CHAPTER 82 SECOND-ORDER DIFFERENTIAL EQUATIONS
OF THE FORM
a
d2 y
dy
f ( x)
+b
+ cy =
2
dx
dx
EXERCISE 308 Page 855
1. Find the general solution of: 2
2
dy
d2 y
+5
– 3y = 6
dx
d x2
dy
d2 y
+5
– 3y = 6 in D-operator form is:
dx
d x2
6
( 2 D 2 + 5 D − 3) y =
2 m 2 + 5m − 3 =
0
Auxiliary equation is:
i.e.
(2m – 1)(m + 3) = 0
m=
from which,
1
2
and m = –3
1
Hence, the complementary function, C.F., =
u Ae 2 + Be − 3 x
x
Let the particular integral, P.I., v = k
6
( 2 D 2 + 5D − 3)( k ) =
then
D(k) = 0
and
2 (k )
D=
D=
(0) 0
Hence,
0 + 0 – 3k = 6
from which,
k = –2
Hence, P.I., v = –2
1
and the general solution, y = u + v = Ae 2 + Be − 3 x − 2
2. Find the general solution of: 6
6
x
dy
d2 y
+4
– 2y = 3x – 2
dx
d x2
d2 y
dy
+4
− 2 y = 3 x − 2 in D-operator form is:
d x2
dx
Auxiliary equation is:
i.e.
6=
m 2 + 4m − 2 0
( 6 D 2 + 4 D − 2 ) y =3x − 2
i.e.
=
3m 2 + 2m − 1 0
(3m – 1)(m + 1) = 0
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© 2014, John Bird
m=
from which,
1
3
and m = –1
1
Hence, the complementary function, C.F., =
u Ae 3 + Be − x
x
Let the particular integral, P.I., v = ax + b
( 6 D 2 + 4 D − 2 )( ax + b ) = 3x − 2
then
D(ax + b) = a
and
D 2 (ax + b=
) D(a=
) 0
Hence,
0 + 4a – 2ax – 2b = 3x – 2
from which,
–2a = 3
and
and
4a – 2b = –2
a= −
3
2
and b = 2a + 1 = –2
3
Hence, P.I., v = − x – 2
2
1
and the general solution, y = u + v = Ae 3 + Be − x − 2 −
3. Find the particular solution of: 3
3
x
3
x
2
dy
dy
d2 y
+
– 4y = 8; when x = 0, y = 0 and
=0
2
dx
dx
dx
d2 y
dy
+
– 4y = 8 in D-operator form is:
dx
d x2
8
( 3D 2 + D − 4 ) y =
3m 2 + m − 4 =
0
Auxiliary equation is:
i.e.
(3m + 4)(m – 1) = 0
m= −
from which,
4
3
and m = 1
Hence, the complementary function, C.F.,=
u Ae
−
4
x
3
+ Be x
Let the particular integral, P.I., v = k
8
( 3D 2 + D − 4 )( k ) =
then
D(k) = 0
and
Hence,
from which,
2 (k )
D=
D=
(0) 0
0 + 0 – 4k = 8
k = –2
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© 2014, John Bird
Hence, P.I., v = –2
and the general solution, y = u + v = Ae
4
x
3
−
+ Be x − 2
When x = 0, y = 0, hence,
0=A+B–2
or
2=A+B
(1)
dy
4 −4x
=
− A e 3 + B ex
dx
3
When x = 0,
dy
= 0, hence,
dx
0= −
4
A+ B
3
Equation (1) – (2) gives:
2=
7
A
3
From (1),
2=
6
+B
7
Hence, the particular solution is:
y=
4. Find the particular solution of: 9
9
(2)
from which, A =
from which, B =
6 − 4x 8 x
e 3 + e −2
7
7
or
y=
6
7
8
7
2  −4x
x
 3e 3 + 4e  − 2
7

dy
4
d2 y
dy
– 12
+ 4y = 3x – 1; when x = 0, y = 0 and
=–
2
dx
3
dx
dx
d2 y
dy
− 12
+ 4 y = 3 x − 1 in D-operator form is:
2
dx
dx
( 9 D 2 − 12 D + 4 ) y =3x − 1
9m 2 − 12m + 4 =
0
Auxiliary equation is:
i.e.
(3m – 2)(3m – 2) = 0
m=
from which,
Hence, C.F., =
u
2
3
twice
2
( Ax + B ) e 3 x
Let the particular integral, P.I., v = ax + b
( 9 D 2 − 12 D + 4 )( ax + b ) = 3x − 1
then
D(ax + b) = a
Hence,
and
D 2 (ax + b=
) D(a=
) 0
0 – 12a + 4ax + 4b = 3x – 1
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© 2014, John Bird
from which,
4a = 3
and
and
–12a + 4b = –1
Hence, P.I., v =
a=
3
4
i.e. –9 + 4b = –1
and
b=2
3
x+2
4
2
and the general solution, y = u + v = ( Ax + B ) e 3 +
x = 0 and y = 0, hence,
0=B+2
x
3
x+2
4
from which,
B = –2
dy
3
 2 2x   2 
=+
( Ax B)  e 3  +  e 3  ( A) +
dx
4
3
  
