B.Sc, students differential equations

1. HIGHER ORDER LINEAR DIFFERENTIAL
EQUATIONS WITH CONSTANT COEFFICIENTS,
WHEN Q(X) = ex
.v(x)
6 janvier 2023
Dr.K.V.Vidyasagar, lecturer in Mathematics, SVLNS Government
Degree College, Bheemunipatnam
HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIEN
6 janvier 2023 1 / 12

2. Contents
♣ 1 INTRODUCTION
♣ 2 Problem.1
♣ 3 Problem.2
♣ 4 Problem.3
♣ 5 Problem.4
HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIEN
6 janvier 2023 2 / 12

3. 1 INTRODUCTION
Procedure to compute P.I of f(D)y = Q when Q = xk
* Step 1 : Write f(D) = cDm
(1+ψ(D)) where c is a constant and
Dm
is the least power of D in f(D).
* Step 2 : Expand [1+ψ(D)]−1
and consider only the terms that
contain D,D2
,··· ,Dk
.
* Step 3 : Apply the operators and simplify. It may be noted that
in some cases 1+ψ(D) may be of the form (1+ϕD)n
and in
this case we expand (1+ϕD)−n
.
Procedure to compute P.I of f(D)y = Q when Q = xV where V
is function of x :
* Step 1 : Compute
1
f(D)
xV using the formula
1
f(D)
xV = x
1
f(D)
V −
1
f(D)
f′
(D)
1
f(D)
V.
HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIEN
6 janvier 2023 3 / 12

4. 1 INTRODUCTION
Procedure to compute P.I of f(D)y = Q when Q = eax
V where V is a
function of x
* Step 1 : Compute P.I =
1
f(D)
eax
·v by using the formula
1
f(D)
eax
V = eax
·
1
f(D +a)
v.
HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIEN
6 janvier 2023 4 / 12

5. 1 INTRODUCTION
Procedure to compute P.I of f(D)y = Q when Q = xV where V is
function of x :
* Step 1 : Compute
1
f(D)
xV using the formula
1
f(D)
xV = x
1
f(D)
V −
1
f(D)
f′
(D)
1
f(D)
V.
HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIEN
6 janvier 2023 5 / 12

6. HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH
CONSTANT COEFFICIENTS, WHEN Q(X) = ex .v(x)
PROBLEM.1
Given equation is
D2
−5D +6
y = xe4x
. then the Auxillary
equation is m2
−5m +6 = 0 ⇒ m = 2,3. C.F. = c1e2x
+c2e3x
P.I. =
1
D2 −5D +6
xe4x
= e4x 1
(D +4)2 −5(D +4)+6
x =
e4x 1
D2 +3D +2
x =
e4x
2
1
1+ 3D+D2
2
x =
e4x
2
1+
3D +D2
2
−1
x =
e4x
2
1−
3D +D2
2
+...
x
=
e4x
2
x −
3
2
=
e4x (2x −3)
4
. ∴ The solution is
y = c1e2x
+c2e3x
+
e4x (2x −3)
4
.
HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIEN
6 janvier 2023 6 / 12

7. HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH
CONSTANT COEFFICIENTS, WHEN Q(X) = ex .v(x)
PROBLEM.2
Given equation in operator form is
D2
−6D +13
y = 8e3x
sin2x The A.E. is
m2
−6m′
+13 = 0 ⇒ m =
6±
√
36−52
2
= 3±2i
∴ C.F. = e3x
(c1 cos2x +c2 sin2x)
P.I. = 8
1
D2 −6D +13
e3x
sin2x
= 8e3x 1
(D +3)2 −6(D +3)+13
sin2x
= 8e3x 1
D2 +4
sin2x =
8e3x 1
D2 +22
sin2x = 8e3x
−
x
2(2)
cos2x
∴ The general solution is
y = C.F.+P.I ⇒ y = e3x
(c1 cos2x +c2 sin2x)−2xe3x
cos2x
HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIEN
6 janvier 2023 7 / 12

8. HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH
CONSTANT COEFFICIENTS, WHEN Q(X) = ex .v(x)
PROBLEM.3
Given equation is
D2
−4D +3
y = 2xe3x
+3ex
cos2x
A.E is m2
−4m +3 = 0 ⇒ (m −1)(m −3) = 0 ⇒ m = 1, 3.
C.F. = c1ex
+c2e3x
P.I. =
1
D2 −4D +3
2xe3x
+3ex
cos2x
= 2
1
D2 −4D +3
xe3x
+3
1
D2 −4D +3
ex
cos2x
= 2e3x 1
(D +3)2 −4(D +3)+e
x +
3ex 1
(D +1)2 −4(D +1)+3
cos2x
= 2e3x 1
D2 +2D
x +3ex 1
D2 −2D
cos2x
= 2e3x 1
2D 1+ D
2
x +3ex 1
−22 −2D
cos2x
HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIEN
6 janvier 2023 8 / 12

9. HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH
CONSTANT COEFFICIENTS, WHEN Q(X) = ex .v(x)
PROBLEM.3
= e3x 1
D
1+
D
2
−1
.x −
3
2
ex 1
D +2
cos2x
=e3x 1
D
1−
D
2
+
D2
4
−···
x −
3ex
2
D −2
D2 −4
cos2x
= e3x
1
D
−
1
2
+
D
4
−...
x −
3ex
2
D −2
−22 −4
cos2x
= e3x
x2
2
−
x
2
+
1
4
+
3ex
16
(−2sin2x −2cos2x)
= e3x
x2 −x
2
+
e3x
4
−
3
8
ex
(sin2x +cos2x)
∴ The general solution is y = C.F +P.I
⇒ y = c1ex
+c2e2x
+
x2 −x
2
e3x
−
3
8
ex
(sin2x +cos2x).
HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIEN
6 janvier 2023 9 / 12

10. HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH
CONSTANT COEFFICIENTS, WHEN Q(X) = ex .v(x)
PROBLEM.4
Given equation is
D2
−4D +4
y = x2
sinhx +e2x
+3
A.E is m2
−4m +4 = 0 ⇒ (m −2)2
= 0 ⇒ m = 2,2.
C.F = (c1 +c2x)e2x
P.I =
1
D2 −4D +4
x2
sinhx +e2x
+3
=
1
(D −2)2
x2 (ex −e−x )
2
+
1
(D −2)2
e2x
+
1
(D −2)2
(3)
=
ex
2
1
(D −1)2
x2
−
e−x
2
1
(D −3)2
x2
+
x2
2
e2x
+
3
4
=
ex
2
(1−D)−2
x2
−
e−x
18
1−
D
3
−2
x2
+
x2
2
e2x
+
3
4
HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIEN
6 janvier 2023 10 / 12

11. HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH
CONSTANT COEFFICIENTS, WHEN Q(X) = ex .v(x)
PROBLEM.4
=
ex
2
1+2D +3D2
+...
x2
−
e−x
18
1+
2D
3
+
3c2
9
+...
x2
+
x2
2
e2x
+
3
4
=
ex
2
x2
+4x +6
−
e−x
18
x2
+
4
3
x
+
x2
2
e2x
+
3
4
∴ The solution is
y = (c1 +c2x)e2x
+
ex
2
e2x
+
3
4
x2
+4x +6
−
e−x
18
x2
+
4
3
x +
2
3
+
x2
2
e2x
+
3
4
HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIEN
6 janvier 2023 11 / 12

12. Thank you
Thank You
HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIEN
6 janvier 2023 12 / 12