^{3}\cos
x,y_{5}=x^{4}\cos x\\
m_{6,7,8,9,10}=-i  y_{6}=\cos x,y_{7}=x\cos x,y_{8}=x^{2}\cos x,y_{9}%
=x^{3}\cos x,y_{10}=x^{4}\cos x\\
y=y_{h}\\
y=\sum _{i=1}^{10}C_{i}y_{i}
\end{array} </math>
|}
| style="width: 5px;text-align: right;white-space: nowrap;" | 
|}

Now, let's try to solve the non-homogeneous equation using the classical methods and then the formula we recommend.   

'''(1) By using the method of consecutive integration'''

<math display="inline">y_{P}=e^{ix}\int \int \int \int \int e^{-2ix}\int \int \int \int \int e^{ix}\left( 2\sin x-3\cos x\right) \left( dx\right) ^{10}</math>  

As can be seen, the solution of <math display="inline">y_{P}</math> with this method will be found by integrating ten times in succession. It is clear that integrating in this way is not quick and easy.

'''(2) By using the method of variation of parameters'''

Let us  <math display="inline">y_{P}=\sum _{i=1}^{10}L_{i}y_{i}</math>. <math display="inline">  L_{i}</math> functions are defined as:    

<math display="inline">\sum _{i=1}^{10}L_{i}^{\prime }y_{i}=0,\sum _{i=1}^{10}L_{i}^{\prime } y_{i}^{\left( 1\right) }=0,...,\sum _{i=1}^{10}L_{i}^{\prime }y_{i}^{\left( 8\right) }=0,\sum _{i=1}^{10}L_{i}^{\prime }y_{i}^{\left( 9\right) }=2\sin  x-3\cos x</math> 

Solving above linear equation system, with the derivative functions of <math display="inline">L_{i}^{\prime }, (1\leq i\leq 10)</math>   using this derivative functions and following integration <math display="inline">L_{i}</math>  yields  

<math display="inline"> L_{i}=\int L_{i}^{\prime }dx,  (1\leq i\leq 10)</math>.   

It can not be said that these two processes will be quick and easy.

'''(3) By using the method of undetermined coefficients'''

Let us <math display="inline">y_{P}=\left( \sum _{i=0}^{9}A_{i}x^{i}\right) \cos x+\left( \sum _{i=0}^{9}B_{i}x^{i}\right) \sin x</math>. In this equation, replaces  <math display="inline">y,D_{x^{2}}y,...,D_{x^{10}}y</math> and its derivatives. With the identification of the two sides, <math display="inline">A_{i},B_{i}(1\leq i\leq 10)</math> coefficients are determined.  It can also not be said that these two processes will be quick and easy.

'''(4) Finally, finding the particular solution by using recommended method in eq.17'''

Since <math display="inline">n=5</math>, the method recommended for the odd value in equation (17) is used. Therefore,  

<math display="inline"> y_{P}=\frac{x^{n}}{\left( -1\right) ^{\frac{n-1}{2}}2^{n}a^{n}n!}\left( -k_{1}\cos ax+k_{2}\sin ax\right)</math>

For <math display="inline">a=1,k_{1}=2,k_{2}=-3,n=5</math> yields  

<math display="inline">y_{P}=\frac{1}{3840}x^{5}\left( -2\cos x-3\sin x\right)</math>.

'''Proof:'''

<math display="inline">y=y_{P}</math>  

<math display="inline">\begin{array}{ll}(D^{2}+1)^{5}y & =\left( D^{10}+5D^{8}+10D^{6}+10D^{4}+5D^{2}+1\right) y\\                  & =D_{x^{10}}y+5D_{x^{8}}y+10D_{x^{6} }y+10D_{x^{4}}y+5D_{x^{2}}y+y  \end{array}</math>

<math display="inline">y=\frac{1}{3840}x^{5}\left( -2\cos x-3\sin x\right)</math> 