x = 0 and
dy
4
4 2
3
B + A+
= − , hence, − =
3 3
4
dx
3
−
and since B = –2,
4
4
3
=− + A +
3
3
4
3
4
from which,
A= −
Hence,
 3
 2x 3
y =  − x − 2 e3 + x + 2
4
 4

i.e.
3  2x
3

y = −  2 + x  e3 + 2 + x
4 
4

5. The charge q is an electric circuit at time t satisfies the equation: L
dq
1
d2 q
+R
+ q =E
2
dt
C
dt
where L, R, C and E are constants. Solve the equation given L = 2 H, C = 200 × 10 −6 F and
E = 250 V, when (a) R = 200 Ω and (b) R is negligible. Assume that when t = 0, q = 0 and
dq
=0
dt
(a) L
1
d2 q
dq 1

E
+R
+ q=
E in D-operator form is:  L D 2 + R D +  q =
C
d t2
dt C

L m2 + R m +
The auxiliary equation is:
−R ± R2 −
and
m=
2L
Hence, C.F., =
u
1
=
0
C
4L
4(2)
−200 ± 2002 −
200 ×10−6 = −200 ± 0 = −50
C =
4
4
( At + B ) e−50t
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© 2014, John Bird
Let the particular integral, P.I., v = k
1

2
E
 LD + RD +  ( k ) =
C

then
D(k) = 0
2 (k )
D=
D=
(0) 0
and
Hence,
1
(k ) = E
C
Hence, P.I., v =
1
20
1
k = CE = ( 200 ×10−6 )( 250 ) =
0.05 or
20
and
and the general solution, y = u + v = ( At + B ) e −50 t +
t = 0 and q = 0, hence,
0=B+
1
20
1
20
from which,
B=–
1
20
dq
= ( At + B) ( −50 e − 50 t ) + ( e − 50 t ) ( A)
dt
t = 0 and
dq
= 0 , hence,
dt
 1 
i.e. 0 = −50  −  + A
 20 
−50B + A
0=
Hence,
1 
1
 5
q =  − t −  e − 50 t +
20
 2 20 
i.e.
q=
−R ± R2 −
(b) When R = 0, m=
2L
i.e. A = −
5
2
1 5
1 
−  t +  e − 50 t
20  2 20 
4L
4(2)
−0 ± 02 −
200 ×10−6 = 0 ± j 200= 0 ± j 50
C =
4
4
and
q = (A cos 50 t + B sin 50t) +
q = 0 and t = 0, hence,
0=A+
1
20
i.e.
A=–
1
20
1
20
dq
=
( −50 A sin 50t + 50 B cos 50t )
dt
t = 0 and
Thus,
i.e.
dq
= 0 , hence, 0 = 50B
dt
q=–
=
q
i.e.
B=0
1
1
cos 50t +
20
20
1
(1 − cos 50t )
20
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© 2014, John Bird
6. In a galvanometer the deflection θ satisfies the differential equation
Solve the equation for θ given that when t = 0, θ =
d2 θ
dθ
8 in D-operator form is:
+4
+ 4θ =
2
dt
dt
d2 θ
dθ
8.
+4
+ 4θ =
2
dt
dt
dθ
=2
dt
8
( D 2 + 4 D + 4 )θ =
m 2 + 4m + 4 =
0
Auxiliary equation is:
i.e.
(m + 2)(m + 2) = 0
from which,
Hence, C.F., =
u
m = –2
twice
( At + B ) e− 2t
Let the particular integral, P.I., v = k
8
( D2 + 4D + 4) k =
then
D(k) = 0
and
2 (k )
D=
D=
(0) 0
Hence,
4k = 8
from which,
k=2
Hence, P.I., v = 2
and the general solution, θ = u + v = ( At + B ) e − 2 t + 2
t = 0 and θ = 2, hence,
2=B+2
from which,
B=0
dθ
= ( At + B) ( −2 e −2t ) + ( e −2t ) ( A)
dt
x = 0 and
dθ
= 2 , hence,
dt
2=
−2B + A
Hence,
θ = 2 t e− 2t + 2
i.e.
θ = 2 ( t e −2t + 1)
from which,
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A=2
© 2014, John Bird
EXERCISE 309 Page 857
1. Find the general solution of:
d2 y d y
–
– 6y = 2e x
dx
d x2
d2 y d y
–
– 6y = 2e x .
d x2
dx
2 ex
( D2 − D − 6) y =
in D-operator form is:
m2 − m − 6 =
0
Auxiliary equation is:
i.e.
