Linear Differential Equation with Constant Co-efficient

21MAB101T - UNIT-3
Ordinary Differential Equation

Dr. M. Suresh,
Assistant Professor, Department of Mathematics,
SRM Institute of Science and Technology,
Kattankulathur - 603203.

October 22, 2024

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 1/43

Linear Differential Equation with Constant Co-efficient

General Form
The general form of the nth order linear differential equation with
constant coefficients is

d ny d n −1 y d n −2 y dy
a0 n
+ a1 n − 1
+ a2 n − 2
+ · · · + an−1 + an y = R (x ) (1)
dx dx dx dx

where a0 , a1 , a2 , · · · , an are constants with a0 6= 0

If R (x ) = 0, then the equation (1) becomes

d ny d n −1 y d n −2 y dy
a0 + a 1 + a 2 + · · · + an−1 + an y = 0 (2)
dx n dx n−1 dx n−2 dx

which is called the homogeneous equation corresponding to (1).

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 2/43

Linear Differential Equation with Constant Co-efficient

General Solution
The general solution of (1) is
y =Complementary Function + Particular Integral
y = C.F + P.I

Note
The general solution of an ordinary linear differential equation is
also known as complete solution.

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 3/43

Linear Differential Equation with Constant Co-efficient

Notation
d
D=
dx
d2
D2 = 2
dx
d3
D3 = 3
dx
d4
D4 = 4
dx

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 4/43

Linear Differential Equation with Constant Co-efficient

Reform of General Equation

The equation (1) can be written as

a0 D n y + a1 D n−1 y + a2 D n−2 y + · · · + an−1 Dy + an y = R (x )(3)

(a0 D n + a1 D n−1 + a2 D n−2 + · · · + an−1 D + an )y = R (x )(4)

Procedure to find C.F

Replace R (x ) = 0 in (4), we get the homogeneous equation

(a0 D n + a1 D n−1 + a2 D n−2 + · · · + an−1 D + an )y = 0 (5)

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 5/43

Linear Differential Equation with Constant Co-efficient

Step: 1
Replace D = m in (5), we get

a0 mn + a1 mn−1 + a2 mn−2 + · · · + an−1 m + an = 0 (6)

This equation (6) is called the auxiliary equation(A.E) of (5).

Step: 2
Solve (6), we get the roots,
Let the roots are m1 , m2 , m3 , · · · , mn .

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 6/43

Linear Differential Equation with Constant Co-efficient

Step: 3 Case(i-a)
Let us consider two roots m1 and m2 are real and not equal, then
the complementaty function is

C .F = Ae m1 x + Be m2 x

Case(i-b)
Let us consider three roots m1 , m2 and m3 are real and not equal,
then the complementaty function is

C .F = Ae m1 x + Be m2 x + Ce m3 x

Similarly, we can extend upto n roots

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 7/43

Linear Differential Equation with Constant Co-efficient

Case(ii-a)
Let us consider two roots m1 and m2 are real and equal [m1 = m2 ],
then the complementaty function is

C .F = (A + Bx )e mx

Case(ii-b)
Let us consider three roots m1 , m2 and m3 are real and equal
[m1 = m2 = m3 = m], then the complementaty function is

C .F = (A + Bx + Cx 2 )e mx

Similarly, we can extend upto n roots

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 8/43

Linear Differential Equation with Constant Co-efficient

Case(ii-c)
Let us consider three roots m1 , m2 and m3 . Here all the roots are
real, two are equal and third is unequal[m1 = m2 = m 6= m3 ], then
the complementaty function is
C .F = (A + Bx )e mx + Ce m3 x

Case(ii-d)
Let us consider four roots m1 , m2 , m3 and m4 . Here all the roots
are real, two are equal and another two are unequal
[m1 = m2 = m 6= m3 6= m4 ], then the complementaty function is
C .F = (A + Bx )e mx + Ce m3 x + De m4 x

