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

Latest revision as of 11:29, 20 September 2017

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 [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 [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 [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

(1)

where ${\textstyle n\in {Z}^{+},a,k_{1},k_{2}\in {R},D^{2}={\frac {d^{2}}{dx^{2}}}}$ . The general solution of a linear differential equation as follows:

(2)

where ${\textstyle y_{h}}$ represents the solution of the homogeneous equation and ${\textstyle y_{p}}$ 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. ${\textstyle (D^{2}+a^{2})^{n}y_{h}=0}$ is obtained by the following:

(3)

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

(4)

where ${\textstyle F(D)}$ is real coefficient polynomial of ${\textstyle D}$ . 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 ${\textstyle F(D)=\sum _{j=0}^{n}a_{j}D^{j},a_{j}\in {R}}$ and ${\textstyle F(D)y=e^{iax},i^{2}=-1,F(ia)\neq 0,}$ particular solution is

(5)

Proof.

Let ${\textstyle y=e^{iax}.}$ Then,

${\textstyle Dy=iae^{iax}}$

${\textstyle D^{2}y=\left(ia\right)^{2}e^{iax}}$

${\textstyle D^{3}y=\left(ia\right)^{3}e^{iax}}$

…

${\textstyle D^{n}y=\left(ia\right)^{n}e^{iax}.}$

Hence

${\textstyle a_{0}y=a_{0}e^{iax}}$

${\textstyle a_{1}Dy=a_{1}\left(ia\right)e^{iax}}$

${\textstyle a_{2}D^{2}y=a_{2}\left(ia\right)^{2}e^{iax}}$

${\textstyle a_{3}D^{3}y=a_{3}\left(ia\right)^{3}e^{iax}}$

…

${\textstyle a_{n}D^{n}y=a_{n}\left(ia\right)^{n}e^{iax}.}$

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

${\textstyle a_{0}y+a_{1}Dy+a_{2}D^{2}y+a_{3}D^{3}y+\ldots +a_{n}D^{n}y=a_{0}e^{iax}+a_{1}\left(ia\right)e^{iax}+a_{2}\left(ia\right)^{2}e^{iax}+}$ ${\textstyle a_{3}\left(ia\right)^{3}e^{iax}+\ldots +a_{n}\left(ia\right)^{n}e^{iax}}$

we get

If we denote

and

and also

we get the particular solution as below,

Lemma 1.2. If, ${\textstyle F(D)=\sum _{j=0}^{n}a_{j}D^{j},a_{j}\in {R}}$ and ${\textstyle F(D)y=e^{-iax},}$ ${\textstyle i^{2}=-1,}$ ${\textstyle F(-ia)\neq 0}$ , particular solution is

(6)

Proof.

Let ${\textstyle y=e^{-iax}}$ . Therefore,

${\textstyle Dy=-iae^{-iax}}$

${\textstyle D^{2}y=\left(-ia\right)^{2}e^{-iax}}$

${\textstyle D^{3}y=\left(-ia\right)^{3}e^{-iax}}$

…

${\textstyle D^{n}y=\left(-ia\right)^{n}e^{-iax}.}$

Hence

${\textstyle a_{0}y=a_{0}e^{-iax}}$

${\textstyle a_{1}Dy=a_{1}\left(-ia\right)e^{-iax}}$

${\textstyle a_{2}D^{2}y=a_{2}\left(-ia\right)^{2}e^{-iax}}$

${\textstyle a_{3}D^{3}y=a_{3}\left(-ia\right)^{3}e^{-iax}}$

…

${\textstyle a_{n}D^{n}y=a_{n}\left(-ia\right)^{n}e^{-iax}.}$

Here, if these equations are adjoined to each other,

${\textstyle a_{0}y+a_{1}Dy+a_{2}D^{2}y+\ldots +a_{n}D^{n}y=a_{0}e^{-iax}+a_{1}\left(-ia\right)e^{-iax}+a_{2}\left(-ia\right)^{2}e^{-iax}+\ldots +a_{n}\left(-ia\right)^{n}e^{-iax}}$

We get

If we denote

and

and also

We get the particular solution as below,

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

(7)

Proof.