(m – 3)(m + 2) = 0
from which,
m=3
and
m = –2
Hence, C.F., =
u A e3 x + B e − 2 x
Let the particular integral, P.I., v = k e x
2 ex
( D 2 − D − 6 )( k e x ) =
then
D( k e x ) = k e x
Hence,
and D 2 (k e x ) = k e x
2 ex
( k e x ) − ( k e − x ) − 6k e x =
i.e.
−6k e x =
2 ex
i.e.
k= −
1
3
1
hence, the particular integral, v = − e x
3
1
and the general solution, y = u + v = A e3 x + B e − 2 x − e x
3
2. Find the general solution of:
d2 y
dy
–3
– 4y = 3e − x .
d x2
dx
d2 y
dy
−3
− 4y =
3e − x in D-operator form is:
d x2
dx
Auxiliary equation is:
i.e.
from which,
3e − x
( D 2 − 3D − 4 ) y =
m 2 − 3m − 4 =
0
(m – 4)(m + 1) = 0
m=4
and
m = –1
Hence, C.F., =
u A e4 x + B e− x
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© 2014, John Bird
As e − x appears in the C.F. and in the right-hand side of the differential equation, let the particular
integral, P.I., v = kx e − x
3e − x
( D 2 − 3D − 4 )( kx e− x ) =
then
D( kx e − x ) = ( kx )( − e − x ) + ( e − x )( k ) =
− kx e − x + k e − x
and D 2 (kx e − x ) =( −kx )( − e − x ) + ( e − x )( −k ) − k e − x =kx e − x − 2k e − x
Hence,
3e − x
( kx e− x − 2k e− x ) − 3 ( −kx e− x + k e− x ) − 4kx e− x =
kx e − x − 2k e − x + 3kx e − x − 3k e − x − 4kx e − x =
3e − x
i.e.
−5k e − x =
3e − x
i.e.
i.e.
–5k = 3
and
k= −
3
5
3
hence, the particular integral, v = − x e − x
5
3
and the general solution, y = u + v = A e 4 x + B e − x − x e − x
5
3. Find the general solution of:
d2 y
+ 9y = 26e 2 x
d x2
d2 y
+ 9y = 26e 2 x
d x2
in D-operator form is:
26 e 2 x
( D2 + 9) y =
Auxiliary equation is:
m2 + 9 =
0
and
m2 = − 9
i.e.
m=
−9 =± j 3
Hence, C.F.,
=
u A cos 3 x + B sin 3 x
Let the particular integral, P.I., v = k e 2 x
then
D( k e 2 x ) = 2 k e x
Hence,
i.e.
26 e 2 x
( D 2 + 9 )( k e2 x ) =
and D 2 (2k e 2 x ) = 4k e 2 x
26 e 2 x
( 4k e x ) + 9k e x =
13k e 2 x = 26 e 2 x
i.e.
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k=2
© 2014, John Bird
hence, the particular integral, v = 2 e 2 x
and the general solution, y = u + v = A cos 3 x + B sin 3 x + 2 e 2 x
4. Find the general solution of: 9
9
dy
d2 y
–6
+ y = 12e t /3
2
dt
dt
t
d2 y
dy
−6
+y=
12 e 3 in D-operator form is:
d t2
dt
t
12 e 3
( 9 D 2 − 6 D + 1) y =
9m 2 − 6m + 1 =0
Auxiliary equation is:
i.e.
(3m – 1)(3m – 1) = 0
from which,
m=
Hence, C.F., =
u
t
1
twice
3
t
( At + B ) e 3
t
As e 3 and t e 3 appears in the C.F. and in the right-hand side of the differential equation, let the
t
particular integral, P.I., v = k t 2 e 3
12 e 3
( 9 D 2 − 6 D + 1)  k t 2e 3  =
t
then

t

t
t
t
t
k
1 t   t 
t

D( k t 2 e 3 ) = ( kt 2 )  e 3  +  e 3  ( 2kt ) = t 2 e 3 + 2kt e 3 = kt e 3  + 2 
3
3   
3

t

1 t   t  
 2 3t  
 1   t
and D 2  kt
=
e   kt e 3    +  + 2  ( kt )  e 3  +  e 3  ( k ) 

 
 3   3

3    
=
t
t
k t k 2 t k t 2
t e 3 + t e 3 + t e 3 + kt e 3 + 2k e 3
3
9
3
3
t
t
t
t
t
t
4 t k
 k