Similarly, we can extend upto n roots

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 9/43
Linear Differential Equation with Constant Co-efficient

Case(ii-e)
Let us consider four roots m1 , m2 , m3 and m4 . Here all the roots
are real, m1 = m2 = ma 6= m3 = m4 = mb , then the
complementaty function is

C .F = (A + Bx )e ma x + (C + Dx )e mb x

Similarly, we can extend upto n roots

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 10/43

Linear Differential Equation with Constant Co-efficient

Case(iii-a)
If the roots are complex,say α ± i β, [m1 = α + i β, m2 = α − i β].
Then the complementaty function is

C .F = e αx (A cos βx + B sin βx )

Case(iii-b)
If the roots are complex,say α1 ± i β 1 , [m1 = α1 + i β 1 ,
m2 = α1 − i β 1 ] and α2 ± i β 2 , [m3 = α2 + i β 2 , m4 = α2 − i β 2 ].
Then the complementaty function is

C .F = e α1 x x (A cos β 1 x + B sin β 1 x ) + e α2 x (C cos β 2 x + D sin β 2 x )

Similarly, we can extend upto n roots

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 11/43

Linear Differential Equation with Constant Co-efficient

Type:1 R(x)=0, Example:1

d 2y dy
Solve the differential equation 2
− 4 + 3y = 0.
dx dx
d 2y dy
Solution : Given − 4 + 3y = 0
dx 2 dx
d2 d
Use the notation, 2
= D 2, = D, we get
dx dx

D 2 y − 4Dy + 3y =0
(D 2 − 4D + 3)y = 0

Replace D = m, we get the auxiliary equation

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 12/43

Linear Differential Equation with Constant Co-efficient

m2 − 4m + 3 = 0

m2 − 3m − m + 3 = 0

m (m − 3) − (m − 3) = 0

(m − 3)(m − 1) = 0
⇒ m1 = 3 & m2 = 1

The complementary function is

C .F = Ae m1 x + Be m2 x

C .F = Ae 3x + Be x

⇒ y = C .F = Ae 3x + Be x
Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 13/43
Linear Differential Equation with Constant Co-efficient

Example:2
Solve (D 2 − 2D + 2)y = 0.

Solution : Given (D 2 − 2D + 2)y = 0

The auxiliary equation is

m2 − 2m + 2 = 0

Here, a = 1, b = −2, c = 2
p
−b ± (b2 − 4ac )
m =
2ap
−(−2) ± ((−2)2 − 4(1)(2))
=
2(1)
p
2 ± (4 − 8)
=
2
Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 14/43
Linear Differential Equation with Constant Co-efficient

√
2± −4
m =
2√
2±i 4 √
= i= −1
2
2 ± i2
=
2
2(1 ± i )
=
2
m = (1 ± i )
∴ α = 1&β=1

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 15/43

Linear Differential Equation with Constant Co-efficient

The complementary function is

C .F = e αx (A cos βx + B sin βx )

C .F = e x (A cos x + B sin x )

⇒ y = C .F = e x (A cos x + B sin x )

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 16/43

Linear Differential Equation with Constant Co-efficient

Example:3
Solve (4D 2 − 4D + 1)y = 0.

Solution : Given (4D 2 − 4D + 1)y = 0

The auxiliary equation is

4m2 − 4m + 1 =0
4m2 − 2m − 2m + 1 =0
2m (2m − 1) − (2m − 1) =0
(2m − 1)(2m − 1) = 0
1 1
⇒ m1 = & m2 =
2 2

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 17/43

Linear Differential Equation with Constant Co-efficient

The complementary function is

C .F = (A + Bx )e mx
1
C .F = (A + Bx )e 2 x

x
⇒ y = C .F = (A + Bx )e 2

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 18/43

Linear Differential Equation with Constant Co-efficient

Type:2 R(x)=e ax
Procedure to find particular Integral
1 ax
P.I = e
f (D )
Replace D = a, we get
1 ax
P.I = e , if f (a) 6= 0
f (a )
Suppose, f (a) = 0, then multiple 0 x 0 in the numerator and
differentiating the denominator alone W.R.T 0 D 0 , we get
x
P.I = e ax
f 0 (D )
Replace D = a, we get