Let ${\textstyle F(D)y=e^{iax}}$ .

Then, ${\textstyle \left(D-ia\right)^{r}\left(D+ia\right)^{r}F_{1}(D)y=e^{iax}}$ and ${\textstyle \left(D-ia\right)^{r}y={\frac {e^{iax}}{\left(D+ia\right)^{r}F_{1}(D)}}.}$ From Lemma 1.1.

${\textstyle \left(D-ia\right)^{r}y={\frac {e^{iax}}{\left(2ia\right)^{r}F_{1}(ia)}}}$ ${\textstyle y_{p}={\frac {1}{\left(2ia\right)^{r}F_{1}(ia)}}\left({\frac {e^{iax}}{\left(D-ia\right)^{r}}}\right)}$

${\textstyle y_{p}={\frac {1}{\left(2ia\right)^{r}F_{1}(ia)}}\left({\frac {e^{iax}\int e^{-iax}e^{iax}dx}{\left(D-ia\right)^{r-1}}}\right)={\frac {1}{\left(2ia\right)^{r}F_{1}(ia)}}\left({\frac {e^{iax}x}{\left(D-ia\right)^{r-1}}}\right)}$

${\textstyle y_{p}={\frac {1}{\left(2ia\right)^{r}F_{1}(ia)}}\left({\frac {e^{iax}\int e^{-iax}e^{iax}xdx}{\left(D-ia\right)^{r-2}}}\right)={\frac {1}{\left(2ia\right)^{r}F_{1}(ia)}}\left({\frac {e^{iax}{\frac {x^{2}}{2!}}}{\left(D-ia\right)^{r-2}}}\right)}$

…

${\textstyle y_{p}={\frac {1}{\left(2ia\right)^{r}F_{1}(ia)}}\left({\frac {e^{iax}{\frac {x^{r-2}}{(r-2)!}}}{\left(D-ia\right)^{2}}}\right)}$

${\textstyle y_{p}={\frac {1}{\left(2ia\right)^{r}F_{1}(ia)}}\left({\frac {e^{iax}\int e^{-iax}e^{iax}{\frac {x^{r-2}}{(r-2)!}}dx}{\left(D-ia\right)}}\right)={\frac {1}{\left(2ia\right)^{r}F_{1}(ia)}}\left({\frac {e^{iax}{\frac {x^{r-1}}{(r-1)!}}}{\left(D-ia\right)}}\right)}$

${\textstyle y_{p}={\frac {1}{\left(2ia\right)^{r}F_{1}(ia)}}\left(e^{iax}\int e^{-iax}e^{iax}{\frac {x^{r-1}}{(r-1)!}}dx\right)={\frac {1}{\left(2ia\right)^{r}F_{1}(ia)}}\left(e^{iax}\int {\frac {x^{r-1}}{(r-1)!}}dx\right)}$

${\textstyle y_{p}={\frac {1}{\left(2ia\right)^{r}F_{1}(ia)}}\left(e^{iax}\int e^{-iax}e^{iax}{\frac {x^{r-1}}{(r-1)!}}dx\right),}$

we obtain the particular solution as below,

(8)

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

(9)

Proof.

Let ${\textstyle F(D)y=e^{-iax}}$ Therefore,

${\textstyle \left(D-ia\right)^{r}\left(D+ia\right)^{r}F_{1}(D)y=e^{-iax}}$

${\textstyle {\begin{array}{l}\left(D+ia\right)^{r}y={\frac {e^{-iax}}{\left(D-ia\right)^{r}F_{1}(D)}}.\\\end{array}}}$

From Lemma 1.2.