Hence, 9  t e 3 + t 2 e 3 + 2k e 3  − 6  t 2 e 3 + 2kt e 3  + kt 2 e 3 =
12 e 3
9
3
 3

t
i.e.
t
t
i.e.
t
t
t
t
k=
12 2
=
18 3
t
12t e 3 + kt 2 e 3 + 18k e 3 − 2kt 2 e 3 − 12kt e 3 + kt 2 e 3 =
12 e 3
t
18k e 3 = 12 e 3
i.e.
and
t
t
2
hence, the particular integral, v = kt 2 e 3 = t 2 e 3
3
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© 2014, John Bird
t
t
2
and the general solution, y = u + v = ( At + B ) e 3 + t 2 e 3
3
5. Find the particular solution of: 5
5
d2 y
dy
1
dy
+9
– 2y = 3e x ; when x = 0, y = and
=0
4
d x2
dx
dx
d2 y
dy
+9
− 2y =
3 e x in D-operator form is:
2
dx
dx
3e x
( 5D 2 + 9 D − 2 ) y =
5m 2 + 9m − 2 =
0
Auxiliary equation is:
i.e.
(5m – 1)(m + 2) = 0
from which,
m=
1
1
5
and
m = –2
Hence, C.F., =
u A e 5 + B e −2 x
x
Let P.I., v = k e x
3e x
( 5D 2 + 9 D − 2 )( k e x ) =
then
D( k e x ) = k e x
2 (k e x )
and D=
D=
( k ex ) k ex
5k e x + 9 k e x − 2 k e x =
3e x
Hence,
12k e x = 3e x
i.e.
hence, the particular integral, v =
i.e.
k=
1
4
1 x
e
4
1
1
x
and the general solution, y = u + v = A e 5 + B e −2 x + e x
4
1
x = 0 and y = , hence,
4
i.e.
1
1
=A+B+
4
4
0=A+B
(1)
d y 1 1x
1
= A e 5 − 2 B e −2 x + e x
dx 5
4
x = 0 and
dy
= 0, hence,
dx
i.e.
2 × (1) gives:
0=
1
1
A − 2B +
5
4
1 1
A − 2B
− =
4 5
(2)
0 = 2A + 2B
(3)
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© 2014, John Bird
−
(2) + (3) gives:
1 11
=A
4 5
Hence,
y= −
i.e.
=
y
6. Find the particular solution of:
A= −
i.e.
5
5
and from (1), B =
44
44
5 1 x 5 −2 x 1 x
e5 + e + e
44
44
4
5  −2 x 1 x  1 x
 e − e5  + e
44 
 4
dy
dy
d2 y
–6
+ 9y = 4e3t; when t = 0, y = 2 and
=0
dx
dx
d t2
d2 y
dy
–6
+ 9y = 4e3t in D-operator form is:
d t2
dx
4 e3 t
( D2 − 6D + 9) y =
m 2 − 6m + 9 =
0
Auxiliary equation is:
i.e.
(m – 3)(m – 3) = 0
from which,
m = 3 twice
Hence, C.F., =
u
( At + B ) e3t
As e3t and t e3t both appear in the C.F., let the particular integral, P.I., v = k t 2 e3t
4 e3 t
( D 2 − 6 D + 9 )( k t 2 e3t ) =
then
D( k t 2 e3t ) = ( kt 2 )( 3e3t ) + ( e3t )( 2kt ) = 3kt 2 e3t + 2kt e3t = kt e3t ( 3t + 2 )
2 e3 t
and D 2 ( kt=
)
( kt e3t )( 3) + ( 3t + 2 ) ( kt )( 3e3t ) + ( e3t )( k )
= 3kt e3t + 9kt 2 e3t + 3kt e3t + 6kt e3t + 2k e3t
Hence, ( 3kt e3t + 9kt 2 e3t + 3kt e3t + 6kt e3t + 2k e3t ) – 6[ kt e3t ( 3t + 2 ) ] + 9 ( k t 2 e3t ) = 4 e3t
i.e.
3kt e3t + 9kt 2 e3t + 3kt e3t + 6kt e3t + 2k e3t −18k t 2 e3t − 12k t e3t + 9 kt 2 e3t = 4 e3t
i.e.
2 k e 3t = 4 e 3t
and k = 2
hence, the particular integral, v = 2 t 2 e3t
and the general solution, y = u + v = ( At + B ) e3t + 2t 2 e3t
When t = 0, y = 2, hence,
2=B+0
i.e.
B=2
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© 2014, John Bird
dy
=+
( At B )( 3e3t ) + ( e3t )( A) + ( 2t 2 )( 3e3t ) + ( e3t )( 4t )
dt
When t = 0,
dy
= 0, hence,
dt
0 = 3B + A + 0 + 0
from which, A = –6
(since B = 2)