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 19/43

Linear Differential Equation with Constant Co-efficient

Procedure to find particular Integral

x
P.I = e ax , if f 0 (a) 6= 0
f 0 (a )
Suppose, f 0 (a) = 0, do the above process again

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 20/43

Linear Differential Equation with Constant Co-efficient

Example:1
Solve (D 2 + 6D + 5)y = e 2x .

Solution : Given (D 2 + 6D + 5)y = e 2x

The auxiliary equation is

m2 + 6m + 5 =0
m2 + 5m + m + 1 =0
m (m + 5) + (m + 1) =0
(m + 5)(m + 1) =0
⇒ m1 = − 5 & m2 = − 1

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 21/43

Linear Differential Equation with Constant Co-efficient

The complementary function is

C .F = Ae m1 x + Be m2 x

C .F = Ae −5x + Be −x

Now to find the particular integral

1 ax
P.I = e
f (D )
1
P.I = e 2x
D 2 + 6D + 5
Replace D = a = 2, we get
1
P.I = e 2x
(2)2 +6∗2+5

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 22/43

Linear Differential Equation with Constant Co-efficient

1
P.I = e 2x
4 + 12 + 5
1 2x
P.I = e [f (a) 6= 0]
21
The general solution is

y = Complementary function + particular integral =C.F + P.I

1 2x
y = Ae −5x + Be −x + e
21

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 23/43

Linear Differential Equation with Constant Co-efficient

Example:2
Solve (D 2 + 4D + 5)y = 2 cos hx.

Solution : Given
e x + e −x
 
(D 2 + 4D + 5)y = 2 cos hx = 2 = e x + e −x
2
The auxiliary equation is

m2 + 4m + 5 = 0

Here, a = 1, b = 4, c = 5
p
−b ± (b2 − 4ac )
m =
2a
p
−(4) ± (4)2 − 4(1)(5)
=
2(1)

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 24/43

Linear Differential Equation with Constant Co-efficient

p
−4 ± (16 − 20)
m =
√2
−4 ± −4
=
2√
−4 ± i 4 √
= i = −1
2
−4 ± i2
=
2
2(−2 ± i )
=
2
m = (−2 ± i )
∴ α = −2 & β = 1

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 25/43

Linear Differential Equation with Constant Co-efficient

The complementary function is

C .F = e αx (A cos βx + B sin βx )

C .F = e −2x (A cos x + B sin x )

Now to find the particular integral

1
P.I = R (x )
f (D )
1
P.I = (e x + e −x )
D 2 + 4D + 5
1 1
P.I = (e x ) + 2 (e −x )
D2 + 4D + 5 D + 4D + 5
P.I = P.I1 + P.I2

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 26/43

Linear Differential Equation with Constant Co-efficient

Consider the P.I1 , Consider the P.I2 ,

Replace D = a = 1, we get Replace D = a = −1, we get
1 1
P.I1 = (e x ) P.I2 = (e −x )
12
+4∗1+5 (−1)2 + 4 ∗ (−1) + 5
1 1
P.I1 = (e x ) P.I2 = (e −x )
1+4+5 1−4+5
1 x 1
P.I1 = e P.I2 = e −x
10 2
ex e −x
P.I = P.I1 + P.I2 = +
10 2
The general solution is

y = Complementary function + particular integral =C.F + P.I

ex e −x
y = e −2x (A cos x + B sin x ) + +
10 2
Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 27/43
Linear Differential Equation with Constant Co-efficient

Type:3 R(x)=sin ax / cos ax

Procedure to find particular Integral
1
P.I = sin ax / cos ax
f (D 2 )
Replace D 2 = −a2 , we get
1
P.I = e ax , if f (−a2 ) 6= 0
f (−a2 )
Suppose, f (−a2 ) = 0, then multiple 0 x 0 in the numerator and
differentiating the denominator alone W.R.T 0 D 0 , we get
x
P.I = e ax
f 0 (D 2 )
Replace D = −a2 , we get

