y
r,
p=(

)
x

dx
r dt
V =- a2
DifferentialEquations

O
3
2
DIFFERENTIAL EQUATIONS–2
2.1 LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT
COEFFICIENTS
A linear differential equation is that in which the dependent variable and its
derivatives occur only in the first degree and are not multiplied together.
Thus
dny
dx
n
 p1
d yn1
dx
n 1
 p2
dy n  2
dx
n2
.........  pn1
dy
 pn y  X
dx
where p1, p2, p3 ......... pn and X are functions of x only.
A differential equation of the form
dny
dx n
 k1
d yn1
dx n1
 k2
d yn 2
dx n 2
 ............  kn y  X
...(1)
where k1, k2, k3 .......... kn are constants, is called a linear differential
equation with constant coefficients.
Let
d
 D,
dx
d2
dx
2
 D2 ,
d3
dx
3
 D3 ...........
dn
dx
n
 Dn
where D is a differential operator.
Then equation (1) becomes
 D n  k1 Dn1  k2 Dn2  k3 Dn3  ............  kn  y  X
or
f(D) Y = X
where
f(D) = Dn + k1Dn–1 + k2 Dn–2 + .......... + kn
....(2)
Differential Equations–2
Example 1: Solve the following differential equations.
)1
Solve
d2y
dx
2
7
dy
 44 y  0
dx
Solution: The given equation in symbolic form is
(D2 – 7D – 44) y = 0

Its AE is D2 – 7D – 44 = 0
(D + 4) (D – 11) = 0

D = – 4,
D = 11

Its solution is
y = c1e–4x + c2 e11x
)2
Solve (D4 – 5D2 + 4) y = 0
Solution: The given equation is (D4 – 5D2 + 4) y = 0
Its, AE is D4 – 5D2 + 4 = 0
D4 – 4D2 – D2 + 4 = 0
D2 (D2 – 4) – 1 (D2 – 4) = 0
(D2 – 4) (D2 – 1) = 0

(D – 2) (D + 2) (D – 1) (D + 1) = 0

D = 2, –2, 1, –1 are the roots.
Hence its solution is
y = c1e2x + c2e–2x + c3ex + c4e–x.
)3
Solve
d3y
dx
3
2
d2y
dx
2
4
dy
 8y  0
dx
Solution: The given equation is
(D3 – 2D2 – 4D + 8) y = 0
Its

A.E. is D3 – 2D2 – 4D + 8 = 0
D2 (D – 2) – 4(D – 2) = 0
(D – 2) (D – 2) (D + 2) = 0


D = 2, 2, –2 are the roots
Its solution is
y = (c1x + c2) e2x + c3 e–2x
47
48
)4
Engineering Mathematics–II
Solve
d4y
dx 4
 m4 y  0
Solution: The given equation is
(D4 – m4) y = 0

Its AE is D4 – m4 = 0

(D2 – m2) (D2 + m2) = 0

(D – m) (D + m) (D – mi) (D + mi) = 0

D = m, –m, mi and –mi

Its solution is
y = c1emx + c2e–mx + (c3 cos x + c4 sin x)
)5
Solve (D4 + m4) y = 0
Solution: The given equation is
(D4 + m4) y = 0
Its AE is D4 + m4 = 0
D4 + 2m2 D2 + m4 – 2m2D2 = 0

(D2 + m2)2 – 2m2 D2 = 0

(D2 + m2 +

D 2  2 mD  m2  0,

D=–
m
mi

2
2
2 mD) = 0
D 2  2 mD  m 2  0
2 m  2 m 2  4m 2
;
2
= 

2 mD) (D2 + m2 –
D =
;
=
2m  2 m 2  4 m 2
2
m
mi

2
2
Its solution is
y = e

mx
2
mx
mx 

 c2 sin
 c1 cos
e
2
2

mx
2
mx
mx 

 c4 sin
 c3 cos

2
2

Differential Equations–2
49
EXERCISE
Solve the following differential equations.
1.
(D2 + 1)2 (D2 + D + 1)y = 0
Ans.:
y = (c1x + c2) cos x + (c3x + c4) sin x
+
d3y
2.
dx
Ans.:
3
dx
Ans.:
3
dx
Ans.:
2
dx
Ans.:
 11
dy
 6y  0
dx
3
d2y
dx
2
3
dy
y0
dx
4
dy
 y0
dx
(2 3 x
(2 3
y = c1 e)
 c2e)
d4y
5.
dx
2
y = (c1x2 + c2 x + c3) ex
d2y
4.
d2y

3
3 
x  c6 sin
x
2
2 
y = c1e–x + c2 e–2x + c3 e–3x
d3y
3.
6
1
 x
e 2  c5 cos
4
4
d3y
dx
3
5
d2y
dx
2
 36
x
dy
 36 y  0
dx
y = (c1x + c2) e–2x + c3 e3x + c4 e–3x
2.3 INVERSE OPERATOR
1
f ( D)
1
Definition: f ( D )
X is that function of x, free from arbitrary constants which
when operated upon by f(D) gives X.
 1

Thus f(D) 
X X .
 f (D) 
50
Engineering Mathematics–II

f(D) and
1
are inverse operators.
f (D)
Note the following important results:
1.
1
X is the particular integral of f(D) y = X.
f ( D)
2.
1
X   X dx .
D
3.
1
X  eax  X e ax dx .
Da
2.4 RULES FOR FINDING THE PARTICULAR INTEGRAL
Consider the differential equation
f (D) y = X

PI =
1
X
f ( D)
Type I(A): When X is of the form eax, where a is any constant.
PI =
1
1 ax
1 ax
X
e 
e . Provided f(a)  0.
)
f (D
)
f (D
)
f (a
Rule: In f(D), put D = a and PI will be calculated, provided f(a)  0.
Type I(B): When X is of the form eax but f(D) = 0 has got ‘a’ as its root
(Failure case of type 1).
PI =

1 ax 1 ax
e  e
f)
(D
0
 f ( a)  0
Our method fails.
Now, since ‘a’ is a root of f(D), (D – a) must be a factor of f(D) which
therefore can be written as (D – a)  (D)

PI =
1
e ax
( D  a) ( D)
Differential Equations–2
51
eax
1
1
=
(a ) ( D  a  a )
e ax 1
1
=
(a ) D
=
e ax
.x
(a )
 1

   
 D

Similarly, if f(D) = (D – a)p  (D) then
PI =
eax
1
. p 1
(a ) D
e ax x p
.
=
()
a p!
Rule: Put D = a in those factors of f(D) which do not vanish for D = a and
then make the question as PI of a product of eax and 1 which is calculated.
Note: In case of sinh ax and cosh ax we take
)i
sinh ax =
eax  e ax
2
ii) cosh ax =
eax  e  ax
2
Example 1: Solve (D2 + 3D + 5) y = e2x
Solution: Given equation is
(D2 + 3D + 5) y = e2x
AE is D2 + 3D + 5 = 0