Hence the particular solution is: y = ( −6t + 2 ) e3t + 2t 2 e3t
i.e.
y = 2 e3t (1 − 3t + t 2 )
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© 2014, John Bird
EXERCISE 310 Page 859
1. Find the general solution of: 2
2
d2 y d y
–
– 3y = 25 sin 2x
dx
d x2
d2 y d y
25sin 2 x in D-operator form is:
−
− 3y =
d x2 d x
25sin 2 x
( 2 D 2 − D − 3) y =
2m 2 − m − 3 =
0
Auxiliary equation is:
i.e.
(2m – 3)(m + 1) = 0
from which,
m=
3
3
2
and
m = –1
Hence, C.F., =
u A e 2 + B e− x
x
Let P.I., v = A sin 2x + B cos 2x
Hence,
25sin 2 x
( 2 D 2 − D − 3)( A sin 2 x + B cos 2 x ) =
D(A sin 2x + B cos 2x) = 2A cos 2x – 2B sin 2x
−4 A sin 2 x − 4 B cos 2 x
D 2 ( A sin 2 x + B cos 2 x ) =
D ( 2 A cos 2 x − 2 B sin 2 x ) =
Hence, 2 ( −4 A sin 2 x − 4 B cos 2 x ) − ( 2 A cos 2 x − 2 B sin 2 x ) − 3 ( A sin 2 x + B cos 2 x ) =
25sin 2 x
i.e.
–8A + 2B – 3A = 25
and
–8B – 2A – 3B = 0
i.e.
–11A + 2B = 25
(1)
and
–2A – 11B = 0
(2)
2 × (1) gives:
–22A + 4B = 50
(3)
11 × (2) gives:
–22A – 121B = 0
(3) – (4) gives:
125B = 50
from which,
B=
50 2
=
125 5
4
= 25
5
Substituting in (1) gives:
–11A +
from which,
–11A = 25 –
Thus, P.I., v = −
(4)
4 125 − 4 121
=
=
5
5
5
and
A=
121
11
= −
5(−11)
5
11
2
1
sin 2 x + cos 2 x =
− (11sin 2 x − 2 cos 2 x )
5
5
5
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© 2014, John Bird
3
y = u + v = Ae 2 + Be − x −
and
x
2. Find the general solution of:
1
(11sin 2 x − 2 cos 2 x )
5
d2 y
dy
–4
+ 4y = 5 cos x.
d x2
dx
d2 y
dy
–4
+ 4y = 5 cos x in D-operator form is:
dx
d x2
5cos x
( D2 − 4D + 4) y =
m 2 − 4m + 4 =
0
Auxiliary equation is:
i.e.
(m – 2)(m – 2) = 0
from which,
m = 2 twice
Hence, C.F., =
u ( Ax + B) e 2 x
Let P.I., v = A sin x + B cos x
Hence,
5cos x
( D 2 − 4 D + 4 )( A sin x + B cos x ) =
D(A sin x + B cos x) = A cos x – B sin x
D 2 ( A sin x + B cos x ) =
D ( A cos x − B sin x ) =
− A sin x − B cos x
Hence,
5cos x
( − A sin x − B cos x ) − 4 ( A cos x − B sin x ) + 4 ( A sin x + B cos x ) =
Equating coefficients of sin x gives:
– A + 4B + 4A = 0
and equating coefficients of cos x gives:
– B – 4A + 4B = 5
i.e.
3A + 4B = 0
(1)
and
–4A + 3B = 5
(2)
3 × (1) gives:
9A + 12B = 0
(3)
4 × (2) gives:
–16A +12B = 20
(4)
(4) – (3) gives:
Substituting in (1) gives:
– 25A = 20
–
12
A + 4B = 0
5
from which,
from which,
A= −
B=
20
4
=
−
25
5
12 3
=
20 5
4
3
Thus, P.I., v = − sin x + cos x
5
5
1270
© 2014, John Bird
4
3
y = u + v = ( Ax + B)e 2 x − sin x + cos x
5
5
and
3. Find the general solution of:
d2 y
+ y = 4 cos x.
d x2
d2 y
+y=
4 cos x in D-operator form is:
d x2
m 2 + 1 =0
Auxiliary equation is:
m 2 = −1
i.e.
4 cos x
( D 2 + 1) y =
and
−1 = 0 ± j 1
m=
Hence, C.F.,
=
u A cos x + B sin x
Since cos x occurs in the C.F. and in the right-hand side of the differential equation,
let P.I., v = x(C sin x + D cos x)
4 cos x
( D 2 + 1)  x ( C sin x + D cos x ) =
Hence,

