Comparing method of undetermined coefficients with modification rule and method of variation of parameters

J. Math. Comput. Sci. 8 (2018), No. 5, 542-552 https://doi.org/10.28919/jmcs/3687

ISSN: 1927-5307

COMPARING METHOD OF UNDETERMINED COEFFICIENTS WITH MODIFICATION RULE AND METHOD OF VARIATION OF PARAMETERS

JUSTIN KISAKALI1,∗, LEOPARD C. MPANDE2

1_{Business College of Education, Dar es Salaam Campus, P. O. Box 1968 Dar es Salaam, Tanzania} 2_{Tumaini University Makumira, P. O. Box 55 Arusha, Tanzania}

Copyright c2018 Kisakali and Mpande. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract. In this research article we have used the method of undetermined coefficients with modification rule

and method of variation of parameters to solve linear ordinary differential equations and found that the method of

variation of parameters is the easiest method to use.

Keywords: method of undetermined coefficients; method of variation of parameters; linear ordinary differential

equations.

2010 AMS Subject Classification:35A24.

1.

Introduction

Solving nonhomogeneous second or higher order linear ordinary differential equation (ODE)

involve finding first the solution of the homogeneous differential equation and then finding the

solution of the nonhomogeneous part [8, 9]. There are basically two methods for determining

the solution of the particular integral, the method of undetermined coefficients and variation of

∗_{Corresponding author}

E-mail address: [email protected]

Received February 25, 2018

parameters [5, 7, 3, 4, 1, 6, 2]. In this review article we have determined the appropriate method

to use in case where the particular solution is part of the homogeneous solution.

2.

Material and Methods

A general linear nth-order nonhomogeneous ODE is

(1) an(x)y(n)+an−1(x)y(n−1)+· · ·+a1(x)y0+a0(x)y=G(x).

where the coefficientsa_n(x),a_n−1(x),· · ·, a₁(x),a₀are all constants or variables with the same degree as their derivatives (Euler-Cauchy equations).

The general form of second order nonhomogeneous ODE with constant coefficients is

(2) a₂y00+a₁y0+a₀y=G(x)

The general form of second order nonhomogeneous ODE with variable coefficients

(Euler-Cauchy equation) is

(3) a2x2y00+a1xy0+a0y=G(x), x>0

Solution of the homogeneous ODE

The general homogeneous ODE is

(4) a_n(x)y(n)+a_n−1(x)y(n−1)+· · ·+a₁(x)y0+a₀(x)y=0

Theorem

Fundamental Theorem for the Homogeneous Linear ODE (4)

For a homogeneous linear ODE (4), sums and constant multiples of solutions on some open

intervalIare again solutions onI.

IfG(x) =0 in Equation (2), then the homogeneous equation has the following solutions

(com-plementary function):

y_c=Aem1x_+Bem2x_{, y}

for distinct, identical and complex roots of the auxiliary equations, respectively.

IfG(x) =0 in Equation (3), then the homogeneous equation has the following solutions:

yc=Axm1+Bxm2, yc=Axm+Bxmlnxandyc=xα[Acos(βlnx) +Bsin(βlnx)]

for distinct, identical and complex roots of the auxiliary equations, respectively.

2.1. Determination of the particular solution by Method of Undetermined Coefficients.

The particular solution, y_p of the nonhomogeneous part is assumed to take the same form as

G(x).Theypand its derivative are substituted in the differential equation and the unknown

con-stants are determined. If a term in the choice ofy_pis part of the complementary function, then

y_pis modified by multiplying byx. If the newy_pand its derivatives does not work we multiply

byx2.

Generally, the following general rule is applied

If anyy_pi,i=1,2,3,· · ·,ncontains terms that duplicate terms iny_c, then thaty_pi must be

mul-tiplied byxn, wherenis the smallest positive integer that eliminates that duplication [9, 7].

2.2. Determination of the particular solution by Method of Variation of Parameters.

Solv-ing second order linear equations by method of variation of parameters.

We know that the complementary function for the homogeneous Equation (2) is

(5) y_c=Ay₁(x) +By₂(x)

wherey1andy2are linearly independent solutions. Let’s replace the constants (parameters) A and B in Equation (5) by arbitrary functionsu₁(x)andu₂(x),respectively. We want to determine a particular solution on the nonhomogeneous equationa₂y00+a₁y0+a₀y=G(x)of the form (6) y_p=u₁(x)y₁(x) +u₂(x)y₂(x)

