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# Differential Equations-1 Computer Science Engineering (CSE) Notes | EduRev

## Computer Science Engineering (CSE) : Differential Equations-1 Computer Science Engineering (CSE) Notes | EduRev

``` Page 1

UNIT

Differential Equations-I
5.1 INTRODUCTION
We have studied methods of solving ordinary differential equations of first order and first degree, in
chapter-7 (Ist semester). In this chapter, we study differential equations of second and higher orders.
Differential equations of second order arise very often in physical problems, especially in connection
with mechanical vibrations and electric circuits.
5.2
LINEAR DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER
ORDER WITH CONSTANT COEFFICIENTS
A differential equation of the form
dy
dx
a
dy
dx
a
dy
dx
n
n
n
n
n
n
++
1
1
1
2
2
2
–
–
–
–
+ ... + a
n
y = X ...(1)
where X is a function of x and a
1
, a
2
..., a
n
are constants is called a linear differential equation of
n
th
order with constant coefficients. Since the highest order of the derivative appearing in (1) is n,
it is called a differential equation of n
th
order and it is called linear.
Using the familiar notation of differential operators:
D =
d
dx
,
D
2
=
d
dx
2
2
,
D
d
dx
D
d
dx
n
n
n
3
3
3
== ...,
Then (1) can be written in the form
{D
n
+ a
1
D
n–1
+ a
2
D
n–2
+ ... a
n
} y = X
i.e., f (D) y = X ...(2)
where f (D)= D
n
+ a
1
D
n–1
+ a
2
D
n–2
+ ... a
n
.
Here f (D) is a polynomial of degree n in D
If x = 0, the equation
f (D) y =0
is called a homogeneous equation.
If x ? 0 then the Eqn. (2) is called a non-homogeneous equation.
214
Page 2

UNIT

Differential Equations-I
5.1 INTRODUCTION
We have studied methods of solving ordinary differential equations of first order and first degree, in
chapter-7 (Ist semester). In this chapter, we study differential equations of second and higher orders.
Differential equations of second order arise very often in physical problems, especially in connection
with mechanical vibrations and electric circuits.
5.2
LINEAR DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER
ORDER WITH CONSTANT COEFFICIENTS
A differential equation of the form
dy
dx
a
dy
dx
a
dy
dx
n
n
n
n
n
n
++
1
1
1
2
2
2
–
–
–
–
+ ... + a
n
y = X ...(1)
where X is a function of x and a
1
, a
2
..., a
n
are constants is called a linear differential equation of
n
th
order with constant coefficients. Since the highest order of the derivative appearing in (1) is n,
it is called a differential equation of n
th
order and it is called linear.
Using the familiar notation of differential operators:
D =
d
dx
,
D
2
=
d
dx
2
2
,
D
d
dx
D
d
dx
n
n
n
3
3
3
== ...,
Then (1) can be written in the form
{D
n
+ a
1
D
n–1
+ a
2
D
n–2
+ ... a
n
} y = X
i.e., f (D) y = X ...(2)
where f (D)= D
n
+ a
1
D
n–1
+ a
2
D
n–2
+ ... a
n
.
Here f (D) is a polynomial of degree n in D
If x = 0, the equation
f (D) y =0
is called a homogeneous equation.
If x ? 0 then the Eqn. (2) is called a non-homogeneous equation.
214
DIFFERENTIAL EQUATIONS–I 215
5.3
SOLUTION OF A HOMOGENEOUS SECOND ORDER LINEAR
DIFFERENTIAL EQUATION
We consider the homogeneous equation
dy
dx
p
dy
dx
qy
2
2
++ = 0 ...(1)
where p and q are constants
(D
2
+ pD + q) y = 0 ...(2)
The Auxiliary equations (A.E. ) put D = m
m
2
+ pm + q = 0 ...(3)
Eqn. (3) is called auxiliary equation (A.E.) or characteristic equation of the D.E. eqn. (3) being
quadratic in m, will have two roots in general. There are three cases.
Case (i): Roots are real and distinct
The roots are real and distinct, say m
1
and m
2
i.e., m
1
? m
2
Hence, the general solution of eqn. (1) is
y = C
1
e
m
1
x
+ C
2
e
m
2
x
where C
1
and C
2
are arbitrary constant.