And,
D =
CF =
PI =
3  9  20 3
11
i


2
2
2
3
 x
2
e
 c1 cos

1
2
D  3D  5
Put D = 2 in f (D)
11
11 
x  c2 sin
x
2
2 
e2 x
52
Engineering Mathematics–II

1
1
e 2 x  e2 x
465
15
G.S. = CF + PI

3
 x
e 2  c1 cos
=
y =
Example 2: Solve

d2y
dx
11
11  1 2 x
x  c2 sin
x  e
2
2  15
2
dy
 4 y  2sinh 2 x.
dx
4
Solution: Given equation is
(D2 + 4D + 4) y = 2 sinh 2x
=
2(e 2 x  e 2 x )
 e 2 x  e 2 x
2
AE is D2 + 4D + 4 = 0

(D + 2)2 = 0

CF = (c1x + c2) e–2x
And PI for
e2x
D = –2, –2 are the roots.
1
1
e2 x 
=
2
)
(D  2
)
(2  2
Also PI for e–2x =


e2 x 
2
e2 x
16
...(1)
1
1
e 2 x  e 2 x
0
( D  2)
2
Our method fails.
Then to find PI put D – 2 for D in f(D) and take out e–2x


1

PI = e 2 x 
1
2
 ( D  2  2) 
2 x
= e

1
D
1  e2 x
2
x2
2!
GS = CF + PI1 – PI2
y = (c1x + c2) e–2x +
e2 x
x2
 e 2 x
.
16
2!
....(2)
Differential Equations–2
Example 3: Solve (D3 + 3D2 + 3D + 1) y = e–x.
Solution: Given equation is
(D3 + 3D2 + 3D + 1) y = e–x

AE is D3 + 3D2 + 3D + 1 = 0

(D + 1)3 = 0

x2
CF = (c1
PI =
And
 D = –1, –1, –1 are the roots.
+ c2x + c3) e–x
1
( D  1)
3
e x

1
x 
(1)
= e 
3
 ( D  1  1) 
= e–x
1
D
1  e x
3
x3
3!
G.S. = CF + PI

x3
e x  e  x
y = (c1 x 2  c2 x  c3 )
3!
Example 4: Solve (D2 + 4D + 3)y = e–3x.
Solution: Given equation is
(D2 + 4D + 3) y = e–3x
AE is
D2 + 4D + 3 = 0
D = –1,
(D + 1) (D + 3) = 0

D = –3
CF = c1e–x + c2e–3x
PI =
1
e 3 x
( D  1) ( D  3)
put D = –3 in f(D) then
PI =

1 3 x
e
0

method fails
e 3 x
e 3 x 1
e 3 x


x
1
1


PI =
2
(3  1) ( D  3  3)
( 2) D
G.S. = CF + PI
53
54
Engineering Mathematics–II
e 3 x
x
2
y = c1e  x  c2e 3x 
Type II(A): When X is of the form sin ax or cos ax.
PI =
1
sin ax
f (D )
2
or
1
cos ax
f ( D2 )
1
1
sin ax or
sin ax
=
2
)
f ( a
)
f (a2
In this case put the quantity – a2 in the place of D2, we can not put any
thing for D as is imaginary. D3 on substitution shall become D2 . D = –a2D,
D4 will be D2 . D2 = (–a2) (–a2).
In otherwords f(D) shall be reduced to a linear factor of the form lD – m
or D1 + m. Then multiply both numerator and denominator by a conjugate
factor lD + m or lD – m respectively, the denominator shall be reduced to
l2 D2 – m2 in which again put D2 = –a2 and it will become a constant i.e.,
l2 (–a2) – m2 = –a2 l2 – m2.
PI =

la cos ax  m sin ax
 a 2l 2  m 2
Type II B: (Failure case)
When X is of the form sin ax or cos ax but f(D) becomes zero when we
put D2 = –a2.
PI =
1
sin ax
f (D )
or
1
sin ax
0
1
cos ax
0
2
1
cos ax
f ( D2 )
Put D2 = –a2

PI =

Our method fails
Then
PI =
and
PI =
1
2
D a
2
1
2
D a
2
or
sin ax 
x
sin ax dx
2
cos ax 
x
cos ax dx
2
Differential Equations–2
55
Example 1: Solve (D2 + D + 1) y = sin 2x.
Solution: Given equation is
(D2 + D + 1) y = sin 2x
AE is
D2 + D + 1 = 0
D =

And
Put
1  1  4
1
3
 
i
2
2 2
CF = e

1
x
2
 c1 cos

1
PI =
2
D  D 1
2
D = – z2
=
=
3
3 
x  c2 sin
x
2
2 
sin 2 x
1
1
sin 2 x 
sin 2 x
D 3
4  D  1
( D  3)
D2  9
sin 2 x
Put again D2 = –22
=
D(sin 2)x  (3sin 2)x
4  9
2 cos 2 x  3sin 2 x
13
G.S. = CF + PI
=

y= e
1
 x
2
 c1 cos

3
3  1
x  c2 sin
x   (2cos 2 x  3sin 2 x )
2
2  13
Example 2: Solve (D2 – 5D + 6) y = cos 3x
Solution: Given equation is
(D2 – 5D + 6) y = cos 3x

AE is D2 – 5D + 6 = 0

(D – 2) (D – 3) = 0
 D = 2, 3 are the roots.
2x
3x
CF = c1e + c2 e
56
Engineering Mathematics–II
And
Put
1
PI =
2
D  5D  6
2
D = –32
cos3 x
=
1
cos3 x
9  5D  6
=
1
cos3x
5D  3
=
(5D  3)
25 D 2  9
cos3x
Put D2 = –32 again
=
=
 5 D(cos3)x  3(cos3)x

25  ( )9  9
 15sin 3 x  3cos 3x
234
(15sin 3x  3cos3x )
234
G.S. = C.F. + P.I.
=

2x
3x
y = c1e  c2e 
1
(15sin 3x  3cos x)
234
Example 3: Solve (D2 + 4)y = cos 2x.
Solution: Given equation is
(D2 + 4) y = cos 2x

And
AE is D2 + 4 = 0
 D = ± 2i
C.F. = c1 cos 2x + c2 sin 2x
PI =
Put
1
2
D 4
2
D = –22
=
cos 2 x
1
cos 2 x
0

Our method fails.
Differential Equations–2

P.I. =
1
2
D 4
cos 2 x 
=

57
x
cos 2 x dx
2
x sin 2 x x
 sin 2 x
2 2
4
G.S. = C.F. + P.I.
y = c1 cos 2x + c2 sin 2x +
x
sin 2 x .
4
Example 4: Solve (D2 + 4) (D2 + 1) y = cos 2x + sin x.
Solution: Given equation is
(D2 + 4) (D2 + 1) y = cos 2x + sin x
AE is (D2 + 4) (D2 + 1) = 0