${\textstyle {\begin{array}{l}\left(D+ia\right)^{r}y={\frac {e^{-}iax}{\left(-2ia\right)^{r}F_{1}(-ia)}}\\y_{p}={\frac {1}{\left(-2ia\right)^{r}F_{1}(-ia)}}\left({\frac {e^{-}iax}{\left(D+ia\right)^{r}}}\right)\\y_{p}={\frac {1}{\left(-2ia\right)^{r}F_{1}(-ia)}}\left({\frac {e^{-}iaxe^{i}axe^{-}iaxdx}{\left(D+ia\right)^{r}-1}}\right)={\frac {1}{\left(-2ia\right)^{r}F_{1}(-ia)}}\left({\frac {e^{-}iaxx}{\left(D+ia\right)^{r}-1}}\right)\\y_{p}={\frac {1}{\left(-2ia\right)^{r}F_{1}(-ia)}}\left({\frac {e^{-}iaxe^{i}axe^{-}iaxxdx}{\left(D+ia\right)^{r}-2}}\right)={\frac {1}{\left(-2ia\right)^{r}F_{1}(-ia)}}\left({\frac {e^{-}iax{\frac {x^{2}}{2!}}}{\left(D+ia\right)^{r}-2}}\right)\\\ldots \\y_{p}={\frac {1}{\left(-2ia\right)^{r}F_{1}(-ia)}}\left({\frac {e^{-}iax{\frac {x^{r}-2}{(r-2)!}}}{\left(D+ia\right)^{2}}}\right)\\y_{p}={\frac {1}{\left(-2ia\right)^{r}F_{1}(-ia)}}\left({\frac {e^{-}iaxe^{i}axe^{-}iax{\frac {x^{r}-2}{(r-2)!}}dx}{\left(D+ia\right)}}\right)={\frac {1}{\left(-2ia\right)^{r}F_{1}(-ia)}}\left({\frac {e^{-}iax{\frac {x^{r}-1}{(r-1)!}}}{\left(D+ia\right)}}\right),\end{array}}}$

we obtain the particular solution as below,

${\textstyle y_{p}={\frac {1}{\left(-2ia\right)^{r}F_{1}(-ia)}}\left(e^{-iax}\int e^{iax}e^{-iax}{\frac {x^{r-1}}{(r-1)!}}dx\right)={\frac {x^{r}}{r!}}{\frac {e^{-iax}}{\left(-2ia\right)^{r}F_{1}(-ia)}}}$ .

2. A Simple Approach to the Particular Solution, y_p

Particular solution of considered differential equations which is linear, constant coefficient, non-homogeneous and higher order differential equation is written as

(10)

Using lemmas 1.1.,1.2.,1.3.,1.4. the Euler Identity by writing instead of $\sin ax\,\cos ax$ in (10), it becomes the following:

{\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}}

(11)

The particular solution is obtained in two ways depending on being even or odd of $n$ . If $n$ is even and $n=2m\left(m\in {Z}^{+}\right)$

{\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}}

(12)

Therefore, particular solution (10) for even values of n is obtained below:

(13)

Using these result, we obtain the following general solution of (1) for even values of n as follows:

(14)

If n is odd and ${\textstyle n=2m+1\left(m\in {Z}^{+}\right)}$ , hence, particular solution (10) is

{\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}}

(15)

Therefore, particular solution (10) for odd values of n is obtained below:

(16)

Using these result, we obtain the following general solution of (1) for odd values of n as follows:

(17)

3. Examples

Example 1

Consider such a differential equation ${\textstyle (D^{2}+1)^{5}y=2\sin x-3\cos x}$ . Solution of this equation: firstly, we solve the homogen equation,

{\begin{array}{l}(D^{2}+1)^{5}y=0\\(m^{2}+1)^{5}=0\\m_{1,2,3,4,5}=iy_{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}=-iy_{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}}

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

${\textstyle y_{P}=e^{ix}\int \int \int \int \int e^{-2ix}\int \int \int \int \int e^{ix}\left(2\sin x-3\cos x\right)\left(dx\right)^{10}}$

As can be seen, the solution of ${\textstyle y_{P}}$ 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 ${\textstyle y_{P}=\sum _{i=1}^{10}L_{i}y_{i}}$ . ${\textstyle L_{i}}$ functions are defined as:

${\textstyle \sum _{i=1}^{10}L_{i}^{\prime }y_{i}=0,\sum _{i=1}^{10}L_{i}^{\prime }y_{i}^{\left(1\right)}=0,...,\sum _{i=1}^{10}L_{i}^{\prime }y_{i}^{\left(8\right)}=0,\sum _{i=1}^{10}L_{i}^{\prime }y_{i}^{\left(9\right)}=2\sin x-3\cos x}$

Solving above linear equation system, with the derivative functions of ${\textstyle L_{i}^{\prime },(1\leq i\leq 10)}$ using this derivative functions and following integration ${\textstyle L_{i}}$ yields

${\textstyle L_{i}=\int L_{i}^{\prime }dx,(1\leq i\leq 10)}$ .