Case (ii): Roots are equal
The roots are equal i.e., m
1
= m
2
= m.
Hence, the general solution of eqn. (1) is
y =(C
1
+ C
2
x) e
mx
where C
1
and C
2
are arbitrary constant.
Case (iii): Roots are complex
The Roots are complex, say a ± iß
Hence, the general solution is
y = e
ax
(C
1
cos ß x + C
2
sin ß x)
where C
1
and C
2
are arbitrary constants.
Note. Complementary Function (C.F.) which itself is the general solution of the D.E.
	


1. Solve
dy
dx
–5
dy
dx
6y 0
2
2
+= .
Solution. Given equation is (D
2
– 5D + 6) y = 0
A.E. is m
2
– 5m + 6 = 0
i.e., (m–2) (m– 3)= 0
i.e., m = 2, 3
? m
1
= 2, m
2
= 3
? The roots are real and distinct.
Page 3

UNIT

Differential Equations-I
5.1 INTRODUCTION
We have studied methods of solving ordinary differential equations of first order and first degree, in
chapter-7 (Ist semester). In this chapter, we study differential equations of second and higher orders.
Differential equations of second order arise very often in physical problems, especially in connection
with mechanical vibrations and electric circuits.
5.2
LINEAR DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER
ORDER WITH CONSTANT COEFFICIENTS
A differential equation of the form
dy
dx
a
dy
dx
a
dy
dx
n
n
n
n
n
n
++
1
1
1
2
2
2
–
–
–
–
+ ... + a
n
y = X ...(1)
where X is a function of x and a
1
, a
2
..., a
n
are constants is called a linear differential equation of
n
th
order with constant coefficients. Since the highest order of the derivative appearing in (1) is n,
it is called a differential equation of n
th
order and it is called linear.
Using the familiar notation of differential operators:
D =
d
dx
,
D
2
=
d
dx
2
2
,
D
d
dx
D
d
dx
n
n
n
3
3
3
== ...,
Then (1) can be written in the form
{D
n
+ a
1
D
n–1
+ a
2
D
n–2
+ ... a
n
} y = X
i.e., f (D) y = X ...(2)
where f (D)= D
n
+ a
1
D
n–1
+ a
2
D
n–2
+ ... a
n
.
Here f (D) is a polynomial of degree n in D
If x = 0, the equation
f (D) y =0
is called a homogeneous equation.
If x ? 0 then the Eqn. (2) is called a non-homogeneous equation.
214
DIFFERENTIAL EQUATIONS–I 215
5.3
SOLUTION OF A HOMOGENEOUS SECOND ORDER LINEAR
DIFFERENTIAL EQUATION
We consider the homogeneous equation
dy
dx
p
dy
dx
qy
2
2
++ = 0 ...(1)
where p and q are constants
(D
2
+ pD + q) y = 0 ...(2)
The Auxiliary equations (A.E. ) put D = m
m
2
+ pm + q = 0 ...(3)
Eqn. (3) is called auxiliary equation (A.E.) or characteristic equation of the D.E. eqn. (3) being
quadratic in m, will have two roots in general. There are three cases.
Case (i): Roots are real and distinct
The roots are real and distinct, say m
1
and m
2
i.e., m
1
? m
2
Hence, the general solution of eqn. (1) is
y = C
1
e
m
1
x
+ C
2
e
m
2
x
where C
1
and C
2
are arbitrary constant.
Case (ii): Roots are equal
The roots are equal i.e., m
1
= m
2
= m.
Hence, the general solution of eqn. (1) is
y =(C
1
+ C
2
x) e
mx
where C
1
and C
2
are arbitrary constant.
Case (iii): Roots are complex
The Roots are complex, say a ± iß
Hence, the general solution is
y = e
ax
(C
1
cos ß x + C
2
sin ß x)
where C
1
and C
2
are arbitrary constants.
Note. Complementary Function (C.F.) which itself is the general solution of the D.E.
	


1. Solve
dy
dx
–5
dy
dx
6y 0
2
2
+= .