D = ± 2i
and
D=±i
CF = (c1 cos 2x + c2 sin 2x) + (c3 cos x + c4 sin x)
Now, PI for cos 2x =
1
2
( D  4) ( D 2  1)
cos 2 x
1
1
= (4  4) ( 4  1) cos 2 x  0 cos 2 x

PI
1
cos 2 x
= (4  1)
D2  4
=
1 x
3 2

cos 2 x dx  
x sin 2 x
x
  sin 2 x
6
2
12
...(1)
( put D 2  2 2 in those factors which do not vanish)
And PI for sin x =
1
2
( D  4) ( D 2  1)
sin x
put D2 = –12 in those factors which do not vanish.
x
1
= (1  4) 2

sin x dx  
1 x
(  cos x )
3 2
x
=  cos x
6
....(2)
58
Engineering Mathematics–II
G.S. = CF + PI1 + PI2
y = c1 cos 2x + c2 sin 2x + c3 cos x

+ c4 sin x –
x
x
sin 2 x  cos x
12
6
When X is of the form xm.
Type III:
PI =
1
xm
f ( D)
To find the PI of this type one should remember the following
expansions.
)i
(1 – D)–1 = 1 + D + D2 + D3 + ..........
)
ii
(1 + D)–1 = 1 – D + D2 – D3 + ..........
iii
)
(1 – D)–2 = 1 + 2D + 3D2 + 4D3 + .........
iv
)
(1 + D)–2 = 1 – 2D + 3D2 – 4D3 + ...........
)v
(1 – D)–3 = 1 + 3D + 6D2 + 10D3 + .........
vi
)
(1 + D)–3 = 1 – 3D + 6D2 – 10D3 + .........
Example 1:
Solve 2
d2y
dx 2
5
dy
 2 y  5  2x .
dx
Solution: The given equation is
(2D2 + 5D + 2) y = 5 + 2x
AE is
2D2 + 5D + 2 = 0
(2D + 1) (D + 2) = 0

C.F. = c1 e
P.I. =

1
x
2
2D  5D  2
1
, D = –2
2
(5  2 x )
1  (2 D 2  5D) 

= 1 
2 
2

1
= 2
D=–
 c2 e 2 x
1
2

1
(5  2 x)
 (2 D 2  5 D)

 ........ (5  2 x )
1 
2


Differential Equations–2
=
1
2
5 

5  2 x  2 . 2
= x
G.S. = CF + PI

y = c1e
1
 x
2
 c2 e2 x  x
Example 2: Solve (D2 + 2D + 1) y = x2 + ex – sin x + 2x.
Solution: Given equation is
(D2 + 2D + 1) y = x2 + ex – sin x + 2x
AE is
D2 + 2D + 1 = 0
2
 D = –1, –1
(D + 1) = 0

C.F. = (c1x + c2) e–x
1)
x2 =
PI for
1
(1  D)
2
x2
= (1 + D)–2 (x2)
= (1 – 2D + 3D2 – 4D3 ...... ) x2
= x2 – 4x + 6
ii)
ex =
PI for
Put D = 1
iii) PI for

sin x =
Put D2 = –12, =
=
iv) PI for
2x =
=
1
2
D  2D  1
ex
1
1
ex  ex
1 2 1
4
1
2
D  2D  1
sin x
1
1

sin x
1  2 D  1 2 D
1
1
sin x dx   cos x

2
2
1
2
D  2D  1
1
2
D  2D  1
2x
e x log 2
1


 

D


59
Differential Equations–2
y = c1e ax  c2e ax  c3 cos ax  c4 sin ax 
61
1  4 24 
1
x  4 4
sin bx
4
a 
a  b  a4
Type IV: When X is of the form eaxV, where V is any function of x.
1 ax
e V
PI = f ( D)

1
ax 
V
= e 
 f ( D  a)

Rule: It means that take out eax and in f(D) write D + a for every D so that
f(D) becomes f(D + a) and then operate
1
with V alone by previous
f ( D  a)
method.
Example 1: Solve (D2 – 4D + 3) y = e2x sin 3x.
Solution: Given equation is
(D2 – 4D + 3) y = e2x sin 3x

A.E. is
D2 – 4D + 3 = 0
 D = 1, 3
(D – 1) (D – 3) = 0

And
C.F. = c1
PI =
ex
+ c2
e3x
1
2
D  4D  3

2x 
= e 
 ( D  )2
e 2 x sin 3 x

sin 3x

2
 4( D  )2  3 
1
1


 sin 3 x
= e2x  2
 D  4D  4  4D  8  3 
2x  1 
= e  2  sin 3 x
 D  1
put D2 = –32
2x  1 
 sin 3x
= e 
 9  1 
62
Engineering Mathematics–II
= 
1 2x
e sin 3x
10
G.S. = CF + PI

y = c1ex + c2 e3x –
1 2x
e sin 3x
10
Example 2: Solve (D2 – 5D + 6) y = x (x + ex)
Solution:
Given equation is
(D2 – 5D + 6) y = x (x + ex)
= x2 + xex
AE is
D2 – 5D + 6 = 0
(D – 2) (D – 3) = 0


D = 2, 3
CF = c1e2x + c2e3x
And
i)
PI for
x2 =
1
2
D  5D  6
1
=
6
x2
1
 ( D 2  5D )

2
1 
 (x )
6


1  ( D 2  5 D) ( D 2  5D)

= 6 1 
6
6

1
= 6
2

 ........... ( x 2 )

 D 2 5D 25 2


 D .......... ( x 2 )
1 
6
6 36


=
1
6
 2 2 10 x 50 
 x  6  6  36 
=
1
6
 2 10 x 19 
 x  6  18 
64
Engineering Mathematics–II

1

x  
x sin x
= e 
2
D  1  1) 
 ( 
1
x
= e
D2
( x sin x)
x
= e
1
D
x
= e
1
 x( cos )x  ( sin )x
D
x
= e


x sin x dx

( x cos x  sin x) dx
= e–x [–x sin x – (–1) (–cos x) – cos x]
= e–x [–x sin x – 2 cos x]
= – e–x (x sin x + 2 cos x)
G.S = CF + PI

y = (c1x + c2) e–x – e–x (x sin x + 2 cos x)
d2y
Example 4: Solve
dx
2
2
dy
 y  xe x cos x
dx
Solution: Given equation is
(D2 _ 2D + 1) y = xex cos x
AE is
D2 – 2D + 1 = 0
(D – 1)2 = 0

And,

D = 1, 1
CF = (c1x + c2) ex
PI =
1
( D  1)
2
xe x cos x

1
 1 

x  
x cos x  e x  2  x cos x
= e 
2
 1  1) 
D 
 ( D

x
= e .
1
x cos x dx
D
Differential Equations–2
x
= e
1
 x sin x  1 ( cos x)
D
x
= e