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

(3) By using the method of undetermined coefficients

Let us ${\textstyle y_{P}=\left(\sum _{i=0}^{9}A_{i}x^{i}\right)\cos x+\left(\sum _{i=0}^{9}B_{i}x^{i}\right)\sin x}$ . In this equation, replaces ${\textstyle y,D_{x^{2}}y,...,D_{x^{10}}y}$ and its derivatives. With the identification of the two sides, ${\textstyle A_{i},B_{i}(1\leq i\leq 10)}$ 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 ${\textstyle n=5}$ , the method recommended for the odd value in equation (17) is used. Therefore,

${\textstyle y_{P}={\frac {x^{n}}{\left(-1\right)^{\frac {n-1}{2}}2^{n}a^{n}n!}}\left(-k_{1}\cos ax+k_{2}\sin ax\right)}$

For ${\textstyle a=1,k_{1}=2,k_{2}=-3,n=5}$ yields

${\textstyle y_{P}={\frac {1}{3840}}x^{5}\left(-2\cos x-3\sin x\right)}$ .

Proof:

${\textstyle y=y_{P}}$

${\textstyle {\begin{array}{ll}(D^{2}+1)^{5}y&=\left(D^{10}+5D^{8}+10D^{6}+10D^{4}+5D^{2}+1\right)y\\&=D_{x^{10}}y+5D_{x^{8}}y+10D_{x^{6}}y+10D_{x^{4}}y+5D_{x^{2}}y+y\end{array}}}$

${\textstyle y={\frac {1}{3840}}x^{5}\left(-2\cos x-3\sin x\right)}$

${\textstyle D_{x^{2}}y={\frac {1}{3840}}x^{3}\left(20x\sin x-60\sin x-30x\cos x-40\cos x+2x^{2}\cos x+3x^{2}\sin x\right)}$

${\textstyle D_{x^{4}}y=\left(-{\frac {1}{3840}}\right)x\left({\begin{array}{c}240\cos x+360\sin x+720x\cos x-480x\sin x-240x^{2}\cos x\\-60x^{3}\cos x+2x^{4}\cos x-360x^{2}\sin x+40x^{3}\sin x+3x^{4}\sin x\end{array}}\right)}$

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

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

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

${\textstyle D_{x^{10}}y+5D_{x^{8}}y+10D_{x^{6}}y+10D_{x^{4}}y+5D_{x^{2}}y+y={2}\sin x-3\cos x}$

Example 2

Consider such a differential equation

${\textstyle (D^{2}+1)^{10}y=2\sin x-3\cos x}$ .

Since ${\textstyle n=10}$ , the method recommended for the even value in equation (14) is used. Therefore,

${\textstyle y_{P}={\frac {x^{n}}{\left(-1\right)^{\frac {n}{2}}2^{n}a^{n}n!}}\left(k_{1}\sin ax+k_{2}\cos ax\right)}$

For ${\textstyle a=1,k_{1}=2,k_{2}=-3,n=10}$ gives

${\textstyle y_{P}=\left(-{\frac {1}{3715\,891\,200}}\right)x^{10}\left(2\sin x-3\cos x\right)}$

Proof:

${\textstyle {\begin{array}{l}(D^{2}+1)^{10}y&=\left({\begin{array}{c}D^{20}+10D^{18}+45D^{16}+120D^{14}+210D^{12}+{252}D^{10}\\+210D^{8}+120D^{6}+45D^{4}+10D^{2}+1\end{array}}\right)y\\&=D_{x^{20}}y+10D_{x^{18}}y+45D_{x^{16}}y+120D_{x^{14}}y+210D_{x^{12}}y+{252}D_{x^{10}}y\\&+210D_{x^{8}}y+120D_{x^{6}}y+45D_{x^{4}}y+10D_{x^{2}}y+y\end{array}}}$