<math display="inline">D_{x^{2}}y = \frac{1}{3840}x^{3}\left( 20x\sin x-60\sin  x-30x\cos x-40\cos x+2x^{2}\cos x+3x^{2}\sin x\right)</math>

<math display="inline">D_{x^{4}}y =\left( -\frac{1}{3840}\right) x\left( \begin{array}{c} 240\cos x+360\sin x+720x\cos x-480x\sin x-240x^{2}\cos x\\ -60x^{3}\cos x+2x^{4}\cos x-360x^{2}\sin x+40x^{3}\sin x+3x^{4}\sin x \end{array} \right)</math>

<math display="inline">D_{x^{6}}y =\frac{1}{3840}\left( \begin{array}{c} 1440\sin x-2160\cos x+3600x\cos x+5400x\sin x+3600x^{2}\cos x\\ -600x^{3}\cos x-90x^{4}\cos x+2x^{5}\cos x-2400x^{2}\sin x\\ -900x^{3}\sin x+60x^{4}\sin x+3x^{5}\sin x \end{array} \right)</math>

<math display="inline">D_{x^{8}}y =\left( -\frac{1}{3840}\right) \left( \begin{array}{c} 13\,440\sin x-20\,160\cos x+16\,800x\cos x+25\,200x\sin x\\ +10\,080x^{2}\cos x-1120x^{3}\cos x-120x^{4}\cos x+2x^{5}\cos x\\ -6720x^{2}\sin x-1680x^{3}\sin x+80x^{4}\sin x+3x^{5}\sin x \end{array} \right)</math>

<math display="inline">D_{x^{10}}y= \frac{1}{3840}\left( \begin{array}{c} 60\,480\sin x-90\,720\cos x+50\,400x\cos x+75\,600x\sin x\\ +21\,600x^{2}\cos x-1800x^{3}\cos x-150x^{4}\cos x+2x^{5}\cos x\\ -14\,400x^{2}\sin x-2700x^{3}\sin x+100x^{4}\sin x+3x^{5}\sin x \end{array} \right) </math>

<math display="inline">D_{x^{10}}y+5D_{x^{8}}y+10D_{x^{6}}y+10D_{x^{4}}y+5D_{x^{2}} y + y  ={2}\sin x-3\cos x</math>

=== Example 2 ===

Consider such a differential equation  

<math display="inline">(D^{2}+1)^{10}y=2\sin x-3\cos x</math> denkleminin çözümü.

Since <math display="inline">n=10</math>, the method recommended for the even value in equation (14) is used. Therefore,  

<math display="inline">y_{P}=\frac{x^{n}}{\left( -1\right) ^{\frac{n}{2}}2^{n}a^{n}n!}\left( k_{1}\sin ax+k_{2}\cos ax\right) </math>  

For <math display="inline">a=1,k_{1}=2,k_{2}=-3,n=10</math> gives  

<math display="inline">y_{P}=\left( -\frac{1}{3715\,891\,200}\right) x^{10}\left( 2\sin x-3\cos  x\right)</math>

'''Proof''':

<math display="inline">\begin{array}{l}(D^{2}+1)^{10}y & =\left( \begin{array}{c} D^{20}+10D^{18}+45D^{16}+120D^{14}+210D^{12}+{252}D^{10}\\ +210D^{8}+120D^{6}+45D^{4}+10D^{2}+1 \end{array} \right) y\\ & =D_{x^{20}}y+10D_{x^{18}}y+45D_{x^{16}}y+120D_{x^{14}}y+210D_{x^{12}}y+{252}D_{x^{10}}y\\
&+210D_{x^{8}}y+120D_{x^{6}}y+45D_{x^{4}}y+10D_{x^{2}}y+y \end{array}</math>

<math display="inline">y=-\frac{1}{2^{18}3^{4}5^{2}7}x^{10}\left( 2\sin x-3\cos x\right)  </math>

<math display="inline">D_{x^{2}}y=-\frac{1}{2^{18}3^{4}5^{2}7}x^{8}\left( \begin{array}{c} 180\sin x-270\cos x+40x\cos x\\ +60x\sin x+3x^{2}\cos x-2x^{2}\sin x \end{array} \right)</math>

