Unit - 1
Linear Differential Equations (LDE) and Applications
The linear homogeneous differential equation of the nth order with constant coefficients can be written as
y(n) (x) + a1y(n-1)(x)+.....+an-1y’(x)+any(x) =0
Where a1, a2 ...... an are constants which may be real or complex.
Using the linear differential operator L (D), this equation can be represented as
L (D) y(x) =0,
Where
L (D) = Dn + a1D(n-1)+⋯+an-1D+an.
Example 1:
Solve the differential equation y’’’+2y’’-y’-2y=0
Solution:
The corresponding characteristic equation is,
Solving it, we find the roots
=0
=-1
The general solution for the differential equation is
Y(x)= 
Where 
are arbitrary constants.
Example 2:
Solve the equation y’’’+11y’-5y=0
Solution:
The characteristic equation of the give D.E is
Here one of the root is 
then factorising the term 
from the equation we obtain
(
(
(
=0
(
(
=0
Thus the equation has two roots 
Hence the general equation of the D.E is
Y(x)=(
Where 
are arbitrary constants.
- A function which satisfies the L.D.E F(D)y=0 is known as complementary function of L.D.E
- A function which satisfies the L.D.E F(D)y=Q is known as complementary function of L.D.E
- The general solution of L.D.E for F(D)y = Q is given by
Y= C.F + P.I
Rules for finding complementary function-
To solve the equation- 
Where 
(co-efficient equated to zero) ……. (1)
Is called auxiliary equation.
Suppose 
are the roots.
First case- If all the roots are real and different, then equation (1) will be-
…………….. (2)
Then the complete solution is- 
Second case- If two roots are equal (
The complete solution is-
Third case- If one pair of roots be imaginary that means 
and 
, then the complete solution is-
Fourth case- If two points of imaginary roots are equal that means 
and 
, then the complete solution is-
Example-1: Solve 
Sol. The given equation can be written in the symbolic form as-
So that A.E is 
Hence the complete solution is-
Example 2:
Determine the complementary function for the following D.E
at t=0,1
Solution:
We can write the given homogeneous equation as follows
The auxiliary equation is 
By factorising the auxiliary equations we get,
(m-3)(m-4)=0
And the two real solutions are m=3,4
Hence the complementary function is,
Since f(t) = 2 then 
= c
12c=2
c=1/6 and 
= 1/6
Therefore the general solution is
X(t) = 
Example 3:
Solve 
-
Given y=4 ,
when x=0
Solution :
The auxiliary equation is
(m-1)(m-2)=0
m=1,2
Therefore complimentary function is,
The particular integral is given by y=
Substituting the above values in given equation we get,
-
= 0-3
by comparing the co-efficients we get 2
=1 
= ½
-3
+ 2
= 4+x
= 11/4
Therefore P.I is,
11/4+(1/2)x
General solution is, y = A
+ 11/4 +1/2x
Rules to find the particular integral-
Consider the equation 
So that
Case-1: when 
When 
, then the above method fails.
So that in this case-
If f’(a) = 0 then
Case-2: when 
When 
, then the above method fails.
So that in this case-
If f’(a) = 0 then
Case-3: when 
P.I. = 
Case-4: when 
here V is the function of x.
Case-5: When X is any function of x.