${\textstyle y=-{\frac {1}{2^{18}3^{4}5^{2}7}}x^{10}\left(2\sin x-3\cos x\right)}$

${\textstyle D_{x^{2}}y=-{\frac {1}{2^{18}3^{4}5^{2}7}}x^{8}\left({\begin{array}{c}180\sin x-270\cos x+40x\cos x\\+60x\sin x+3x^{2}\cos x-2x^{2}\sin x\end{array}}\right)}$

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

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

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

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

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

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

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

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

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

${\textstyle {\begin{array}{l}D_{x^{20}}y+10D_{x^{18}}y+45D_{x^{16}}y+120D_{x^{14}}y+210D_{x^{12}}y+{252}D_{x^{10}}y+210D_{x^{8}}y\\+120D_{x^{6}}y+45D_{x^{4}}y+10D_{x^{2}}y+y={2}\sin x-3\cos x.\end{array}}}$

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

[1] R K Edward and B Saff and A D Snider. (1996) Fundamentals of Differential Equations. Addison-Wesley

[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

[3] Shawagfeh, N T. (1996) Analytic Approximate Solution for a Nonlinear Oscillator Equation, Volume 31. Computers & Mathematics with Applications 135--141

[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

[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

[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

[7] Barrio R. (2005) Performance of the Taylor series method for ODEs/DAEs, Volume 163. Applied Mathematics and Computation 525--545

[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

[9] Wang, W and Li,Z. (2006) A mechanical algorithm for solving ordinary differential equation, Volume 172. Applied Mathematics and Computation 568--583

[10] Wang, J. (2011) How to solve the polynomial ordinary differential equations, Volume 218. Applied Mathematics and Computation 2421--2438

[11] D Zwillinger. (1997) Handbook of Differential Equations. Academic Press

[12] M Braun. (1997) Differential Equations and Their Application. Springer-Verlag