D  x ( C sin x + D cos x )=
( x )( C cos x − D sin x ) + (1)( C sin x + D cos x )
D 2 [v] =
( x )( −C sin x − D cos x ) + ( C cos x − D sin x ) + ( C cos x − D sin x )
4 cos x
( D 2 + 1) v =
( x )( −C sin x − D cos x ) + ( C cos x − D sin x ) + ( C cos x − D sin x )
Hence, since
then
+ x(C sin x + D cos x)
= 4 cos x
from which,
and
2C = 4
from which, C = 2
– 2D = 0
from which, D = 0
Hence, P.I., v = 2x sin x
y = u + v = A cos x + B sin x + 2 x sin x
and
4. Find the particular solution of the differential equation:
dy
d2 y
dy
–3
– 4y = 3 sin x; when x = 0, y = 0 and
=0
2
dx
dx
dx
d2 y
dy
−3
− 4y =
3sin x in D-operator form is:
2
dx
dx
3sin x
( D 2 − 3D − 4 ) y =
m 2 − 3m − 4 =
0
Auxiliary equation is:
1271
© 2014, John Bird
i.e.
(m – 4)(m + 1) = 0
from which,
m=4
and
m = –1
Hence, C.F., =
u A e4 x + B e− x
Let P.I., v = A sin x + B cos x
Hence,
3sin x
( D 2 − 3D − 4 )( A sin x + B cos x ) =
D(A sin x + B cos x) = A cos x – B sin x
D 2 ( A sin x + B cos x ) =
D ( A cos x − B sin x ) =
− A sin x − B cos x
Hence, (–A sin x – B cos x) – 3(A cos x – B sin x) – 4(A sin x + B cos x) = 3 sin x
i.e.
–A + 3B – 4A = 3
and
–B – 3A – 4B = 0
i.e.
–5A + 3B = 3
(1)
and
–3A – 5B = 0
(2)
3 × (1) gives:
–15A + 9B = 9
(3)
5 × (2) gives:
–15A – 25B = 0
(4)
(3) – (4) gives:
34B = 9
Substituting in (1) gives:
 9 
–5A + 3   = 3
 34 
from which,
5A =
Thus, P.I., v = −
and
from which,
B=
27
27 − 102 −75
–3 =
=
34
34
34
9
34
and
A=
−75
15
= −
34(5)
34
15
9
sin x + cos x
34
34
y = u + v = A e4 x + B e− x −
x = 0 when y = 0, hence,
0=A+B+
9
34
15
9
sin x + cos x
34
34
i.e.
A+B=–
9
34
(5)
15
34
(6)
dy
15
9
= 4 A e 4 x − B e − x − cos x − sin x
dx
34
34
x = 0, when
dy
15
= 0, hence, 0 = 4A – B −
dx
34
i.e.
1272
4A – B =
© 2014, John Bird
4 × (5) gives:
36
34
4A + 4B = –
(7) – (6) gives:
5B = –
B= −
and
A−
Substituting in (5) gives:
51
9
=
−
170
34
from which,
36 15
51
–
=−
34 34
34
51
51
=
−
34(5)
170
51 9 51 − 45
6
−
=
=
170 34
170
170
A=
Hence,
y=
6 4 x 51 − x 15
9
e −
e − sin x + cos x
170
170
34
34
i.e.
y=
1
1
( 6 e4 x − 51e− x ) − (15sin x − 9 cos x )
170
34
5. A differential equation representing the motion of a body is
(7)
d2 y
+ n 2 y = k sin pt, where k, n
d t2
and p are constants. Solve the equation (given n ≠ 0 and p 2 ≠ n2) given that when t = 0,
y=
dy
=0
dt
d2 y
+ n 2 y = k sin pt in D-operator form is:
d t2
k sin pt
( D 2 + n2 ) y =
m2 + n2 =
0
Auxiliary equation is:
i.e.
m2 = −n2
from which,
m=
−n 2 =
± jn
Hence, C.F.,
=
u A sin nt + B cos nt
Let P.I., v = C sin pt + D cos pt
Hence,
k sin pt
( D 2 + n2 )( C sin pt + D cos pt ) =
D(C sin pt + D cos pt) = Cp cos pt – Dp sin pt
D 2 ( C sin pt + Dcos pt ) =
D(Cp cos pt − Dp sin pt ) =
−Cp 2 sin pt − Dp 2 cos pt
Hence, ( −Cp 2 sin pt − Dp 2 cos pt ) + n 2 (C cos pt + D sin pt) = k sin pt
i.e.
–C p 2 + D n 2 = k
and
–D p 2 + C n 2 = 0
(1)
1273
© 2014, John Bird
i.e.
C n2 – D p 2 = 0
p 2 × (1) gives:
–C p 4 + D p 2 n 2 = p 2 k
n 2 × (2) gives:
(4)
C( n 4 – p 4 ) = p 2 k
p 2n2k
= D p2
n4 − p 4
Substituting in (2) gives:
from which, C =
from which,
D=
p2k
n4 − p 4
n2k
n4 − p 4
p2k
n2k
sin pt +
cos pt
n4 − p 4
n4 − p 4
y = u + v = A sin nt + B cos nt +
and
(3)
C n4 – D n2 p 2 = 0
(3) + (4) gives:
Thus, P.I., v =
(2)
t = 0 when y = 0, hence,
0=B+
n2k
n4 − p 4
p2k
n2k
sin pt +
cos pt
n4 − p 4
n4 − p 4
i.e.
B=–
n2k
n4 − p 4
dy
p3k
pn 2 k
= An cos nt − Bn sin nt +
cos pt −
sin pt
dx
n4 − p 4
n4 − p 4
t = 0, when
p3k
dy
= 0, hence, 0 = An +
n4 − p 4
dx
i.e.
A=–
p3k / n
n4 − p 4
p3k / n
n2k
p2k
n2k
sin nt −
cos nt +
sin pt +
cos pt
n4 − p 4
n4 − p 4
n4 − p 4
n4 − p 4
Hence,
y= −
i.e.
y=
k  p3