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 28/43

Linear Differential Equation with Constant Co-efficient

Procedure to find particular Integral

x
P.I = e ax , if f 0 (−a2 ) 6= 0
f 0 (−a2 )
Suppose, f 0 (−a2 ) = 0, do the above process again

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 29/43

Linear Differential Equation with Constant Co-efficient

Example:1
Solve (D 2 − 4D + 3)y = sin 3x cos 2x.

Solution : Given
1 1 1
(D 2 − 4D + 3)y = [sin 5x + sin x ] = sin 5x + sin x
2 2 2
To find the complementary function,

solve (D 2 − 4D + 3)y = 0

Auxiliary equation is

m2 − 4m + 3 = 0

⇒ (m − 1)(m − 3) = 0
⇒ m = 1 or 3
Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 30/43
Linear Differential Equation with Constant Co-efficient

The roots are real and different

∴ C.F. = Ae x + Be 3x
 
1 1 1
P.I = sin 5x + sin x
(D 2 − 4D + 3) 2 2
1 1 1 1
P.I = 2
sin 5x + 2
sin x
2 (D − 4D + 3) 2 (D − 4D + 3)
P.I = P.I1 + P.I2

Consider the P.I1

1 1
P.I1 = 2
sin 5x
2 (D − 4D + 3)
1 1
= sin 5x [replacing D 2 by − 52 = −25]
2 (−52 − 4D + 3)
1
= sin 5x
2(−4D − 22)
Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 31/43
Linear Differential Equation with Constant Co-efficient

1
P.I1 = − sin 5x
4(2D + 11)
1 2D − 11
= − sin 5x
4(2D + 11) 2D − 11
2D − 11
= − sin 5x
4(4D 2 − 121)
2D (sin 5x ) − 11 sin 5x
= −
4(4(−52 ) − 121)
(2 cos 5x · 5 − 11 sin 5x )
= −
4[−100 − 121]
(10 cos 5x − 11 sin 5x )
=
4[221]
10 cos 5x − 11 sin 5x
P.I1 =
884
Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 32/43
Linear Differential Equation with Constant Co-efficient

1 1
P.I .2 = 2
sin x
2 (D − 4D + 3)
1 1
= · 2
sin x
2 (−1 − 4D + 3)
1 1
= · sin x
2 (2 − 4D )
1 1
= · sin x
4 (1 − 2D )
1 1 (1 + 2D )
= · sin x
4 (1 − 2D ) (1 + 2D )
1 + 2D
= sin x
4(1 − 4D 2 )
(1 + 2D ) sin x
=
4[1 − 4(−1)2 ]

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 33/43

Linear Differential Equation with Constant Co-efficient

sin x + 2D (sin x )
P.I2 =
4[1 − 4(−1)2 ]
(sin x + 2 cos x )
=
4(1 + 4)
sin x + 2 cos x
P.I2 =
20
P.I . = P.I1 + P.I2
10 cos 5x − 11 sin 5x sin x + 2 cos x
P.I = +
884 20

∴ The general solution is

y = C.F. + P.I.
10 cos 5x − 11 sin 5x sin x + 2 cos x
⇒ y = A ex + Be 3x + +
884 20
Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 34/43
Linear Differential Equation with Constant Co-efficient

Example:2
Solve (D 2 + 1)y = sin2 x.

Solution : Given (D 2 + 1)y = sin2 x

1 − cos 2x 1 cos 2x
⇒(D2 + 1)y = = −
2 2 2
To find the complementary function,
solve: (D 2 + 1)y = 0

The auxiliary equation is:

m2 + 1 = 0

m 2 = −1

m = ±i

The roots are complex numbers with α = 0 and β = 1.