Solution. Given equation is (D
2
– 5D + 6) y = 0
A.E. is m
2
– 5m + 6 = 0
i.e., (m–2) (m– 3)= 0
i.e., m = 2, 3
? m
1
= 2, m
2
= 3
? The roots are real and distinct.
216 ENGINEERING MATHEMATICS—II
? The general solution of the equation is
y = C
1
e
2x
+ C
2
e
3x
.
2. Solve
dy
dx
–
dy
dx
–4
dy
dx
4y 0
3
3
2
2
+= .
Solution. Given equation is (D
3
– D
2
– 4D + 4) y = 0
A.E. is m
3
– m
2
– 4m + 4 = 0
m
2
(m – 1) – 4 (m – 1) = 0
(m – 1) (m
2
– 4) = 0
m = 1, m = ± 2
m
1
= 1, m
2
= 2, m
3
= – 2
? The general solution of the given equation is
y = C
1
e
x
+ C
2
e
2x
+ C
3
e
–2x
.
3. Solve
dy
dx
–
dy
dx
–6y 0
2
2
= .
Solution. The D.E. can be written as
(D
2
– D – 6) y =0
A.E. is m
2
– m – 6 = 0
? (m – 3) (m + 2) = 0
? m = 3, – 2
? The general solution is
y = C
1
e
3x
+ C
2
e
–2x
.
4. Solve
dy
dx
8
dy
dx
16y 0
2
2
++ = .
Solution. The D.E. can be written as
(D
2
+ 8D + 16) y =0
A.E. is m
2
+ 8m + 16 = 0
? (m + 4)
2
=0
(m + 4) (m + 4) = 0
m = – 4, – 4
? The general solution is
y =(C
1
+ C
2
x) e
–4x
.
5. Solve
dy
dx
wy 0
2
2
2
+= .
Solution. Equation can be written as
(D
2
+ w
2
) y =0
A.E. is m
2
+ w
2
=0
Page 4

UNIT

Differential Equations-I
5.1 INTRODUCTION
We have studied methods of solving ordinary differential equations of first order and first degree, in
chapter-7 (Ist semester). In this chapter, we study differential equations of second and higher orders.
Differential equations of second order arise very often in physical problems, especially in connection
with mechanical vibrations and electric circuits.
5.2
LINEAR DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER
ORDER WITH CONSTANT COEFFICIENTS
A differential equation of the form
dy
dx
a
dy
dx
a
dy
dx
n
n
n
n
n
n
++
1
1
1
2
2
2
–
–
–
–
+ ... + a
n
y = X ...(1)
where X is a function of x and a
1
, a
2
..., a
n
are constants is called a linear differential equation of
n
th
order with constant coefficients. Since the highest order of the derivative appearing in (1) is n,
it is called a differential equation of n
th
order and it is called linear.
Using the familiar notation of differential operators:
D =
d
dx
,
D
2
=
d
dx
2
2
,
D
d
dx
D
d
dx
n
n
n
3
3
3
== ...,
Then (1) can be written in the form
{D
n
+ a
1
D
n–1
+ a
2
D
n–2
+ ... a
n
} y = X
i.e., f (D) y = X ...(2)
where f (D)= D
n
+ a
1
D
n–1
+ a
2
D
n–2
+ ... a
n
.
Here f (D) is a polynomial of degree n in D
If x = 0, the equation
f (D) y =0
is called a homogeneous equation.
If x ? 0 then the Eqn. (2) is called a non-homogeneous equation.
214
DIFFERENTIAL EQUATIONS–I 215
5.3
SOLUTION OF A HOMOGENEOUS SECOND ORDER LINEAR
DIFFERENTIAL EQUATION
We consider the homogeneous equation
dy
dx
p
dy
dx
qy
2
2
++ = 0 ...(1)
where p and q are constants
(D
2
+ pD + q) y = 0 ...(2)
The Auxiliary equations (A.E. ) put D = m
m
2
+ pm + q = 0 ...(3)
Eqn. (3) is called auxiliary equation (A.E.) or characteristic equation of the D.E. eqn. (3) being
quadratic in m, will have two roots in general. There are three cases.
