Solving Linear Differential Equations with Constant Coefficients

International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056

Volume: 11 Issue: 01 | Jan 2024 www.irjet.net p-ISSN: 2395-0072

Solving Linear Differential Equations with Constant Coefficients

Prof. Ms. Pooja Pandit Kadam1 , Prof. Swati Appasaheb Patil2, Prof. Akshata Jaydeep Chavan3

Assistant Professor, Dept. Of Sciences & Humanities, Rajarambapu institute of Technology, Maharashtra, India

Assistant Professor, Dept. Of Sciences & Humanities, Rajarambapu institute of Technology, Maharashtra, India

Assistant Professor, Dept. Of Engineering Sciences , SKN Sinhgad Institute of Technology & Science, Kusgoan India

Abstract - In this paper, we are discussing about the different method of finding the solution of the linear differential equations with constant coefficient. For more understandingofdifferentmethodwehaveincludeddifferent examples to find Particular solution of function of linear differential equations with constant coefficient. Particular solution is combination of complementary function and particular integral.

Key Words: ParticularIntegral,ComplementaryFunction, Auxiliaryequation

1. INTRODUCTION

Thispaperincludesoverviewofthedefinitionanddifferent typesoffindingsolutionofthelineardifferentialequations withconstantcoefficient.Herewementioneddifferentrules for finding particular solution. We applied the linear differentialequationinvariousscienceandengineeringfield. It is applied to electrical circuit, chemical kinetics, Heat transfer, control system, mechanical systems etc. First we see the definitions of linear differential equation with constant coefficient then we see general solution with complementaryfunctionandparticularintegral.

1.1 Linear Differential Equation

Definition:

A general linear differential equation of nth order with constantcoefficientsis givenby:

WhereK‘sareconstantandF(x)isfunctionof x aloneor constant.

Or

1.2 Solving Linear Differential Equations with Constant Coefficients:

Complete solution of equation f(D)y=F(x) is given by C.F+P.I. WhereC.F.denotescomplementary functionand P.I.isparticularintegral.WhenF(x)=0,thenthesolutionof equationF(D)y=0isgivenbyy=C.F.

Otherwise the solution of the F(D)y=X is satisfies the particularintegral.

Thecompletesolutionofthedifferentialequationis

Y=C.F.+P.I.

i.e.Y=ComplementaryFunction+Particularintegral

2. Methods of Finding Complementary Function

ConsidertheequationF(D)y=F(x)

Step1:PutD=m,Auxiliaryequationisgivenbyf(m)=0

Step2:Solvetheauxiliaryequationandfindtherootsof theequation.Letsrootsare m1,m2,m3….mn .

Step3:After wegetcomplementaryfunctionaccordingto givenrulesbelow:

Table -1: Methods of finding C.F.

MethodsoffindingC.F RootsofA.E. Complementaryfunction (C.F.)

IfthenrootsofA.E.are realanddistinctsay

Iftwoormorerootsare equal.i.e. .

Ifrootsareimaginary m1=α+iβ,m2=α-iβ

Ifrootsareimaginary andrepeatedtwice m1=m2= α+iβ, m3=m4= α-iβ,

+

+

Exaple1]Solvethedifferentialequation:

Ans : The given differential equation is

Now put D=m then auxiliary equation is

Here roots are 0,-1,-1 then C.F. is

SinceF(x)=0thensolutionisgivenbyY=C.F.

Y =

Exaple2]Solvethedifferentialequation

Ans:

Thegivendifferentialequationis

NowputD=mthenauxiliaryequationis

Hererootsare3&5thenwehaveC.F.is C.F.=

SinceF(x)=0thensolutionisgivenbyY=C.F. Y=

3 Short Methods for Finding Particular integral

IfX≠0,inequation(1)then

ClearlyP.I.=0ifF(x)=0

Followingarethemethodsforfindingparticularintegral:

Table-2: Rules for finding P.I.

Typesof function Whattodo Corresponding P.I.

Incaseoffailure,i.e.f(a)=0

X=(sinax+b) or(cosax+b)

Andsoon

PutD2=-a2

D3=D.D2 =-a.D

D4=(D2)2 =a4

…soon inf(D)

This will form a linear expression in the denominator. Now rationalize the denominator to substitute. Operate on the numerator term by termbytaking

If case fail ,then do as type1

X=xn Take [F(D)]-1

xm

X=eax

PutD=a inf(D)

Iff’(a)=0

X=eaxg(x) Whereg(x) isany functionofx

Makef(D)is f(D+a)

3.1 Solved Example

Expand[F(D)]-1using binomialexpansionsandif (D-a)remainsinthe denominatorthentake rationalizationof denominator.TreatDin thenumeratoras derivativeofthe correspondingfunction

Wehavetoneed

Example 1] Solve the differential equation

Solution:

The given differential equation is

The auxiliary equation is

Here roots are 1+3i and 1-3i Then

Now we see P.I.

ThencompletesolutionisY=C.F.+P.I.

Ex.2]Solvethedifferentialequation

Solution:

Thedifferentialequationis (D2-1)y=5x-2

Theauxiliaryequationis m2-1=0

m=±1

Therootsare+1,-1

ThentheC.F.is C.F.=

Then

=

ThencompletesolutionisY=C.F.+P.I.

ThusY

3. CONCLUSIONS

In this paper we conclude that the solution of the differential equation is linear combination of the homogenousandparticularsolutions.Thebriefstudyofthe linear differential equation with constant coefficient providesstudentsunderstandabledatainsummariesform to apply this data in the field of physics, biology, and engineering. They also used into the electric circuits, mechanical vibrations, population growths and chemical reactions.

REFERENCES

[1] B.S. Grewal “Advanced Engineering Mathematics” by KhannaPublishers,44th Edition.

[2] “Higher Engineering Mathematics” by B.V.raamna,Tata McGraw-HillPublication,NewDelhi.

[3]“AppliedEngineeringMathematics”byH.K.Dass,Dr.Rama Verma byS.Chand.

[4]“HigherEngineeringMathematics”byB.V.raamna,Tata McGraw-HillPublication,NewDelhi.