=
=
GS =
y =

( x sin x  cos x) dx
ex [x (– cos x) + sin x + sin x]
ex (–x cos x + 2 sin x)
CF + PI
(c1x + c2) ex + ex (–x cos x + 2 sin x)
d2y
x
Example 5: Solve e
dx
2
 2e x
dy
 e x y  x2
dx
Solution: Given equation is
d2y
dx

2
2
dy
 y  x 2 e x
dx
AE is D2 + 2D + 1 = 0

And,
(D + 1)2 = 0
CF = (c1x + c2) e–x
PI =
1
( D  1)
2
 D = –1, 1
x 2 e x
x  1  2
= e  2x
D 
= e x
=
e x
3
1
D


x 3dx 
x 2 dx  e x
1
D
 x3 
 
 3 
e x x4
.
3
4
e x x 4
12
GS = CF + PI
=

y = (c1x + c2) e–x +
1 x 4
e x
12
65
Differential Equations–2

G.S. = CF + PI
y = c1 cos 2x + c2 sin 2x +
Example 2: Solve
d2y
dx
2
1
(3x sin x – 2 cos x)
9
 y  x 2 sin x
Solution: Given equation is
(D2 + 1) y = x2 sin x


AE is D2 + 1 = 0
 D=±i
CF = c1 cos x + c2 sin x
PI =
=
1
2
D 1
1
2
D 1
x 2 sin x
imaginary part of x2 eix

 2
1
= Imaginary part of eix 
x
2
 ( D  i )  1 
1
 2
ix 
= Imaginary part of e  2
x
 D  2 Di  1  1 
1
= Imaginary part of e
2 Di
ix
ix
= Imaginary part of e
eix
= Imaginary part of
2 Di
= Imaginary part of
eix
2 Di
 D
1  2i 
1
( x2 )
 2
1  D D2
1   2 ......... ( x )
2 Di  2i 4i

 2 2x 2 
 x  2i  4 


1
 2
 x  ix  2 


eix  x3 ix 2 x 
= Imaginary part of 2i  3  2  2 


67
68
Engineering Mathematics–II
= Imaginary part of
eix
(2 x3  3 x  ix 2 )
12i
= Imaginary part of
1
(cos x + i sin x) (2x3 – 3x + i 3x2)
12i
= Imaginary part of
1
[(cos x) (2x3 – 3x) – (sin x) 3x2
12i
+ i (2x3 – 3x) sin x + 3x2 cos x]
1
(2x3 – 3x) cos x – 3x2 sin x
12
Taking imaginary part only.
GS = CF + PI
= 

y = c1 cos x + c2 sin x –
Example 3: Solve
d2y
dx
2
1
(2x3 – 3x) cos x – 3x2 sin x.
12
 a 2 y  sec ax
Solution: Given equation is
(D2 + a2) y = sec ax

AE is
D2 + a2 = 0

CF =
And,
PI=
 D = ± ai
c1 cos ax + c2 sin ax
1
2
D  a2
sec ax
1
= ( D  ai) ( D  ai) sec ax
Resolve into partial fractions
Now,
=
1  1
1 
sec ax


2ai  D  ai D  ai 
=
1  1
1

sec ax 
sec ax 

D  ai
2ai  D  ai

70
Engineering Mathematics–II
PI =
Now
1
D2  a2
tan ax 
1
tan ax
( D  ai)
( D  ai)
=
1  1
1 

tan ax, by partial fractions

2ai  D  ai D  ai 
=
1  1
1

tan ax 
tan ax 

D  ai
2ai  D  ai

1
tan ax  eiax
D  ai
iax
= e

tan ax e iax dx

tan ax (cos ax  i sin ax) dx
iax
= e


sin 2 ax 
 sin ax  i
 dx
cos ax 

iax
= e
 sin ax  i (sec ax  cos ax) dx
i
i sin ax 
iax  cos ax
 log(sec ax  tan ax) 
= e 
a
a
a 

1 iax
ax  i log(sec ax  tan )
ax
=  e (cos ax  i sin )
a

1 iax iax
 i log (sec ax  tan ax) 
=  e e

a
= 
1
1  ieiax log (sec ax  tan ax) 

a
Changing i to –i we get
1
1
tan ax =  1  ieiax log(sec ax  tan ax) 

D  ai
a
 PI =

 

1  1
1

 1  ieiax log(sec x  tan x  1  e iax log(sec ax  tan ax ) 

2ai  a
a

= 
1
a2
log (sec ax + tan ax) . cos ax
Differential Equations–2

71
GS = CF + PI
y = c1 cos ax + c2 sin ax –
1
a2
cos ax log (sec ax + tan ax)
2.5 INITIAL VALUE PROBLEMS
Example 1: Solve y + 4y + 4y = 0 given y (0) = 3, y(0) = 1
Solution: Given equation is
(D2 + 4D + 4) y = 0 with x = 0, y = 3 and x = 0, y = 1
AE is D2 + 4D + 4 = 0
(D + 2)2 = 0


D = –2, –2
The solution is
and
y = (c1x + c2) e–2x
....(1)
y = (c1x + c2) e–2x (–2) + c1e–2x
....(2)
Given y = 3 when x = 0 and y1 = 1 when x = 0
c1 = 7

c2 = 3

The complete solution is
and
y = (7x + 3) e–2x
Example 2: Solve the initial value problem
given y = 0,
d2y
dx
2
4
dy
 5 y  2cosh x  0
dx
dy
 1 at x = 0.
dx
Solution: Given equation is
(D2 + 4D + 5) y = –2 cosh x
=
2(e x  e  x )
 (e x  e  x )
2
AE is D2 + 4D + 5 = 0
4  16  20
 2  i
2
CF = e–2x (c1 cos x + c2 sin x)
D =

Differential Equations–2
= cos a
x
2
=  cos a .
GS = CF + PI


sin x dx  sin a .
x
2

73
cos x dx
x cos x
x sin x
 sin a
2
2
y = c1 cos x + c2 sin x –
1
1
cos a x cos x  sin a x sin x
2
2
1
1
y = c1 sin x  c2 cos x  cos a cos x  cos a . x sin x
2
2
and
1
1
 sin a sin x  sin a . x cos x
2
2

Given y = 0, y = 0 when

0 = c1
1
1
0 = c2  cos a  c2  cos a .
2
2
and

x=0
c1 = 0
The complete solution is
y =
=
1
1
1
cos a sin x  cos ax cos x  sin a x sin x
2
2
2
1
1
cos a (sin x  x cos
)x  sin a x sin x
2
2
Example 4: Solve the initial value problem
y (0) = 0,
d2y
dx
2
5
dy
 6 y  0 given
dx
dy
(0
)  15 .
dx
Solution: Given equation is
(D2 + 5D + 6) y = 0