<math display="inline">D_{x^{4}}y= -\frac{1}{2^{18}3^{4}5^{2}7}x^{6}\left( \begin{array}{c} 10\,080\sin x-15\,120\cos x+5760x\cos x\\ +8640x\sin x+1620x^{2}\cos x-80x^{3}\cos x\\ -3x^{4}\cos x-1080x^{2}\sin x-120x^{3}\sin x+2x^{4}\sin x \end{array} \right)</math>

<math display="inline">D_{x^{6}}y=\frac{1}{2^{18}3^{4}5^{2}7}x^{4}\left( \begin{array}{c} 453\,600\cos x-302\,400\sin x-362\,880x\cos x\\ -544\,320x\sin x-226\,800x^{2}\cos x+28\,800x^{3}\cos x+4050x^{4}\cos x\\ -120x^{5}\cos x-3x^{6}\cos x+151\,200x^{2}\sin x+43\,200x^{3}\sin x\\ -2700x^{4}\sin x-180x^{5}\sin x+2x^{6}\sin x \end{array} \right)</math> 

<math display="inline">D_{x^{8}}y=\left( -\frac{1}{2^{18}3^{4}5^{2}7}\right) x^{2}\left( \begin{array}{c} 3628\,800\sin x-5443\,200\cos x+9676\,800x\cos x\\ +14\,515\,200x\sin x+12\,700\,800x^{2}\cos x-3386\,880x^{3}\cos x\\ -1058\,400x^{4}\cos x+80\,640x^{5}\cos x+7560x^{6}\cos x\\ -8467\,200x^{2}\sin x-160x^{7}\cos x-5080\,320x^{3}\sin x-3x^{8}\cos x\\ +705\,600x^{4}\sin x+120\,960x^{5}\sin x-5040x^{6}\sin x\\ -240x^{7}\sin x+2x^{8}\sin x \end{array} \right) </math>

<math display="inline">D_{x^{10}}y= \frac{1}{2^{18}3^{4}5^{2}7}\left( \begin{array}{c} 10\,886\,400\cos x-7257\,600\sin x-72\,576\,000x\cos x\\ -108\,864\,000x\sin x-244\,944\,000x^{2}\cos x+145\,152\,000x^{3}\cos x\\ +95\,256\,000x^{4}\cos x-15\,240\,960x^{5}\cos x-3175\,200x^{6}\cos x\\ +163\,296\,000x^{2}\sin x+172\,800x^{7}\cos x+217\,728\,000x^{3}\sin x\\ +12\,150x^{8}\cos x-63\,504\,000x^{4}\sin x-200x^{9}\cos x\\ -22\,861\,440x^{5}\sin x-3x^{10}\cos x+2116\,800x^{6}\sin x\\ +259\,200x^{7}\sin x-8100x^{8}\sin x-300x^{9}\sin x+2x^{10}\sin x \end{array} \right)</math>

<math display="inline">D_{x^{12}}y=-\frac{1}{2^{18}3^{4}5^{2}7}\left( \begin{array}{c} 718\,502\,400\cos x-479\,001\,600\sin x-1596\,672\,000x\cos x\\ -2395\,008\,000x\sin x-2694\,384\,000x^{2}\cos x+958\,003\,200x^{3}\cos x\\ +419\,126\,400x^{4}\cos x-47\,900\,160x^{5}\cos x-7484\,400x^{6}\cos x\\ +1796\,256\,000x^{2}\sin x+316\,800x^{7}\cos x+1437\,004\,800x^{3}\sin x\\ +17\,820x^{8}\cos x-279\,417\,600x^{4}\sin x-240x^{9}\cos x\\ -71\,850\,240x^{5}\sin x-3x^{10}\cos x+4989\,600x^{6}\sin x\\ +475\,200x^{7}\sin x-11\,880x^{8}\sin x-360x^{9}\sin x+2x^{10}\sin x \end{array} \right) </math>