Example-1: Find P.I of 
Sol. P.I. = 
Put 
= 
Multiply and divide by 1+4D
= 
= 
Example-2: Find the P.I. Of 
Sol. P.I. = 
Example-3: Find P.I. Of 
Sol. P.I = 
Replace D by D+1
Put 
Example 1: Solve the following linear equations generally
9x+2y=6 9x+2y=6
3x-y=7
6x-2y=14
Solving the above two equations we get,
9x+2y=6
6x-2y=14
15x = 20
x= 20/15 = 4/3
Example 2:
Short cut methods:
(1) 2x-a+4=x+3a-1
Solution: 2x-x=3a-1+a-4 (-a+4 moves to right side)
x= 4a – 5
(2) 5x-7=2x+5
3x = 12 
x= 12/3=4
(3) X-m+3=2m+1
Solution: x= 2m+1+m-3= 3m-2 (-m and +3 to right side)
Consider a second order LDE with constant co-efficients given by
Then let the complimentary function 
is given by
Then the particular integral is
Where u and v are unknown and to be calculated using the formula
u=
Example-1: Solve the following DE by using variation of parameters-
Sol. We can write the given equation in symbolic form as-
To find CF-
It’s A.E. Is 
So that CF is- 
To find PI-
Here 
Now 
Thus PI = 
= 
= 
= 
= 
So that the complete solution is-
Example-2: Solve the following by using the method of variation of parameters.
Sol. This can be written as-
C.F.-
Auxiliary equation is- 
So that the C.F. Will be- 
P.I.-
Here 
Now 
Thus PI = 
= 
= 
So that the complete solution is-
An equation of the form
Here X is the function of x, is called Cauchy’s homogeneous linear equation.
Example-1: Solve 
Sol. As it is a Cauchy’s homogeneous linear equation.
Put 
Then the equation becomes [D(D-1)-D+1]y = t or 
Auxiliary equation-
So that-
C.F.= 
Hence the solution is- 
, we get-
Example-2: Solve 
Sol. On putting 
so that,
and 
The given equation becomes-
Or it can be written as-
So that the auxiliary equation is- 
C.F. = 
Particular integral- 
Where 
It’s a Leibnitz’s linear equation having I.F.= 
Its solution will be-
P.I. = 
= 
So that the complete solution is-
An equation of the form-
Is called Legendre’s linear equation.
Example-3: Solve 
Sol. As we see that this is a Legendre’s linear equation.
Now put 
So that- 
And 
Then the equation becomes- D (D – 1)y+ Dy + y = 2 sin t
Its auxiliary equation is- 
And particular integral-
P.I. = 
Note - 
Hence the solution is - 
Example 1: Solve the following simultaneous differential equations-
....(2)
Solution:
Consider the given equations
....(2)
Consider eq(1),(2)
Dx+2y = et ....(1)
Dx +2x =e-t....(2)
Eliminating ‘x’ from both the equations we get,
1
2
2Dx + 2y = 2et
D 
2Dx +D2y = e-t
y = Ae2t + Be-2t + 
Example 2: Solve the following simultaneous differential equations-
Given that x(0)=1 and y(0)= 0
Solution:
Consider the given equations,
Dy +2x = sin2t
Dx -2y = cos2t
By solving the above equations we get,
(D2 +4)Y =0
X(0) = 1, y(0) = 0
A =0, B=-1
Example-3: Solve the following simultaneous differential equations-
It is given that x = 0 and y = 1 when t = 0.
Sol. Given equations can be written as-
Dx + 2y = - sin t ………. (1) and -2x + Dy = cos t ……… (2)
Eliminate x by multiplying (1) by 2 and (2) by D then add-
Here A.E = 
So that C.F. = 
And P.I. = 
So that- 
…………. (3)
And 
………….. (4)
Substitute (3) in (2), we get-
2x = Dy – cos t = 
………… (5)
When t = 0, x = 0, y = 1. (3) and (5) gives- 
Hence 
Introduction
In “real-world,” there are many physical quantities that can be represented by functions. Involving only one of the four variables e.g., (x, y, z, t). Equations involving highest order derivatives of order one = 1st order differential equations. Examples Function σ(x) = the stress in a uni – axial stretched tapered metal rod .
Function v(x) =the velocity of fluid flowing a straight channel with varying cross-section.
Solution Method of First Order ODEs Solution of Linear (Homogeneous Equation)
Typical form of the equation
The solution u(x) in eq(1) is given by,
U(x) =
Where k= constant to be determined by given condition and the function F(x) has the form:
F(x)= 
...(3)
In which p(x) is given in differential equation.