Latest revision as of 11:29, 20 September 2017

Abstract

1. Introduction

2. A Simple Approach to the Particular Solution, y_p

3. Examples

Example 1

Example 2

4. Concluding remarks

References

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@@ Line 3: / Line 3: @@
 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==
-==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]]].
@@ Line 75: / Line 74: @@
 |}
-==Proof.==
+'''Proof.'''
 Let <math display="inline">y=e^{iax}.</math> Then,
@@ Line 105: / Line 104: @@
 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}+a_{3}\left( ia\right)^{3}e^{iax}+\ldots + a_{n}\left(ia\right)^{n}e^{iax}</math>
+<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
@@ Line 162: / Line 161: @@
 |}
-'''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>,
+'''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
-particular solution is
 {| class="formulaSCP" style="width: 100%; text-align: left;"
@@ Line 176: / Line 173: @@
 |}
-==Proof.==
+'''Proof.'''
 Let <math display="inline">y=e^{-iax}</math>. Therefore,
@@ Line 206: / Line 203: @@
 Here, if these equations are adjoined to each other,
-<math display="inline">a_{0}y+a_{1}Dy+a_{2}D^{2}y+\ldotsa_{n}D^{n}y=a_{0}e^{-iax}+a_{1}\left( -ia\right)e^{-iax}+a_{2}\left(-ia\right)^{2}e^{-iax}+\ldotsa_{n}\left( -ia\right)^{n}e^{-iax}</math>
+<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
@@ Line 263: / Line 260: @@
 |}
-'''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">\begin{array}{l}F(D)=\left(D-ia\right)^{r}\left(D+ia\right) ^{r}F_{1}(D),\\F_{1}(ia)\neq 0,\end{array}</math> and <math display="inline">F(D)y=e^{iax},</math> <math display="inline">i^{2}=-1</math>, particular solution is
+'''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;"
@@ Line 275: / Line 272: @@
 |}
-==Proof.==
+'''Proof.'''
 Let <math display="inline">F(D)y=e^{iax}</math>.
@@ Line 309: / Line 306: @@
 |}
-'''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">\begin{array}{l}F(D)=\left(D-ia\right)^{r}\left(D+ia\right) ^{r}F_{1}(D),\\ F_{1}(-ia)\neq 0,\end{array}</math> and <math display="inline">F(D)y=e^{-iax},</math> <math display="inline">i^{2}=-1,</math> particular solution is
+'''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;"
@@ Line 321: / Line 318: @@
 |}
-==Proof.==
+'''Proof.'''
 Let <math display="inline">F(D)y=e^{-iax}</math> Therefore,
@@ Line 327: / Line 324: @@
 <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)}.\\  \noindent From Lemma 1.2.\\  \end{array}</math>\left(D+ia\right)^ry=\frac{e^-iax}{\left(-2ia\right)^rF_1(-ia)}<math display="inline">\begin{array}{l}\\  \end{array}</math>y_p=\frac{1}{\left(-2ia\right)^rF_1(-ia)}\left(\frac{e^-iax}{ \left(D+ia\right)^r}\right)<math display="inline">\begin{array}{l}\\  \end{array}</math>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)<math display="inline">\begin{array}{l}\\  \end{array}</math>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)<math display="inline">\begin{array}{l}\\  \ldots  \end{array}</math>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)<math display="inline">\begin{array}{l}\\  \end{array}</math>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),<span id="eq-10"></span>
+<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>
-<span id="eq-10"></span>
-<span id="eq-10"></span>
-<span id="eq-10"></span>
-<span id="eq-10"></span>
-<math display="inline">\begin{array}{l}\\  \noindent we obtain the particular solution as below,\\   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)}.
-==2 A Simple Approach to the Particular Solution, ==
+From Lemma 1.2.
-\mathbf{y_{p}}   Particular solution of considered differential equations which is linear, constant coefficient, non-homogeneous and higher order differential equation is written as     \noindent Using lemmas 1.1.,1.2.,1.3.,1.4. the Euler Identity by writing instead of \sin ax   \cos ax in [[#eq-10|10]], it becomes the following:   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).    \noindent The particular solution is obtained in two ways depending on being even or odd of n. If n is even and n=2m\left( m\in {Z}^{+}\right).      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).\\ Therefore, particular solution [[#eq-10|10]] for even values of n is obtained below:  y_{p}=\frac{x^{2m}}{\left( -1\right) ^{m}2^{2m}a^{2m}2m!}\left( k_{1}\sin ax+k_{2}\cos ax\right).    \noindent Using these result, we obtain the following general solution of [[#eq-1|1]] for even values of n as follows:   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).   \noindent If n is odd and n=2m+1\left( m\in {Z}^{+}\right), hence, particular solution [[#eq-10|10]] is   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). \\  \noindent Therefore, particular solution [[#eq-10|10]] for odd values of n is obtained below:   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).  Using these result, we obtain the following general solution of [[#eq-1|1]] for odd values of n as follows:   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 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>