Differentiating Equation (6) we get

(7) y0_p=u0₁y₁+u0₂y₂+u₁y0₁+u₂y0₂

the other condition. From Equation (7), let’s impose the condition that

(8) u0₁y1+u02y2=0

Thus we have

(9) y0_p=u₁y0₁+u₂y0₂

Then we differentiate Equation (9) and get

(10) y00_p=u0₁y0₁+u0₂y0₂+u₁y00₁+u₂y00₂ Substituting Equations (6), (9) and (10) in Equation (2), we get

a₂(u0₁y0₁+u0₂y0₂+u₁y00₁+u₂y00₂) +a₁(u₁y0₁+u₂y0₂) +a₀(u₁y₁+u₂y₂) =G or

(11) u₁(a₂y00₁+a₁y₁0 +a₀y₁) +u₂(a₂y00₂+a₁y0₂+a₀y₂) +a₂(u0₁y0₁+u0₂y0₂) =G Sincey1andy2are solutions of the homogeneous equation, so

a₂y00₁+a₁y0₁+a₀y₁=0 and a₂y00₂+a₁y0₂+a₀y₂=0 and Equation (11) becomes

(12) a₂(u0₁y₁0 +u0₂y0₂) =G

Equations (8) and (12) form a system of two equations in the unknown functionsu0₁andu0₂. Af-ter solving this system using different methods such as Elimination, Substitution, and Cramer’s

Rule or the Wronskian determinant we may be able to integrate to findu₁andu₂ and then the particular solution is given by Equation (6).

Solving third order equations with method of variation of parameters. The complementary

function for 3rd order homogeneous equation is

(13) yc=Ay1(x) +By2(x) +Cy3(x)

a particular solution on the nonhomogeneous equationa₃y000+a₂y00+a₁y0+a₀y=G(x)of the form

(14) y_p=u₁(x)y₁(x) +u₂(x)y₂(x) +u₃(x)y₃(x)

Differentiating Equation (14) we get

(15) y0_p=u1y01+u01y1+u2y02+u02y2+u3y03+u03y3

Sinceu1,u2andu3are arbitrary functions, we impose three conditions on them so as to simplify our calculations. One condition is thaty_pis a solution of the differential equation and we choose

the other condition. From Equation (15), let’s impose the first condition that

(16) u0₁y1+u02y2+u03y3=0 Thus we have

(17) y0_p=u1y01+u2y02+u3y03 Then we differentiate Equation (17) and get

(18) y00_p=u₁y00₁+u0₁y0₁+u₂y00₂+u0₂y0₂+u₃y00₃+u0₃y0₃ Let’s impose the second condition that

(19) u0₁y0₁+u0₂y0₂+u0₃y0₃=0 Thus we have

(20) y00_p=u₁y00₁+u₂y00₂+u₃y00₃ Differentiating Equation (20)

(21) y000_p =u₁y000₁ +u0₁y00₁+u₂y000₂ +u0₂y00₂+u0₃y00₃+u₃y000₃ Substituting Equations (14), (17) and (21) in nonhomogeneous equation gives

a₃(u₁y000₁ +u0₁y00₁+u₂y000₂ +u0₂y00₂+u0₃y00₃+u₃y000₃)+a₂(u₁y00₁+u₂y00₂+u₃y00₃)+a₀(u₁y0₁+u₂y₂0+u₃y0₃) =Gor (22)

Sincey₁, y₂andy₃are solutions of the homogeneous equation, so

a3y0001 +a2y001+a1y01+a0y1=0, a3y0002 +a2y002+a1y02+a0y2=0 and a3y0003 +a2y003+a1y03+a0y3=0

Thus Equation (22) becomes

(23) a₃(u0₁y00₁+u₂0y00₂+u0₃y00₃) =G

Equations (16), (19) and (23) form a system of three equations in the unknown functionsu0₁, u0₂ andu0₃. After solving this system using the Wronskian determinant we may be able to integrate to findu₁, u₂andu₃and then the particular solution is given by Equation (14).

Definition

A general solution of (4) on an open intervalIis a solution of (4) onIof the form

y_c=c₁y₁+c₂y₂+c₃y₃+· · ·+c_ny_n, (c₁,c₂,· · ·,cnare arbitrary)

wherey₁,y₂,· · ·,y_nis a basis (or fundamental system) of solutions of (4) onI. From the definition, by variation of parameters we have

y_p=u₁y₁+u₂y₂+u₃y₃+· · ·+u_ny_n, (u₁,u₂,· · ·,u_nare functions to be determined)

TheWronskianW ofnsolutionsy₁,y2,· · ·,ynis defined as the nth-order determinant

W(y₁,y₂,· · ·,y_n) =

y₁ y₂ y₃ · · · y_n

y0₁ y0₂ y0₃ · · · y0_n

· · · ·

y(₁n−1) y(₂n−1) y(₃n−1) · · · y(nn−1)

The particular solutionypfor the nonhomogeneous ODE (1) is given by

yp(x) = n

∑

k=1 y_k(x)

Z _W

k(x)

W(x)G(x)dx

=y₁(x)

Z _W

1(x)

W(x)G(x)dx+y2(x)

Z _W

2(x)

W(x)G(x)dx+· · ·+yn(x)

Z _W

n(x)

3.

