Solving Linear Differential Equations with Constant Coefficients

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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
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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
understanding of different method we have included different
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: Particular Integral, Complementary Function,
Auxiliary equation.
1. INTRODUCTION
This paper includes overview of the definition and different
types of finding solution of the linear differential equations
with constant coefficient. Here we mentioneddifferent rules
for finding particular solution. We applied the linear
differential equationinvariousscienceandengineeringfield.
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
complementary function and particular integral.
1.1 Linear Differential Equation
Definition:
A general linear differential equation of nth order with
constant coefficients is given by:
Where K ‘s are constant and F(x) is function of x alone or
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. Where C.F. denotes complementary function and
P.I. is particular integral. When F(x) =0, then the solution of
equation F (D)y=0 is given by y=C.F.
Otherwise the solution of the F(D)y=X is satisfies the
particular integral.
The complete solution of the differential equation is
Y= C.F. + P.I.
i.e. Y=Complementary Function + Particular integral
2. Methods of Finding Complementary Function
Consider the equation F(D)y=F(x)
Step 1: Put D=m, Auxiliary equation is given by f(m)=0
Step 2: Solve the auxiliary equation and find the roots of
the equation. Lets roots are m1,m2,m3….mn .
Step 3: After wegetcomplementaryfunctionaccording to
given rules below:
Table -1: Methods of finding C.F.
Methods of finding C.F
Roots of A.E.
Complementary function
(C.F.)
If the n roots of A.E. are
real and distinct say
C1em1X
+ C2em2X
+……
+ Cnemnx
If two or more roots are
equal.i.e.
.
+
…..
If roots are imaginary
m1=α+iβ, m2=α-iβ
If roots are imaginary
and repeated twice
m1=m2= α+iβ,
m3=m4= α-iβ,
+
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

Exaple 1] Solve the differential equation:
Ans : The given differential equation is
Now put D=m then auxiliary equation is
Here roots are 0,-1,-1 then C.F. is
Since F(x) =0 then solution is given by Y=C.F.
Y =
Exaple 2] Solve the differential equation
Ans :
The given differential equation is
Now put D=m then auxiliary equation is
Here roots are 3 & 5 then we have C.F. is
C.F. =
Since F(x) =0 then solution is given by Y=C.F.
Y=
3 Short Methods for Finding Particular integral
If X ≠ 0, in equation (1) then
Clearly P.I. =0 if F(x) =0
Following are the methods for finding particular integral:
Table-2: Rules for finding P.I.
Types of
function
What to do Corresponding P.I.
X = eax
Put D = a
in f(D)
In case of failure, i.e. f(a)=0
If f’(a)=0
And so on
X = (sin ax+b)
or (cos ax+b)
Put D2= -a2
D3=D.D2
=-a.D
D4=(D2)2
=a4
… so on
in f(D)
This will form a linear
expression in the
denominator. Now
rationalize the
denominator to
substitute. Operate on
the numerator term by
term by taking
If case fail ,then do as
type 1
X = xn
Take
[F(D)]-1
xm
Expand [F(D)]-1 using
binomial expansions and if
(D-a) remains in the
denominator then take
rationalization of
denominator. Treat D in
the numerator as
derivative of the
corresponding function
X =eaxg(x)
Where g(x)
is any
function of x
Make f(D) is
f(D+a)
We have to need
3.1 Solved Example
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.

Then complete solution is Y=C.F.+P.I.
Ex.2] Solve the differential equation
Solution:
The differential equation is
(D2-1)y=5x-2
The auxiliary equation is
m2 -1=0
m=±1
The roots are +1,-1
Then the C.F. is
C.F.=
Then
=
Then complete solution is Y=C.F. +P.I.
Thus Y
3. CONCLUSIONS
In this paper we conclude that the solution of the
differential equation is linear combination of the
homogenous and particular solutions. The brief study of the
linear differential equation with constant coefficient
provides students understandable data in summaries form
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
Khanna Publishers, 44th Edition .
[2] “Higher Engineering Mathematics” by B.V.raamna, Tata
McGraw-Hill Publication, New Delhi.
[3] “AppliedEngineeringMathematics”byH.K.Dass,Dr.Rama
Verma by S. Chand.
[4] “Higher Engineering Mathematics” by B.V.raamna,Tata
McGraw-Hill Publication, New Delhi..

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