-==3 Examples==
+we obtain the particular solution as below,
-\noindent \mathbf{Example.1}\\  \noindent Consider such a differential equation  (D^{2}+1)^{5}y=2\sin x-3\cos x.  Solution of this equation: firstly, we solve the homogen equation,  (D^{2}+1)^{5}y=0  (m^{2}+1)^{5}=0   \begin{array}{c} [c]{cc} 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 \end{array}   y=y_{h}  y=\sum _{i=1}^{10}C_{i}y_{i}  Now, let's try to solve the non-homogeneous equation using the classical methods and then the formula we recommend.   \left( \hbox{'''1'''}\right) \mathbf{By using the method of consecutive integration:}
+<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>.
-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}  As can be seen, the solution of y_{P} 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.
-\left( \hbox{'''2'''}\right)  \mathbf{By using the method of variation of parameters:}
+== 2. A Simple Approach to the Particular Solution, y<sub>p</sub>==
-Let us  y_{P}=\sum _{i=1}^{10}L_{i}y_{i}.   L_{i} functions are defined as:    \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  Solving above linear equation system, with the derivative functions of L_{i}^{\prime }, (1\leqslant i\leqslant{10)}   using this derivative functions and following integration L_{i} yields  L_{i}=\int L_{i}^{\prime }dx,  (1\leqslant i\leqslant{10).}   It can not be said that these two processes will be quick and easy.
+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)
+|}
-\left( \hbox{'''3'''}\right)  \mathbf{By using the method of undetermined coefficients}
-Let us 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. In this equation, replaces  y,D_{x^{2}}y,...,D_{x^{10}}y and its derivatives. With the identification of the two sides,   A_{i},B_{i}(1\leqslant i\leqslant{10)} coefficients are determined.  It can also not be said that these two processes will be quick and easy.
+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)
+|}
-\left( \hbox{'''4'''}\right)  \mathbf{Finally, finding the particular solution by using recommended method in eq.17 }
+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)
+|}
-Since n=5, the method recommended for the odd value in equation 17 is used. Therefore,  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)   For a=1,k_{1}=2,k_{2}=-3,n=5 yields  y_{P}=\frac{1}{3840}x^{5}\left( -2\cos x-3\sin x\right) .
+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)
+|}
-\mathbf{Proof:}
+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)
+|}
-y=y_{P}  (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  y=\frac{1}{3840}x^{5}\left( -2\cos x-3\sin x\right)   D_{x^{2}}y\allowbreak  = \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) \allowbreak   D_{x^{4}}y\allowbreak  =\left( -\frac{1}{3840}\right) x\left( \begin{array}{c} [c]{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) \allowbreak  \allowbreak
+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)
+|}
-D_{x^{6}}y\allowbreak =\frac{1}{3840}\left( \begin{array}{c} [c]{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) \allowbreak   D_{x^{8}}y\allowbreak =\left( -\frac{1}{3840}\right) \left( \begin{array}{c} [c]{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) \allowbreak   D_{x^{10}}y=\allowbreak \frac{1}{3840}\left( \begin{array}{c} [c]{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) \allowbreak   D_{x^{10}}y+5D_{x^{8}}y+10D_{x^{6}}y+10D_{x^{4}}y+5D_{x^{2}}y+y\allowbreak  \allowbreak =\allowbreak{2}\sin x-3\cos x
+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)
+|}
-\mathbf{Example.2}
+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)
+|}
-Consider such a differential equation  (D^{2}+1)^{10}y=2\sin x-3\cos x denkleminin \c{c}\& #246;z\& #252;m\& #252;,  Since n=10, the method recommended for the even value in equation 14 is used. Therefore,  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)   For a=1,k_{1}=2,k_{2}=-3,n=10 gives  y_{P}=\left( -\frac{1}{3715\,891\,200}\right) x^{10}\left( 2\sin x-3\cos  x\right)
+==3. Examples==
-\mathbf{Proof:}
+=== Example 1 ===