Results and Discussion

Examples

(1) Find the particular solution: y00−y0=x2ex Solution

y00−y0=0

The solution of the homogeneous equation isy_c=A+Bex

Using the method of undetermined coefficient to determine the particular solution

The trial solution isy_p= (ax2+bx+c)exwe modify by multiplying byx

y_p= (ax3+bx2+cx)exfinding the first and second derivatives and then substituting in the differential equation

y0_p= (ax3+bx2+cx)ex+ (3ax2+2bx+c)ex,

y00_p= (ax3+bx2+cx)ex+ (6ax2+4bx+2c)ex+ (6ax+2b)ex,

(ax3+bx2+cx)ex+ (6ax2+4bx+2c)ex+ (6ax+2b)ex−(ax3+bx2+cx)ex−(3ax2+

2bx+c)ex=x2ex,

3ax2+ (6a+2b)x+2b+c=x2,

By comparing coefficients givesa=1/3, b=−1,andc=2.

Therefore the particular solution is

y_p= (1

3x

3_−x2₊ 2x)ex.

Using the method of variation of parameters to determine the particular solution

The particular solution isy_p=u₁+u₂ex, The system of equations is

Solving the system foru0₁andu0₂and integrating

u0₂=x2 ⇒u2= 1 3x

3_,

u0₁+x2ex=0 ⇒u₁= (−x2+2x−2)ex y_p= (−x2+2x−2)ex+1

3x

3_ex_{= (}1

3x 3_−x2

+2x−2)ex

(2) Find the particular solution: y00−2y0+2y=excosx.

Solution

The solution of the homogeneous equation is

y_c=Aexcosx+Bexsinx

Using the method of undetermined coefficients, the particular solution is modified

y_p=axexcosx+bxexsinx

Finding the first and second derivatives

y0_p=axexcosx+aexcosx−axexsinx+bxexsinx+bexsinx+bxexcosx

y00_p = axexcos x+aexcosx−axexsinx−aexsinx+aexcos x−axexsin x−aexsin x−

axexcosx+bxexsinx+bexsinx+bxexcosx+bexcosx+bexsinx+bxexcosx+bexcosx−

bxexcosx

Substituting in the differential equation gives

−2aexsinx+2bexcosx=excosx

Comparing coefficients givesa=0 andb=1/2.

The particular solution is

y_p= 1

2xe

x_sinx

Using the method of variation of parameters

We have a system of equations

u0₁excosx+u0₂exsinx=0 u0₁(−exsinx+excosx) +u0₂(exsinx+excosx) =excosx Solving the system foru0₁andu0₂by using Wronskian and integrating

u0₁=−cosxsinx ⇒u₁=−1

2sin 2_x,

u0₂=1

2(1+cos 2x) ⇒u2= 1 2x+

1 4sin 2x

yp=−

1 2e

x_sin2_xcosx+1 2xe

x_sinx+1

4e

x_{sinxsin 2x=} 1

2xe

x_sinx

(3) Find the particular solution: x2y00−3xy0+4y=x2lnx, x>0. Solution

The equation is transformed to

y00−4y0+4y=te2t

The homogeneous equation is

y00−4y0+4y=0

The solution of the homogeneous equation isyc= (At+B)e2t

Using the method of undetermined coefficient to determine the particular solution.

The trial solution isyp= (at+b)e2twe modify by multiplying byt.This trial fails thus

we multiply byt2,

y_p= (at3+bt2)e2t finding the first and second derivatives and then substituting in the differential equation

y0_p= (2at3+2bt2)e2t+ (3at2+2bt)e2t,

y00_p= (4at3+4bt2)e2t+ (6at2+4bt)e2t+ (6at2+4bt)e2t+ (6at+2b)e2t,

(6at+2b)e2t=te2t,

By comparing coefficients givesa=1/6 andb=0.Thus

y_p= 1

Therefore the particular solution is

yp=

1 6x

2_(lnx)3_.

Using the method of variation of parameters to determine the particular solution

The particular solution isyp=u1te2t+u2e2t, The system of equations is

u0₁te2t+u0₂e2t=0 2u0₁te2t+u0₁e2t+2u0₂e2t=te2t Solving the system foru0₁andu0₂and integrating

u0₁=t ⇒u₁=1

2t 2_, u0₂=−t2 ⇒u₂=−1

3t 3

y_p=1

2t

3_e2t₋1

3t

3_e2t ₌1

6t 3_e2t

Therefore the particular solution is

y_p=1

6x

2_(lnx)3

4.

Conclusions

We have seen that the method of variation of parameters is easy to implement, saves time and

space during computational. Thus it is the best method to use to solve an ODE when the trial

solution is part of the complementary function.

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

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