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 35/43
Linear Differential Equation with Constant Co-efficient

∴ C.F. = e 0x (A cos x + B sin x )

C .F = A cos x + B sin x

1 1 1 cos 2x
P.I1 = · P.I.2 = ·
D2 +1 2 D2 +1 2
1 1 1 cos 2x
= · 2 e 0x = ·
2 D +1 2 (−22 + 1)
1 1 1 cos 2x
P.I1 = · = P.I.2 = −
2 (0 + 1) 2 6
 
1 cos 2x
∴ P.I. = −
2 6
1 cos 2x
P.I. = −
2 6
Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 36/43
Linear Differential Equation with Constant Co-efficient

∴ The General Solution is y = C.F. + P.I.

1 cos 2x
y = A cos x + B sin x + −
2 6

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 37/43

Linear Differential Equation with Constant Co-efficient

Example:3
Solve (D 2 − 3D + 2)y = 2 cos(2x + 3) + 2e x .

Solution : Given (D 2 − 3D + 2)y = 2 cos(2x + 3) + 2e x

To find the complementary function,

solve:(D 2 − 3D + 2)y = 0

The auxiliary equation is:

m2 − 3m + 2 = 0

(m − 2)(m − 1) = 0

⇒ m = 2, 1

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 38/43

Linear Differential Equation with Constant Co-efficient

Complementary Function:
The roots are real and unequal

∴ C.F. = Ae x + Be 2x

Particular Integral (P.I):

1
P.I = [2 cos(2x + 3) + 2e x ]
D2
− 3D + 2
1 1
= [2 cos(2x + 3)] + 2 [2e x ]
D 2 − 3D + 2 D − 3D + 2
P.I = P.I1 + P.I2

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 39/43

Linear Differential Equation with Constant Co-efficient

1
P.I1 = 2 cos(2x + 3)
D 2 − 3D + 2
2
= cos(2x + 3)
−(2)2 − 3D + 2
2
= cos(2x + 3)
−3D − 2
−2
= cos(2x + 3)
3D + 2
1 3D − 2
= −2 cos(2x + 3)
3D + 2 3D − 2
3D [cos(2x + 3)] − 2 cos(2x + 3)
= −2
(3D )2 − 22
−3 ∗ 2 sin(2x + 3) − 2 cos(2x + 3)
= −2
9D 2 − 4
3 sin(2x + 3) + cos(2x + 3)
P.I1 = 4
9(−22 ) − 4
Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 40/43
Linear Differential Equation with Constant Co-efficient

3 sin(2x + 3) + cos(2x + 3)
P.I1 = 4
9(−4) − 4
3 sin(2x + 3) + cos(2x + 3)
= 4
−40
1
∴ P.I1 = − [3 sin(2x + 3) + cos(2x + 3)]
10
1
P.I2 = 2e x
D 2 − 3D + 2
2
= 2
ex
(1) − 3(1) + 2
2
= ex
1−3+2
2 x
= e
0
2x
= ex
2D − 3
Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 41/43
Linear Differential Equation with Constant Co-efficient

2x
P.I2 = ex
2(1) − 3
2x x
= e
2−3
P.I2 = −2xe x
∴ P.I. = P.I.1 + P.I.2
1
= − [3 sin(2x + 3) + cos(2x + 3)] − 2xe x
10

The general solution is

y = C.F + P.I

1
y = A ex + Be 2x − 10 [3 sin(2x + 3) + cos(2x + 3)] − 2xe x

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 42/43

Linear Differential Equation with Constant Co-efficient

Practice Problems
1. (D 2 + 4D + 13)y = 0

2. (D 2 − 5D + 6)y = e 4x

3. (3D 2 + D − 14)y = 13e 2x

4. (D 3 + 3D + 2)y = e −x + e −2x

5. (D 2 − 3D + 2)y = sin 5x

6. (D 2 + 4)y = cos2 x

Dr. M. SURESH - 21MAB101T- 2024-2025 ODD ODE 43/43