<math display="inline">D_{x^{14}}y=\frac{1}{2^{18}3^{4}5^{2}7}\left( \begin{array}{c} 10\,{897}\,286\,400\cos x-7264\,857\,600\sin x-14\, 529\,715\,200x\cos x\\ -21\,{794}\,572\,800x\sin x-16\,{345}\,929\,600x^{2}\cos  x+4151\,347\,200x^{3}\cos x\\ +1362\,160\,800x^{4}\cos x-121\,080\,960x^{5}\cos x-15\,135\,120x^{6}\cos x\\ +10\,{897}\,286\,400x^{2}\sin x+524\,160x^{7}\cos  x+6227\,020\,800x^{3}\sin x\\ +24\,570x^{8}\cos x-908\,107\,200x^{4}\sin x-280x^{9}\cos x\\ -181\,621\,440x^{5}\sin x-3x^{10}\cos x+10\,090\,080x^{6}\sin x\\ +786\,240x^{7}\sin x-16\,380x^{8}\sin x-420x^{9}\sin x+2x^{10}\sin x \end{array} \right) </math>

<math display="inline">D_{x^{16}}y= \left( -\frac{1}{3715\,891\,200}\right) \left( \begin{array}{c}  87\,{178}\,291\,200\cos x-58\,{118}\,860\,800\sin  x-83\,{026}\,944\,000x\cos x\\ -124\,{540}\,416\,000x\sin x-70\,{053}\,984\,000x^{2}\cos  x+13\,{837}\,824\,000x^{3}\cos x\\ +3632\,428\,800x^{4}\cos x-264\,176\,640x^{5}\cos x-27\,518\,400x^{6}\cos x\\ +46\,{702}\,656\,000x^{2}\sin x+806\,400x^{7}\cos x+20\, 756\,736\,000x^{3}\sin x\\ +32\,400x^{8}\cos x-2421\,619\,200x^{4}\sin x-320x^{9}\cos x\\ -396\,264\,960x^{5}\sin x-3x^{10}\cos x+18\,345\,600x^{6}\sin x\\ +1209\,600x^{7}\sin x-21\,600x^{8}\sin x-480x^{9}\sin x+2x^{10}\sin x \end{array} \right) </math>

<math display="inline">D_{x^{18}}y=\frac{1}{3715\,891\,200}\left( \begin{array}{c} 476\,{367}\,091\,200\cos x-317\,{578}\,060\,800\sin  x-352\,{864}\,512\,000x\cos x\\ -529\,{296}\,768\,000x\sin x-238\,{183}\,545\,600x^{2}\cos  x+38\,{494}\,310\,400x^{3}\cos x\\ +8420\,630\,400x^{4}\cos x-518\,192\,640x^{5}\cos x-46\,267\,200x^{6}\cos x\\ +158\,{789}\,030\,400x^{2}\sin x+1175\,040x^{7}\cos x+57\, 741\,465\,600x^{3}\sin x\\ +41\,310x^{8}\cos x-5613\,753\,600x^{4}\sin x-360x^{9}\cos x\\ -777\,288\,960x^{5}\sin x-3x^{10}\cos x+30\,844\,800x^{6}\sin x\\ +1762\,560x^{7}\sin x-27\,540x^{8}\sin x-540x^{9}\sin x+2x^{10}\sin x \end{array} \right)  </math> 

<math display="inline">D_{x^{20}}y= -\frac{1}{2^{18}3^{4}5^{2}7}\left( \begin{array}{c}  2011\,{327}\,718\,400\cos x-1340\,{885}\,145\,600\sin  x-1218\,{986}\,496\,000x\cos x\\ -1828\,{479}\,744\,000x\sin x-685\,{679}\,904\,000x^{2}\cos  x+93\,{768}\,192\,000x^{3}\cos x\\ +17\,{581}\,536\,000x^{4}\cos x-937\,681\,920x^{5}\cos  x-73\,256\,400x^{6}\cos x\\ +457\,{119}\,936\,000x^{2}\sin x+1641\,600x^{7}\cos  x+140\,{652}\,288\,000x^{3}\sin x\\ +51\,300x^{8}\cos x-11\,{721}\,024\,000x^{4}\sin x-400x^{9}\cos x\\ -1406\,522\,880x^{5}\sin x-3x^{10}\cos x+48\,837\,600x^{6}\sin x\\ +2462\,400x^{7}\sin x-34\,200x^{8}\sin x-600x^{9}\sin x+2x^{10}\sin x \end{array} \right) </math>