AE is D2 + 5D + 6 = 0
(D + 2) (D + 3) = 0

CF is
and
y = c1e–2x + c2 e–3x
dy
= – 2c1e–2x – 3c2 e–3x
dx
 D = –2, –3
74
Engineering Mathematics–II
At
x = 0,

0 = c1 + c2
15 = –2c1 – 3c2

dy
 15
dx
y = 0 and
Solving c2 = –15 and c1 = 15
The complete solution is
y = 15e–2x – 15e–3x
= 15 (e–2x – e–3x)
2.6 SIMULTANEOUS DIFFERENTIAL EQUATIONS OF FIRST
ORDER
The differential equation in which there is one independent variable and two
or more than two dependent variables are called simultaneous linear
differential equations. Here we consider simultaneous linear equation with
constant coefficients only.
Example 1: Solve
dx
 y  sin t ;
dt
dy
 x  cos t
dt
Solution: The simultaneous equation are
Dx + y = sin t
Dy + x = cos t
....(1)
)
....(2
where
d
D
dt
Multiply (1) by D
 D2x + Dy = cos t
....(3)
Dy + x = cos t
....(4)

Subtracting (4) from (3) we get
D2x – x = 0

AE is (D2 – 1) = 0

x = c1 et + c2 e–t
 D = 1, –1
....(1)
Dx + y = sin t
And


d
c1et  c2 e t
dt
y = sin t – c1et – c2 e–t
y = sin t –

Differential Equations–2
2.
Solve (D – 2)2 y = 8 (e2x + sin 2x + x2)
Ans.:
3.
Solve
Ans.:
4.
d4y
dx 4
 y  cos x cosh x
y = c1ex + c2 c–x + c3 cos x + c4 sin x –
Solve
Ans.:
5.
y = (c1x + c2) e2x + 4x2 e2x + cos 2x + 2x2 + 4x + 3.
d2y
dx 2
 y  cosec x
y = c1 cos x + c2 sin x + sin x log sin x – x cos x.
Solve (D2 – 4D + 3) y = sin 3x cos 2x
=
Ans.:
6.
y = c1 ex + c2 e2x +
3 3 x 1
e
 (3cos 2 x  sin 2 x )
10
20
1 2 x 1 2x
y = c1 e2x + c2e–2x   e  xe
4 3
4
Solve
Ans.:
9.
1
1
(10cos 5 x  11sin
)
5 x  (sin x  2cos
)x
884
20
Solve (D2 – 4) y = (1 + ex)2
Ans.:
8.
y = c1ex + c2 e3x +
1
sin 5x  sin x
2
Solve (D2 – 3D + 2) y = 6 e–3x + sin 2x.
Ans.:
7.
1
cos x cosh x .
5
dx
2
3
dy
 2 y  xe3 x  sin 2 x
dx
y = c1ex + c2 e2x +
Solve
Ans.:
d2y
d2y
dx 2
1 3x
1
e (2)
x 3 
(3cos 2 x  sin
)
2x
4
20
 4 y  4 tan 2 x
y = c1 cos 2x + c2 sin 2x – cos 2x log (sec 2x + tan 2x)
77
Differential Equations–2
79
Case VII: If X = eax (a0 + a1x + a2x2 + .......... + an xn) sin bx or cos bx then
the particular integral is of the form.
y = eax {(A0 + A1x + ........ + Anxn) sin bx + (B0 + B1x + .... + Bnxn)
cos bx}
Note: However, when X = tan x or sec x, this method fails, since the
number of terms obtained by differentiating X = tan x or sec x is
infinite.
Example 1: Solve by the method of undetermined coefficients,
y – 3y + 2y = x2 + x + 1
Solution: Given differential equation is
y – 3y + 2y = x2 + x + 1
....(1)
AE is D2 – 3D + 2 = 0
 D = 1, 2
(D – 1) (D – 2) = 0
CF = c1ex + c2 e2x

PI is of the form
y = A0 + A1x + A2x2
y = 0 + A1 + 2A2x
y = 0 + 2A2
Substituting in (1)
2A2 – 3 (A1 + 2A2x) + 2 (A0 + A1x + A2x2) = x2 + x + 1

2A2x2 + (2A1 – 6A2) x + 2A0 – 3A1 + 2A2 = x2 + x + 1
Comparing the coefficients
2A2 = 1,
2A1 – 6A2 = 1,
1
,
2
Solving,
A2 =

PI = 3 + 2x +

GS = CF + PI
2A0 – 3A1 + 2A2 = 1
A1 = 2, A0 = 3
1 2
x
2
1 2

y = c1ex + c2e2x +  3  2 x  x 
2 

80
Engineering Mathematics–II
Example 2: Solving by the method of undetermined coefficients the
equation y – 2y + 5y = e2x.
Solution: Given equation is
y – 2y + 5y = e2x
...(1)
D2 –
AE is
2D + 5 = 0

D = 1 ± 2i

CF = ex (c1 cos 2x + c2 sin 2x)
PI is of the form
y = Ae2x

y = 2Ae2x and y = 4Ae2x
Substituting in (1)
4Ae2x – 2 + 2Ae2x + 5Ae2x = e2x
5A e2x = e2x

5A = 1


A=
1
5
1 2x
e
5
G.S. = CF + PI
PI =


y = ex (c1 cos 2x + c2 sin 2x) +
1 2x
e
5
Example 3: Solve by the method of undetermined coefficients, the
equation y – 5y + 6y = sin 2x.
Solution: Given differential equation is
y – 5y + 6 y = sin 2x
AE is
D2 – 5D + 6 = 0

CF = c1e2x + c2 e3x
PI is of the form
y = A cos 2x + B sin 2x
y = –2A sin 2x + 2B cos 2x
y = –4A cos 2x – 4B sin 2x
Substituting in (1)
....(1)
 D = 2, 3
82
Engineering Mathematics–II


1 1
3
1
PI =   x  sin x  cos x
4 2
10
10
G.S. = CF + PI
1 1
3
1
 x  sin x  cos x
4 2
10
10
y = c1ex + c2 e–2x –
Example 5: Solve by the method of undetermined coefficients the equation
y + 4y = x2 + e–x.
Solution: The given differential equation is
y + 4y = x2 + e–x
....(1)
AE is D2 + 4 = 0

D = ± 2i

CF = c1 cos 2x + c2 sin 2x
Particular integral is of the form
y = A0 + A1x + A2x2 + Be–x

y = A1 + 2A2x – Be–x
y = 2A2 + Be–x
Substituting in (1)
(2A2 + Be–x) + 4(A0 + A1x + A2x2 + Be–x) = x2 + e–x

(4A0 + 2A2) + 4A1x + 4A2x2 + 5Be–x = x2 + e–x
Equating the coefficients, we get
4A0 + 2A2 = 0