2
2
2
 − sin nt − n cos nt + p sin pt + n cos pt 
4
4
n −p  n

i.e.
y=
k
p
 2

2
 p (sin pt − sin nt ) + n (cos pt − cos nt ) 
4
n
(n − p ) 

4
6. The motion of a vibrating mass is given by
d2 y
dy
+8
+ 20y = 300 sin 4t.
2
dt
dt
Show that the general solution of the differential equation is given by:
y = e −4t (A cos 2t + B sin 2t) +
15
(sin 4t – 8 cos 4t)
13
d2 y
dy
+8
+ 20y = 300 sin 4t in D-operator form is:
2
dt
dt
300sin 4t
( D 2 + 8D + 20 ) y =
m 2 + 8m + 20 =
0
Auxiliary equation is:
1274
© 2014, John Bird
i.e.
m=
−8 ± 82 − 4(1)(20) −8 ± −16 −8 ± j 4
= =
2(1)
2
2
from which,
m = –4 ± j2
Hence,
C.F., u e −4t ( A cos 2t + B sin 2t )
=
Let P.I., v = A sin 4t + B cos 4t
Hence,
300sin 4t
( D 2 + 8D + 20 )( A sin 4t + B cos 4t ) =
D(A sin 4t + B cos 4t) = 4A cos x – 4B sin x
D 2 ( A sin x + B cos x ) =
D ( 4 A cos 4t − 4 B sin 4t ) =
−16 A sin 4t − 16 B cos 4t
Hence, (–16A sin 4t – 16B cos 4t) + 8(4A cos x – 4B sin x) + 20(A sin 4t + B cos 4t) = 300 sin 4t
i.e.
–16A – 32B + 20A = 300
and
–16B + 32A + 20B = 0
i.e.
4A – 32B = 300
(1)
and
32A + 4B = 0
(2)
4 × (1) gives:
16A – 128B = 1200
32 × (2) gives:
1024A + 128B = 0
(3) + (4) gives:
Thus, P.I., v =
(4)
1040A = 1200
Substituting in (1) gives:
from which,
(3)
32B =
from which,
A=
1200 15
=
1040 13
60
– 32B = 300
13
60
60 − 3900 −3840
– 300 =
=
13
13
13
and
B=
−3840
120
= −
13(32)
13
15
120
sin 4t −
cos 4t
13
13
and
y = u + v = e −4t ( A cos 2t + B sin 2t ) +
i.e.
y = e −4t ( A cos 2t + B sin 2t ) +
15
120
sin 4t −
cos 4t
13
13
15
( sin 4t − 8cos 4t ) )
13
1275
© 2014, John Bird
7. L
1
dq
d2 q
+R
+ q = V0 sin ωt represents the variation of capacitor charge in an electric
2
C
dt
dt
circuit. Determine an expression for q at time t seconds given that R = 40 Ω, L = 0.02 H,
C = 50 × 10 −6 F, V0 = 540.8 V and ω = 200 rad/s and given the boundary conditions that
t = 0, q = 0 and
L
dq
= 4.8
dt
1
d2 q
dq 1