Case (i): Roots are real and distinct
The roots are real and distinct, say m
1
and m
2
i.e., m
1
? m
2
Hence, the general solution of eqn. (1) is
y = C
1
e
m
1
x
+ C
2
e
m
2
x
where C
1
and C
2
are arbitrary constant.
Case (ii): Roots are equal
The roots are equal i.e., m
1
= m
2
= m.
Hence, the general solution of eqn. (1) is
y =(C
1
+ C
2
x) e
mx
where C
1
and C
2
are arbitrary constant.
Case (iii): Roots are complex
The Roots are complex, say a ± iß
Hence, the general solution is
y = e
ax
(C
1
cos ß x + C
2
sin ß x)
where C
1
and C
2
are arbitrary constants.
Note. Complementary Function (C.F.) which itself is the general solution of the D.E.
	


1. Solve
dy
dx
–5
dy
dx
6y 0
2
2
+= .
Solution. Given equation is (D
2
– 5D + 6) y = 0
A.E. is m
2
– 5m + 6 = 0
i.e., (m–2) (m– 3)= 0
i.e., m = 2, 3
? m
1
= 2, m
2
= 3
? The roots are real and distinct.
216 ENGINEERING MATHEMATICS—II
? The general solution of the equation is
y = C
1
e
2x
+ C
2
e
3x
.
2. Solve
dy
dx
–
dy
dx
–4
dy
dx
4y 0
3
3
2
2
+= .
Solution. Given equation is (D
3
– D
2
– 4D + 4) y = 0
A.E. is m
3
– m
2
– 4m + 4 = 0
m
2
(m – 1) – 4 (m – 1) = 0
(m – 1) (m
2
– 4) = 0
m = 1, m = ± 2
m
1
= 1, m
2
= 2, m
3
= – 2
? The general solution of the given equation is
y = C
1
e
x
+ C
2
e
2x
+ C
3
e
–2x
.
3. Solve
dy
dx
–
dy
dx
–6y 0
2
2
= .
Solution. The D.E. can be written as
(D
2
– D – 6) y =0
A.E. is m
2
– m – 6 = 0
? (m – 3) (m + 2) = 0
? m = 3, – 2
? The general solution is
y = C
1
e
3x
+ C
2
e
–2x
.
4. Solve
dy
dx
8
dy
dx
16y 0
2
2
++ = .
Solution. The D.E. can be written as
(D
2
+ 8D + 16) y =0
A.E. is m
2
+ 8m + 16 = 0
? (m + 4)
2
=0
(m + 4) (m + 4) = 0
m = – 4, – 4
? The general solution is
y =(C
1
+ C
2
x) e
–4x
.
5. Solve
dy
dx
wy 0
2
2
2
+= .
Solution. Equation can be written as
(D
2
+ w
2
) y =0
A.E. is m
2
+ w
2
=0
DIFFERENTIAL EQUATIONS–I 217
m
2
=– w
2
= w
2
i
2
(i
2
= – 1)
m = ± wi
This is the form a ± iß where a = 0, ß = w.
? The general solution is
y = e
0t
(C
1
cos wt + C
2
sin wt)
? y = C
1
cos wt + C
2
sin wt.
6. Solve
dy
dx
4
dy
dx
13y 0
2
2
++ = .
Solution. The equation can be written as
(D
2
+ 4D + 13) y =0
A.E. is   m
2
+ 4m + 13 = 0
m =
–– 41652
2
±
=–2 ± 3i (of the form a ± iß)
? The general solution is
y = e
–2x
(C
1
cos 3x + C
2
sin 3x).
7. Solve
dy
dx
–
dy
dx
0
2
2
= given y'(0) = 0, y(0) = 1.
Solution. Equation is (D
2
– D) y = 0
A.E. is m
2
– m = 0
m (m – 1) = 0
? m = 0, 1
? The general solution is
y = C
1
e
0x
+ C
2
e
x
i.e., y = C
1
+ C
2
e
x
when x = 0, y = 0 (Given)
y (0) = 0
? 0= C
1
+ C
2
...(1)
y'(x)= C
2
e
x
Given, when x = 0, y' = 1
y (0) = 1
? 1= C
2
e
0
? C
2
= 1 ...(2)
From (1) and (2) ? C
1
= –1.
? The general solution is y = e
x
– 1.
