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

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:

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:

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

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

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

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

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,

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

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

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

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

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

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

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

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

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

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,

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:

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

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

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

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

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

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