Solving
4A1 = 0
A0 = –
4A2 = 1
A2 =
1
,
8
1
,
4
A1 = 0
B
1
5
1 1
1
PI =   x 2  e x
8 4
5
G.S. = CF + PI
y = c1 cos 2x + c2 sin 2x –
1 1  1 x
 x  e .
8 4
5
Differential Equations–2
83
Example 6: Solve by the method of undetermined coefficients
y – y – 4y = x + cos 2x
.
Solution:
The given differential equation is
y – y – 4y = x + cos 2x

AE is D2 – D – 4 = 0

D =

....(1)
1  1  16 1  17

2
2
CF = c1
)
(1 17
x
e 2
 c2
)
(1 17
x
e 2
Particular integral is of the form
y = A0 + A1x + B0 cos 2x + B1 sin 2x
y = A1 – 2B0 sin 2x + 2B1 cos 2x
y = –4B0 cos 2x – 4B1 sin 2x
Substituting in (1)
(–4 B0 cos 2x – 4B1 sin 2x) – (A1 – 2B0 sin 2x + 2B1 cos 2x)
– 4 (A0 + A1x + B0 cos 2x + B1 sin 2x) = x + cos 2x.
Equating the coefficients
–4 A0 – A1 = 0
Solving
–4 A1 = 1
A0 
1
,
16
A1  
–8B0 – 2B1 = 1
B0 
2
,
17
B1 
1
4
1
34
2B0 – 8B1 = 0


1 1
2
1
 x  cos 2 x  sin 2 x
16 4
17
34
G.S. = CF + PI
PI =
y =
)
(1 17 x
c1e 2
 c2
)
(1 17 x
e 2

1 1
2
1
 x  cos 2 x  sin 2 x
16 4
17
34
84
Engineering Mathematics–II
Example
d2y
dx 2
7: Solve
by
the
method
of
undetermined
coefficients
 y  sin x .
Solution: Given differential equation is
y + y = sin x
AE is D2 + 1 = 0
...(1)

D+±i
CF = c1 cos x + c2 sin x

Note that ‘sin x’ is common in CF and RHS of the equation and
therefore particular integral is of the form
y = x (A cos x + B sin x)
y = x (–A sin x + B cos x) + (A cos x + B sin x)
y = x (–A cos x – B sin x) + (–A sin x + B cos x)
+ (– A sin x + B cos x)
Substituting in (1)
–2A sin x + 2B cos x = sin x
Equating the coefficients

–2A = 2
and
2B = 0
1
2
and
B=0
A= 
1
PI =  x cos x
2
GS = CF + PI


y = c1 cos x + c2 sin x –
1
x cos x
2
Example 8: Solve by the method of undetermined coefficients the equation
(D2 + 1) y = 4x – 2 sin x.
Solution: The given differential equation is
y + y = 4x – 2 sin x
AE

is
D2 + 1 = 0
CF = c1 cos x + c2 sin x
....(1)
 D=±i
86
Engineering Mathematics–II
Equating the coefficients
2A2 – 3 A1 + 2A0 = 0
A0 =
7
,
4
A1 
2A2 = 1
A2 
1
2
B=–1


3
2
7 3
1
 x  x 2  xe x
4 2
2
G.S. = CF + PI
PI =
x
2x
y = c1e  c2e 
7 3
1
 x  x 2  xe x
4 2
2
Example 10: Solve by the method of undetermined coefficients the
following equations.
)i
(D2 + 1) y = 4x cos x – 2 sin x
Hint: CF = c1 cos x + c2 sin x
....(1)
Since ‘cos x and sin x’ are common in CF and RHS of (1) particular
integral is of the form.
and

)
ii
y1 = x [(A0 + A1x) cos x + (A2 + A3x) sin x]
y2 = PI for 2 sin 2x
= x {B0 cos x + B1 sin x}
y = y1 + y2
(D2 – 1) y = 10 sin2x
10(1  cos 2 x )
 5  5cos 2 x
2
PI is of the form
y = A0 + {B0 cos x + B1 sin x}
Hint: =
iii
)
(D2 – 1) y = e–x (2 sin x + 4 cos x)
Hint: PI is of the form
y = e–x (A cos x + B sin x)
iv
)
(D3 – D) y = 3ex + sin x
....(1)
Hint: D(D2 – 1) = 0
CF = c1 + c2
D = 0,
ex
+ c3
e–x
D=1
D = –1
Differential Equations–2
87
ex is common in CF and RHS of (1)
PI is of the form
y = A0 xex + (B0 sin x + B1 cos x)
)v
(D2 – 5D + 6) y = e2x + sin x
Hint:
PI is of the form
y = Axe2x + (B0 sin x + B1 cos x)
2.8 APPLICATIONS TO LINEAR DIFFERENTIAL EQUATIONS
The applications of linear differential equations to various physical problems
play a dominant role in uniflying seemingly different theories of mechanical
and electrical systems just by renaming the variables. This analogy has an
important practical application. Since electrical circuits are easier to
assembe, less expensive and accurate measurements can be made of
electrical quantities.
A. Simple harmonic mothion (S.H.M.)
A particle is said to execute simple harmonic motion if it moves in a straight
line such that its acceleration is always directed towards a fixed point in the
line is proportional to the distance of the particle from the fixed point.
2
x
O
A
x
P
A
a
Let O be the fixed point in the line AA. Let P be the position of the
particle at any time t where
OP = x
Since the acceleration is always directed towards O i.e., the acceleration
is in the direction opposite to that in which x increases, the equation of
motion of the particle is
d 2x
= – 2x
dt 2
or
(D2 + 2)x = 0 where
d
D
dt
which is the linear differential equation with constant coefficient.
...(1)
88

Engineering Mathematics–II
The solution of equation (1) is
x = c1 cos t + c2 sin t
....(2)
Velocity of the particle at P is
dx
= – c1 sin t + c2 cos t....(3)
dt
If the particle starts from rest at A, where OA = a then from (2) at t = 0,
x = 0.  c1 = a.
dx
0
dt
and from (3), at t = 0,
 c2  0
x = –a  sin t

and

...(4)
x2
dx
2
= a 1  cos t  a 1  2
dt
a
...(5)
dx
=  a 2  x 2
dt
...(6)
Hence equation (4) gives the displacement of the particle from the fixed
point O at any time t.
Equation (6) gives the velocity of the particle at any time t. Equation (6)
also shows that the velocity is directed towards O and decreases as x
increases.
Example 1: In the case of a stretched elastic horizontal string which has
one end fixed and a particle of mass m attached to the other, find the
equation of motion of the particle given that l is natural length of the
string and e is its elongation due to a weight mg. Also find the
displacement of the particle when initially s = s0, v = 0.
Solution: Let OA = l be the elastic horizontal string with the end O fixed
and a particle of mass m attached at A.
l
O
A
T
Let P be the position of the particle at any time t
Let OP = s so that the elongation AP = s – l.
Now, for the elongation e tension = mg
P
Differential Equations–2