-(D^{2}+1)^{10}y=\left( \begin{array}{c} [c]{c} D^{20}+10D^{18}+45D^{16}+120D^{14}+210D^{12}+\allowbreak{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+\allowbreak{252}D_{x^{10}}y    +210D_{x^{8}}y+120D_{x^{6}}y+45D_{x^{4}}y+10D_{x^{2}}y+y  y=-\frac{1}{2^{18}3^{4}5^{2}7}x^{10}\left( 2\sin x-3\cos x\right)   D_{x^{2}}y=\allowbreak -\frac{1}{2^{18}3^{4}5^{2}7}x^{8}\left( \begin{array}{c} [c]{c} 180\sin x-270\cos x+40x\cos x\\ +60x\sin x+3x^{2}\cos x-2x^{2}\sin x \end{array} \right) \allowbreak   D_{x^{4}}y=\allowbreak -\frac{1}{2^{18}3^{4}5^{2}7}x^{6}\left( \begin{array}{c} [c]{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) \allowbreak   D_{x^{6}}y=\allowbreak \frac{1}{2^{18}3^{4}5^{2}7}x^{4}\left( \begin{array}{c} [c]{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) \allowbreak   D_{x^{8}}y=\allowbreak \left( -\frac{1}{2^{18}3^{4}5^{2}7}\right) x^{2}\left( \begin{array}{c} [c]{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) \allowbreak   D_{x^{10}}y=\allowbreak \frac{1}{2^{18}3^{4}5^{2}7}\left( \begin{array}{c} [c]{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) \allowbreak   D_{x^{12}}y=\allowbreak -\frac{1}{2^{18}3^{4}5^{2}7}\left( \begin{array}{c} [c]{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) \allowbreak   D_{x^{14}}y=\allowbreak \frac{1}{2^{18}3^{4}5^{2}7}\left( \begin{array}{c} [c]{c} 10\,\allowbreak{897}\,286\,400\cos x-7264\,857\,600\sin x-14\,\allowbreak  529\,715\,200x\cos x\\ -21\,\allowbreak{794}\,572\,800x\sin x-16\,\allowbreak{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\,\allowbreak{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) \allowbreak   D_{x^{16}}y=\allowbreak \left( -\frac{1}{3715\,891\,200}\right) \left( \begin{array}{c} [c]{c} 87\,\allowbreak{178}\,291\,200\cos x-58\,\allowbreak{118}\,860\,800\sin  x-83\,\allowbreak{026}\,944\,000x\cos x\\ -124\,\allowbreak{540}\,416\,000x\sin x-70\,\allowbreak{053}\,984\,000x^{2}\cos  x+13\,\allowbreak{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\,\allowbreak{702}\,656\,000x^{2}\sin x+806\,400x^{7}\cos x+20\,\allowbreak  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) \allowbreak   D_{x^{18}}y=\allowbreak \frac{1}{3715\,891\,200}\left( \begin{array}{c} [c]{c} 476\,\allowbreak{367}\,091\,200\cos x-317\,\allowbreak{578}\,060\,800\sin  x-352\,\allowbreak{864}\,512\,000x\cos x\\ -529\,\allowbreak{296}\,768\,000x\sin x-238\,\allowbreak{183}\,545\,600x^{2}\cos  x+38\,\allowbreak{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\,\allowbreak{789}\,030\,400x^{2}\sin x+1175\,040x^{7}\cos x+57\,\allowbreak  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)   \allowbreak D_{x^{20}}y=\allowbreak -\frac{1}{2^{18}3^{4}5^{2}7}\left( \begin{array}{c} [c]{c} 2011\,\allowbreak{327}\,718\,400\cos x-1340\,\allowbreak{885}\,145\,600\sin  x-1218\,\allowbreak{986}\,496\,000x\cos x\\ -1828\,\allowbreak{479}\,744\,000x\sin x-685\,\allowbreak{679}\,904\,000x^{2}\cos  x+93\,\allowbreak{768}\,192\,000x^{3}\cos x\\ +17\,\allowbreak{581}\,536\,000x^{4}\cos x-937\,681\,920x^{5}\cos  x-73\,256\,400x^{6}\cos x\\ +457\,\allowbreak{119}\,936\,000x^{2}\sin x+1641\,600x^{7}\cos  x+140\,\allowbreak{652}\,288\,000x^{3}\sin x\\ +51\,300x^{8}\cos x-11\,\allowbreak{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) \allowbreak   D_{x^{20}}y+10D_{x^{18}}y+45D_{x^{16}}y+120D_{x^{14}}y+210D_{x^{12} }y+\allowbreak{252}D_{x^{10}}y+210D_{x^{8}}y+120D_{x^{6}}y+45D_{x^{4} }y+10D_{x^{2}}y+y  =\allowbreak{2}\sin x-3\cos x.
+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;" |
+|}
-==4 Concluding remarks==
+Now, let's try to solve the non-homogeneous equation using the classical methods and then the formula we recommend.
-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.
+'''(1) By using the method of consecutive integration'''
-===BIBLIOGRAPHY===
+<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>.
+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
+[[#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
+[[#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
+[[#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
+[[#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
+[[#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
+[[#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
+[[#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
+[[#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
+[[#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
+[[#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
+[[#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
+[[#citeF-12|[12]]] M Braun. (1997) Differential Equations and Their Application. Springer-Verlag
+</div>
-\end{array}</math>

Latest revision as of 11:29, 20 September 2017

Abstract

1. Introduction

2. A Simple Approach to the Particular Solution, yp

3. Examples

Example 1

Example 2

4. Concluding remarks

References

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2. A Simple Approach to the Particular Solution, y_p