Type 2
Solution Method of First Order ODEs Solution of Linear (Non-Homogeneous Equation)
Typical form of the equation
The appearance of g(x) in eq(4) tends to non-homogenous
The solution u(x) in eq(4) is given by,
U(x) =
Where k= constant to be determined by given condition and the function F(x) has the form:
F(x)= 
Some important terms to be considered to solve the electrical circuit problems:
*If a current i(t) is flowing through a resistor R ohms, then the voltage 
across
*the resistor is given by 
*for an inductor of L Hentry,the voltage-current relationship is given by
*For a capacity of c faraday, the voltage current relationship is given by
Example-1:
Finding the optimal current of an electrical circuit (RL circuits) in which the initial
Condition is i=0 at t=0
Solution:
By Kirchhoff voltage law (KVL) method, we get
The differential equation for the RL circuit will be 
In which initial conditions are i=0 at t=0
The standard form of the equation is,
Dividing the differential equation by L to obtain
The integrating factor is
Multiplying the above equation with standard form gives rise to
By applying integration on both sides we get
Now applying i=0 at t-0 gives us
0=
C=-
NOW
i=
= t 
Therefore by finding current of the RL circuits is i=
Hence we complete the solution by first order differential equation of first order and even several types of networking circuits and fluid mechanics uses this method.
Example-2:
Resistors of R1= 10Ω, R2 = 4Ω and R3 = 8Ω are connected up to two batteries (of negligible resistance) as shown. Find the current through each resistor.
Solution:
Assume currents to flow in directions indicated by arrows
Apply KCL on junctions C and A
Therefore, current in mesh ABC = i1
Current in Mesh CA = i2
Then current in mesh is = i1 - i2
Now, apply KVL on mesh ABC,20V are acting in clockwise direction, equating the sum of IR products, we get
10i1 +4i2 = 20.....(1)
In mesh ACD 12 volts are acting in clock-wise direction, then
8( i1 - i2)-4i2 = 12
8 i1 - 8 i2-4i2 = 12
8 i1 - 12 i2 = 12...(2)
Multiplying eq(1) by 3 we get
30 i1 + 12 i2 = 60
By solving equation 1 and 2 we get,
38i1 = 72
The above equation can be also simplified by elimination or cramer’s ruke
I1 = 72/38 = 1.895 Amperes = current in 10 ohms resistor
Substituting this value in (1) we get,
10(1.895)+4i2 = 20
4i2 = 20-18.95
I2 = 0.263 Amperes=current in 4 ohms resistors
Now ,
i1 - i2 = 1.895-0.263=1.632 Amperes
Note-
L-C circuit-
Suppose we have an electric circuit which has an inductance L and capacitance C.
Let i is the current and q is the charge in the condenser plate at time t, so that the voltage drop across-
And the voltage drop across-
By Kirchhoff’s first law-
………….. (1)
Divide by L and replacing 1/LC =
, equation (1) becomes-
It represents the free electric oscillations of the current having period 
.
L-C-R circuit-
Let us consider the discharge of a condenser C through an inductance L and the resistance R.
Since the voltage drop across L,C and R are respectively.
So that by Kirchhoff’s law- we have
On replacing R/L by 2λ and 1/LC by μ², we get-
Reference Books
1. Erwin Kreyszig, “Advanced Engineering Mathematics”, Wiley India,10th Edition.
2. M.D. Greenberg, “Advanced Engineering Mathematics”, Pearson Education, 2 nd Edition.
3. Peter. V and O‟Neil, “Advanced Engineering Mathematics”, Cengage Learning,7th Edition.
4. S.L. Ross, “Differential Equations”, Wiley India, 3rd Edition.
5. S. C. Chapra and R. P. Canale, “Numerical Methods for Engineers”, McGraw-Hill, 7th Edition.
6. J. W. Brown and R. V. Churchill, “Complex Variables and Applications”, McGraw-Hill Inc, 8th Edition.