Page 5

UNIT

Differential Equations-I
5.1 INTRODUCTION
We have studied methods of solving ordinary differential equations of first order and first degree, in
chapter-7 (Ist semester). In this chapter, we study differential equations of second and higher orders.
Differential equations of second order arise very often in physical problems, especially in connection
with mechanical vibrations and electric circuits.
5.2
LINEAR DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER
ORDER WITH CONSTANT COEFFICIENTS
A differential equation of the form
dy
dx
a
dy
dx
a
dy
dx
n
n
n
n
n
n
++
1
1
1
2
2
2
–
–
–
–
+ ... + a
n
y = X ...(1)
where X is a function of x and a
1
, a
2
..., a
n
are constants is called a linear differential equation of
n
th
order with constant coefficients. Since the highest order of the derivative appearing in (1) is n,
it is called a differential equation of n
th
order and it is called linear.
Using the familiar notation of differential operators:
D =
d
dx
,
D
2
=
d
dx
2
2
,
D
d
dx
D
d
dx
n
n
n
3
3
3
== ...,
Then (1) can be written in the form
{D
n
+ a
1
D
n–1
+ a
2
D
n–2
+ ... a
n
} y = X
i.e., f (D) y = X ...(2)
where f (D)= D
n
+ a
1
D
n–1
+ a
2
D
n–2
+ ... a
n
.
Here f (D) is a polynomial of degree n in D
If x = 0, the equation
f (D) y =0
is called a homogeneous equation.
If x ? 0 then the Eqn. (2) is called a non-homogeneous equation.
214
DIFFERENTIAL EQUATIONS–I 215
5.3
SOLUTION OF A HOMOGENEOUS SECOND ORDER LINEAR
DIFFERENTIAL EQUATION
We consider the homogeneous equation
dy
dx
p
dy
dx
qy
2
2
++ = 0 ...(1)
where p and q are constants
(D
2
+ pD + q) y = 0 ...(2)
The Auxiliary equations (A.E. ) put D = m
m
2
+ pm + q = 0 ...(3)
Eqn. (3) is called auxiliary equation (A.E.) or characteristic equation of the D.E. eqn. (3) being
quadratic in m, will have two roots in general. There are three cases.
Case (i): Roots are real and distinct
The roots are real and distinct, say m
1
and m
2
i.e., m
1
? m
2
Hence, the general solution of eqn. (1) is
y = C
1
e
m
1
x
+ C
2
e
m
2
x
where C
1
and C
2
are arbitrary constant.
Case (ii): Roots are equal
The roots are equal i.e., m
1
= m
2
= m.
Hence, the general solution of eqn. (1) is
y =(C
1
+ C
2
x) e
mx
where C
1
and C
2
are arbitrary constant.
Case (iii): Roots are complex
The Roots are complex, say a ± iß
Hence, the general solution is
y = e
ax
(C
1
cos ß x + C
2
sin ß x)
where C
1
and C
2
are arbitrary constants.
Note. Complementary Function (C.F.) which itself is the general solution of the D.E.
	


1. Solve
dy
dx
–5
dy
dx
6y 0
2
2
+= .
Solution. Given equation is (D
2
– 5D + 6) y = 0
A.E. is m
2
– 5m + 6 = 0
i.e., (m–2) (m– 3)= 0
i.e., m = 2, 3
? m
1
= 2, m
2
= 3
? The roots are real and distinct.
216 ENGINEERING MATHEMATICS—II
? The general solution of the equation is
y = C
1
e
2x
+ C
2
e
3x
.
2. Solve
dy
dx
–
dy
dx
–4
dy
dx
4y 0
3
3
2
2
+= .
Solution. Given equation is (D
3
– D
2
– 4D + 4) y = 0
A.E. is m
3
– m
2
– 4m + 4 = 0
m
2
(m – 1) – 4 (m – 1) = 0
(m – 1) (m
2
– 4) = 0
m = 1, m = ± 2
m
1
= 1, m
2
= 2, m
3
= – 2
? The general solution of the given equation is
y = C
1
e
x
+ C
2
e
2x
+ C
3
e
–2x
.
3. Solve
dy
dx
–
dy
dx
–6y 0
2
2
= .