For the elongation (s – l), tension =
89
mg ( s  l )
e
Since tension is the only horizontal force acting on the particle, its
equation of motion is

m
d 2s
= –T
dt 2
m
mg ( s  l )
d 2s

2 =
e
dt
g
gl
d 2s
 s
2 =
e
e
dt
or
gl
d 2s g
 s =
2
e
e
dt

gl
 2 g
D   s 
e
e

or
....(1) when D 
d
dt
Which is the linear differential equation
Its AE is D 2 
 D i
g
e
 g
CF = c1 cos 
 t  c2 sin
 e

And,

g
0
e
PI =
 g

 t .
 e
gl gl
1
gl 1

e 0t  .
l
g
g
e
e D2 
e g
D2 
e
e
e
1
The complete solution of (1) is
 g
 g
s = c1 cos  e  t  c2 sin  e  t  l




When t = 0, s = s0 from (2) we get
s0 = c1 + 0 + l
Also,
or
c1 = s0 – l
...(2)
Differential Equations–2
b)
Damped Oscillations
O
If the motion of the mass m be subject to an
additional force of resistance, the
oscillations are said to be damped. The
damping force may be constant or
proporational to velocity. The latter type of
damping is important and is usually called
viscous damping.
Now,
if
the
damping
force
A
e
B
be
dx 

proportional to velocity  say  r
 then
dt 

the equation of motion becomes
m
dx
d2x
mg  k (e  x)
r
2 =
dt
dt
= kx  r
Taking
We get
91
r
dx
x
dt
k (e + x)
P
damper
mg
dx
dt
r
k
= 2  and
 2 .
m
m
d 2x
dx
 2
 2 x  0 ....(3)
2
dt
dt
O
Which is the linear differential equation
with constant coefficient.
c)
Forced Oscillations
(Without dumping)
If the point of the support of the spring is also
vibrating with some external periodic force,
then the resulting motion is called the forced
oscillatory motion.
Taking the external periodic force to be
mp cos nt the equation of the motion is
m
A
e
B
x
k (e + x)
P
2
d x
 mg  k ()
e  x  mp cos nt
dt 2
= – kx + mp cos nt
mp cos nt
mg
92
Engineering Mathematics–II
Taking
k
  2 the equation becomes
m
d 2x
 2 x = p cos nt
2
dt
....(4)
Which is linear differential equation.
The solution of equation (4) is
x = c1 cos t + c2 sin t +
d)
p cos nt
n2  2
Forced Oscillations (With damping)
dx 

If in addition, there is a damping force proportional to velocity  say: r

dt 

then the above equation becomes.
O
A
r
dx
dt
e
mp cos nt
B
x
k (e + x)
P
damper
m
mg
dx
d2x
2 = mg – k (e + x) + mp cos nt – r
dt
dt
= – k x + mp cos nt – r
dx
dt
Differential Equations–2
Taking
93
r
k
 2 and
  2 then the equation becomes
m
m
d2x
dx
 2
  2 x  p cos nt
2
dt
dt
Which is a linear differential equation and its solution is


x = e  st c1e
 2  2
 c2 e t
 2  2
nt  2n sin t
  p( ( n)ncos)
 4 n
2
2
2
2
2 2
With the increase of time, the free oscillations die away while the
forced oscillations continue giving the steady state motion.
C. Oscillatory Electrical Circuits
a)
L – C Circuit
Consider an electrical circuit containing an inductance L and capacitance C.
V
C
i
L
Let i be the current and q be the charge in the condenser plate at any
time t then the voltage drop across
L = L
di
d 2q
L 2
dt
dt
and voltage drop across C 
q
C
As there is no applied e.m.f. in the circuit, therefore by Kirchooff’s first
law, we have
L
d 2q q
 0
dt 2 C
94
Engineering Mathematics–II
Dividing by L and taking
1
  2 we get
LC
d 2q
 2 q  0
dt 2
...(1)
Which is linear differential equation and it reprsents free electrical
oscillations of the current having period
b)
2
 2 Lc .

L – C – R Circuit
Consider the discharge of a condenser through an inductance L and the
resistance R. Since the voltage drop across L, C and R are respectively.
C
V
i
R
L
L
d 2q q
dq
,
and R
2
C
dt
dt
By Kirchhoff’s law, we have
L
Taking
d 2q
dq q
R + =0
2
dt C
dt
R
1
 2 and
  2 we get
L
LC
d 2q
dq
 2
 2 q  0
2
dt
dt
Which is the linear differential equation.
....(2)
96
Engineering Mathematics–II
By Kirchhoff’s law the equation is
L
d 2q
dq q
 R   p cos nt
2
dt C
dt
Dividing by L
d 2 q R dq q
p


 cos nt
2
L
dt
LC
L
dt
Taking
R
1
 2 and
  2 , we have
L
LC
d 2q
dq
p
 2   2 q  cos nt
2
dt
L
dt
Which is the linear differential equation.
Note: The L – C – R circuit with a source of alternating e.m.f. is an
electrical equivalent of the mechanical phenomena of forced oscillations with
resistance.
Example 1: In an L – C – R circuit, the charge q on a plate of a condenser
is given by
L
d 2q
dq q
 R   E sin pt
2
dt C
dt
2
The circuit is tunned to resource so that p 
1
. If initially the
LC
current i and the charge q be zero, show that for small values of
current in the circuit at time t is given by
Et
sin pt .
2L
Solution: Given differential equation is
1
 2
 LD  RD   q  E sin pt
C


2
AE is LD  RD 
1
0
C
....(1)
R
. The
L
97
Differential Equations–2
4L
C  R 
2L
 R  R2 

D =
As
R
is small, we have
L
2L
D = 
CF = e


R
i
2L
R2
1

2
4 L LC
1
R

 ip
2L
LC
Rt
2 L (c cos
1
pt  c2 sin pt )
Rt 

= 1 
 (c1 cos pt  c2 sin pt )
2L 

2


R
 rejecting the terms in   etc. 