<math display="inline">\begin{array}{l}D_{x^{20}}y+10D_{x^{18}}y+45D_{x^{16}}y+120D_{x^{14}}y+210D_{x^{12} }y+{252}D_{x^{10}}y+210D_{x^{8}}y\\
+120D_{x^{6}}y+45D_{x^{4} }y+10D_{x^{2}}y+y  ={2}\sin x-3\cos x.\end{array}</math>

==4. Concluding remarks==

When using the classical analytical methods known in the literature for solving the high order linear differential equations, the resulting systems of integral and multivariable linear equations make the numerical calculations of the related physical problems difficult and complex algorithms. However, the method we propose above and proved its validity will remove these difficulties altogether. This simple analytical solution we have presented will help to create a rapid numerical algorithm for computers and thus simplify the numerical solution of high order physical problems.                  

===References===
<div class="auto" style="width: auto; margin-left: auto; margin-right: auto;font-size: 85%;">
<div id="cite-1"></div>
[[#citeF-1|[1]]] R K Edward and B Saff and A D Snider. (1996) Fundamentals of Differential Equations. Addison-Wesley

<div id="cite-2"></div>
[[#citeF-2|[2]]] F W J Olver and D W Lozier and R F Boisvert and C W Clark. (2010) NIST Handbook of Mathematical Functions. Cambridge University Press

<div id="cite-3"></div>
[[#citeF-3|[3]]] Shawagfeh, N T. (1996) Analytic Approximate Solution for a Nonlinear Oscillator Equation, Volume 31. Computers & Mathematics with Applications 135--141

<div id="cite-4"></div>
[[#citeF-4|[4]]] Shawagfeh, N T. (2004) A boundary value technique for boundary value problems for singularly perturbed fourth-order ordinary differential equations, Volume 47. Computers & Mathematics with Applications 1673--1688

<div id="cite-5"></div>
[[#citeF-5|[5]]] Jia, J and Sogabe, T. (2013) On particular solution of ordinary differential equations with constant coefficients, Volume 219. Applied Mathematics and Computation 6761--6767

<div id="cite-6"></div>
[[#citeF-6|[6]]] Ortigueira, M D. (2014) A simple approach to the particular solution of constant coefficient ordinary differential equations, Volume 232. Applied Mathematics and Computation 254--260

<div id="cite-7"></div>
[[#citeF-7|[7]]] Barrio R. (2005) Performance of the Taylor series method for ODEs/DAEs, Volume 163. Applied Mathematics and Computation 525--545

<div id="cite-8"></div>
[[#citeF-8|[8]]] Shanthi, V and Ramanujam, N. (2004) A boundary value technique for boundary value problems for singularly perturbed fourth-order ordinary differential equations, Volume 47. Computers & Mathematics with Applications 1673--1688

<div id="cite-9"></div>
[[#citeF-9|[9]]] Wang, W and  Li,Z. (2006) A mechanical algorithm for solving ordinary differential equation, Volume 172. Applied Mathematics and Computation 568--583

<div id="cite-10"></div>
[[#citeF-10|[10]]] Wang, J. (2011) How to solve the polynomial ordinary differential equations, Volume 218. Applied Mathematics and Computation 2421--2438

<div id="cite-11"></div>
[[#citeF-11|[11]]] D Zwillinger. (1997) Handbook of Differential Equations. Academic Press

<div id="cite-12"></div>
[[#citeF-12|[12]]] M Braun. (1997) Differential Equations and Their Application. Springer-Verlag
</div>