Solution. The D.E. can be written as
(D
2
– D – 6) y =0
A.E. is m
2
– m – 6 = 0
? (m – 3) (m + 2) = 0
? m = 3, – 2
? The general solution is
y = C
1
e
3x
+ C
2
e
–2x
.
4. Solve
dy
dx
8
dy
dx
16y 0
2
2
++ = .
Solution. The D.E. can be written as
(D
2
+ 8D + 16) y =0
A.E. is m
2
+ 8m + 16 = 0
? (m + 4)
2
=0
(m + 4) (m + 4) = 0
m = – 4, – 4
? The general solution is
y =(C
1
+ C
2
x) e
–4x
.
5. Solve
dy
dx
wy 0
2
2
2
+= .
Solution. Equation can be written as
(D
2
+ w
2
) y =0
A.E. is m
2
+ w
2
=0
DIFFERENTIAL EQUATIONS–I 217
m
2
=– w
2
= w
2
i
2
(i
2
= – 1)
m = ± wi
This is the form a ± iß where a = 0, ß = w.
? The general solution is
y = e
0t
(C
1
cos wt + C
2
sin wt)
? y = C
1
cos wt + C
2
sin wt.
6. Solve
dy
dx
4
dy
dx
13y 0
2
2
++ = .
Solution. The equation can be written as
(D
2
+ 4D + 13) y =0
A.E. is   m
2
+ 4m + 13 = 0
m =
–– 41652
2
±
=–2 ± 3i (of the form a ± iß)
? The general solution is
y = e
–2x
(C
1
cos 3x + C
2
sin 3x).
7. Solve
dy
dx
–
dy
dx
0
2
2
= given y'(0) = 0, y(0) = 1.
Solution. Equation is (D
2
– D) y = 0
A.E. is m
2
– m = 0
m (m – 1) = 0
? m = 0, 1
? The general solution is
y = C
1
e
0x
+ C
2
e
x
i.e., y = C
1
+ C
2
e
x
when x = 0, y = 0 (Given)
y (0) = 0
? 0= C
1
+ C
2
...(1)
y'(x)= C
2
e
x
Given, when x = 0, y' = 1
y (0) = 1
? 1= C
2
e
0
? C
2
= 1 ...(2)
From (1) and (2) ? C
1
= –1.
? The general solution is y = e
x
– 1.
218 ENGINEERING MATHEMATICS—II
8. Solve (4D
4
– 4D
3
– 23D
2
+ 12D + 36) y = 0.
Solution. A.E. is 4m
4
– 4m
3
– 23m
2
+ 12m + 36 = 0
If m = 2, 64 – 32 – 92 + 24 + 36 = 0
? m = 2 is a root of inspection.
By synthetic division,
2
4423 1236
8 8 30 36
4415 18 0
––
––
––
i.e.,4m
3
+ 4m
2
– 15m – 18 = 0
If m =2
32 + 16 – 30 – 18 = 0
Again m = 2 is a root.
By synthetic division
2
44
824 18
412 9 0
–15 –18
4m
2
+ 12m + 9 = 0
(2m + 3)
2
=0
m =
–
,
– 3
2
3
2
? The roots of the A.E. are 2, 2,
–
,
–
.
3
2
3
2
Thus, y = (C
1
+ C
2
x) e
2x
+ (C
3
+ C
4
x) e
–3x/2
.
9. Solve (D
5
– D
4
– D + 1) y = 0.
Solution. A.E. is m
5
– m
4
– m + 1 = 0
i.e., m
4
(m – 1) – 1 (m – 1) = 0
(m – 1) (m
4
– 1) = 0
(m – 1) (m
2
– 1) (m
2
+ 1) = 0
(m – 1) (m – 1) (m + 1) (m
2
+ 1) = 0
? The roots of the A.E. are 1, 1, –1, ± i.
Thus y = (C
1
+ C
2
x) e
x
+ C
3
e
–x
+ (C
4
cos x + C
5
sin x).
10. Solve y? + 4y' + 4y = 0 given that y = 0, y' = –1 at x = 1.
Solution. We have (D
2
+ 4D + 4) y =0
A.E. is m
2
+ 4m + 4 = 0
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