L


PI =
And
=
1
LD 2  RD 
E sin pt
1
C
1
 Lp 2  RD 
C
= 
E sin pt

E
R

sin pt dt
 p2 
1
LC
E
cos pt
Rp
Thus the complete solution is
E
 Rt 
cos pt
q = 1    c1 cos pt  c2 sin pt  
L
Rp


i=
...(ii)
 Rt 
dq
= 1   ( c1 sin pt  p  c2 cos pt  p )
L
dt

E
 R
pt  sin pt
    (c1 cos pt  c2 sin)
R
 L
....(iii)
98
Engineering Mathematics–II
When t = 0, q = 0, i = 0
From (ii) 0 = c1 

E
Rp
and from (iii) 0 = c2 p  c1
E
Rp
 c1 
R
2L
 c2 
Rc1
E

2 Lp 2 Lp 2

 Rt   E
sin pt  c2 cos pt  p
i = 1    
L   Rp




=
R
2L
 E
 E
E
cos pt 
sin pt   sin pt

2
2 Lp
 Rp
 R
Et
sin pt
2L
R


is small 

L


D. Deflection of Beams
Example 1: The deflection of a strut of length l with one end (x = 0) built
in and the other supported and subjected to end trust p satisfies the
equation.
d2y
a2R
2

a
y

(l  x)
p
dx 2
Prove that the deflection curve is y 
R
p
 sin ax

 l cos ax  l  x  and

a


al = tan al.
Solution: Given differential equation is
(D2 + a2)y =


And
a2R
(l  x)
p
Its AE is D2 + a2 = 0

....(1)
D=
CF = c1 cos ax + c2 sin ax
PI =
1
a2 R
(l  x)
D2  a 2 p
± ai
Differential Equations–2
w
= 
2 EIa 
= 
101
1
 D2 
2
 1  2  ( x  lx)
a


2 
w  2
 x  lx  2  where p = EIa2
2p 
a 
Thus the complete solution of (1) is
y = c1 cosh ax + c2 sinh ax –
2 
w  2
 x  lx  2 
2p 
a 
...(ii)
At the end O, t = 0 when x = 0

(ii) gives O = c1 
w
pa 2
 c1 
w
pa 2
At the end A, y = 0 when x = l

(ii) gives, O = c1 cosh al + c2 sinh al –

c2 sinh al =

c1 =
w
(1  cosh al )
pa 2
w
pa 2
 c1 
w
pa 2
w
al
tanh
2
2
pa
Substituting in (ii)
y =
w
pa 2
al

 w
 cosh ax  tanh sinh ax  
2

 2p
2 
 2
 x  lx  2 
a 

Which is the deflection of the beam at N.
l

Thus the central deflection = y  at x  
2

w 
al
al
al  wl 2
cosh

tanh
sinh
 1 
y =

2
2
2
pa 2 
 8p
w
=
pa 2
al  wl 2

 1 
sec
h

2

 8p
102
Engineering Mathematics–II
Also the bending moment is maximum at the point of maximum
l

deflection  x   .
2

The maximum bending moment

EI
w
d2y 
l
at x   = py  ( x 2  lx )
2 
2
2
dx 
=
w
a
l

 at x  
2

al 

 sech  1
2


EXERCISES
1.
A particle is executing simple harmonic motion with amplitude 20 cm
and time 4 seconds. Find the time required by the particle in passing
between points which are at distance 15 cm and 5 cm from the centre of
force and are on the same side of it.
Ans.: 0.38 sec.
2.
An elastic string of natural length a is fixed at one end and a particle of
mass m hangs freely from the other end. The modulus of elasticity is
mg. The particle is pulled down a further distance l below its
equilibrium position and released from rest. Show that the motion of the
particle is simple harmonic and find the periodicity.
3.
The differential equation of a simple pendulum is
d 2x
 w02 x = F0 sin nt, where w0 and F0 are constants. If initially x = 0,
dt 2
dx
 0 determine the motion when w0 = n.
dt
Ans.:
4.
x=
F0
2n 2
(sin nt  n cos nt )
An e.m.f. E sin pt is applied at t = 0 to a circuit containing a capacitance
C and inductance L. The current i satisfies the equation L
di 1

dt C

i dt
104
Engineering Mathematics–II
2.9 QUESTION BANK
A. Choose the correct answer from the given alternatives.
1.
2.
3.
4.
5.
A general solution of a linear differential equation of nth order contains.
(a)
one constant
(b) n–constants
(c)
zero constants
(d) two constants
The solution of the differential equation
dy
 0 is
dx
(c) y = (c1 + c2x) e3x
(d) c1x = c2
The solution of the differential equation (D2 + a2) y = 0 is
(a) y = c1eax + c2e–ax
(b) y = c1 cos ax + c2 sin ax
(c) y = (c1x + c2) cos ax
(d) y = ex (c1 cos x + c2 sin x)
The solution of the differential equation
d2y
dx
2
2
dy
 y  0 is
dx
(a) y = c1e–x + c2ex
(b) y = (c1x + c2) ex
(c) y = (c1x – c2) e–x
(d) y = ex (c1 cos x + c2 sin x)
The complementary function of y+ 2y + y = xe–x sin x is
(b) (c1x + c2) e–x
(c1x + c2) ex
(d) c1 + e2 ex
Particular integral of the differential equation (D2 + 5D + 6) y = ex is
(a) ex
(b)
ex
12
(c)
ex
30
(d)
ex
6
Particular integral of the differential equation (D2 + 4) y = sin 2x is
(a)
8.
3
(b) y = c1 + c2e3x
(c)
7.
dx
2
(a) y = c1x + c2e3x
(a) c1ex + c2e–x
6.
d2y
x
sin 2 x
2
(b) 
x
cos 2 x
4
(c)
x
cos 2 x
2
(d)
x
cos 2 x
4
When X = eaxV where V is any function of x then particular integral is
106
Engineering Mathematics–II
18. Solve
19. Solve
20. Solve
d3y
dx
3
d3y
dx3
d2y
dx
2
3
d2y
dx
2
3
dy
 y0
dx
 y 0
2
dy
 10 y  0, y (0)  4, y (0)  1
dx
C. Questions carrying six marks
21. Solve (D3 + 1) y = cos (2x – 1)
22. Solve
d2y
dx
2

dy
 x2  2x  4
dx
23. Solve (D – 2)2y = 8(e2x + sin 2x + x2)
24. Solve (D2 – 4D + 3) y = sin 3x cos 2x.
25. Solve (D2 – 2D + 1) y = xex sin x.
26. Solve
27. Solve
d2y
dx 2
d2y
dx
2
 a 2 y  sec ax
4
dy
 4 y  8 x 2 e2 x sin 2 x
dx
28. Solve the following simultaneous equations
dx
 7 x  y,
dt
dy
 2x  5 y .
dt
29. Solve
dx
dy
 y  sin t ;
 x  cos t given x = 2, y = 0 when t = 0.
dt
dt
30. Solve
dx
dy
 5 x  2 y  5,
 2 x  y  0 given x = 0, y = 0, when
dt
dx
t = 0.
108
Engineering Mathematics–II
(26) y = c1 cos ax + c2 sin ax +
1
1
x sin ax  2 cos ax log cos ax
a
a
(27) y = (c1 + c2x) e2x – e2x [4x cos 2x + (2x2 – 3) sin 2x]
(28) x = e6t [c1 cos t + c2 sin t]
y = e6t [(c1 – c2) cos t + (c1 + c2) sin t]
(29) x = et + e–k, y = e–t – et + sin t
(30)
1
1
x)
(1  6t e3t  )
(1  3t
27
27
y=
2
2
)
(2  3t e 3t  )
(2  3t
27
27
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