V0 sin ωt
V0 sin ωt in D-operator form is:  L D 2 + R D +  q =
+R
+ q=
C
d t2
dt C

L m2 + R m +
The auxiliary equation is:
−R ± R2 −
m=
and
Hence, C.F.,
=
u
2L
1
=
0
C
4L
4(0.02)
−40 ± 402 −
50 ×10−6 = −40 ± 0 = −1000
C =
2(0.02)
0.04
( At + B ) e−1000t
Let P.I., v = A sin ωt + B cos ωt
1

2
V0 sin ωt
 L D + R D +  [ A sin ωt + B cos ωt ] =
C

D(v) = Aω cos ωt – Bω sin ωt
and
D 2 (v ) =
− Aω 2 sin ωt − Bω 2 cos ωt
1

Thus,  L D 2 + R D +  v = 0.02 ( − Aω 2 sin ωt − Bω 2 cos ωt ) + 40 ( Aω cos ωt − Bω sin ωt )
C

1
V0 sin ωt
+
( A sin ωt + B cos ωt ) =
50 ×10−6
i.e. –800A sin 200t – 800B cos 200t + 8000A cos 200t – 8000B sin 200t + 20 000A sin 200t
+ 20 000B cos 200t = 540.8 sin 200t
Hence,
–800A – 8000B + 20 000A = 540.8
and
–800B + 8000A + 20 000B = 0
i.e.
19 200A – 8000B = 540.8
(1)
and
8000A + 19 200B = 0
(2)
8 × (1) gives:
153 600A – 64 000B = 4326.4
1276
(3)
© 2014, John Bird
19.2 × (2) gives:
153 600A + 368 640B = 0
(3) – (4) gives:
(4)
– 432 640B = 4326.4
from which,
B=
Substituting in (1) gives:
4326.4
= −0.01
432640
19 200A – 8000(–0.01) = 540.8
i.e.
19 200A + 80 = 540.8
and
A=
540.8 − 80 460.8
= = 0.024
19200
19200
Hence, P.I., v = 0.024 sin 200t – 0.01 cos 200t
q = u + v = ( At + B ) e −1000 t + 0.024 sin 200t – 0.01 cos 200t
Thus,
When t = 0, q = 0, hence,
0 = B – 0.01
from which,
B = 0.01
dq
= ( At + B )( −1000 e −1000t ) + A e −1000t + (0.024)(200) cos 200t + (0.01)(200) sin 200t
dt
When t = 0,
i.e.
Thus,
dq
= 4.8, hence,
dt
4.8 = – 1000B + A + 4.8
A = 1000B = 1000(0.01) = 10
q = (10t + 0.01) e −1000 t + 0.024 sin 200t – 0.010 cos 200t
1277
© 2014, John Bird
EXERCISE 311 Page 861
1. Find the general solution of: 8
8
dy
d2 y
–6
+ y = 2x + 40 sin x
2
dx
dx
d2 y
dy
−6
+ y = 2 x + 40sin x in D-operator form is:
2
dx
dx
(8D 2 − 6 D + 1) y = 2 x + 40sin x
8m 2 − 6m + 1 =0
Auxiliary equation is:
i.e.
(4m – 1)(2m – 1) = 0
from which,
m=
1
1
Hence, C.F., =
u Ae 4 + Be 2
x
1
4
and
m=
1
2
x
Let P.I., v = ax + b + c sin x + d cos x
Hence, ( 8 D 2 − 6 D + 1) [ ax + b + c sin x + d cos x ] = 2 x + 40sin x
D(v) = a + c cos x – d sin x
Hence,
(8D 2 − 6 D + 1) v
and
D 2 (v) = – c sin x – d cos x
= 8(–c sin x – d cos x) – 6(a + c cos x – d sin x)
+ (ax + b + c sin x + d cos x) = 2x + 40 sin x
i.e.
ax = 2x
from which,
a=2
and
–6a + b = 0,
from which,
b = 12
– 8c + 6d + c = 40
i.e.
–7c + 6d = 40
(1)
–8d – 6c + d = 0
i.e.
–6c – 7d = 0
(2)
6 × (1) gives:
–42c + 36d = 240
(3)
7 × (2) gives:
–42c – 49d = 0
(4)
(3) – (4) gives:
Substituting in (2) gives:
Hence, P.I., v = 2x + 12 −
85d = 240
 48 
– 6c – 7   = 0
 17 
from which,=
d
from which,
240 48
=
85 17
7(48)
56
c=
−
=
−
6(17)
17
56
48
sin x +
cos x
17
17
1278
© 2014, John Bird
1
1
and
y = u + v = A e 4 + B e 2 + 2x + 12 −
or
y = A e 4 + B e 2 + 2x + 12 +
1
1
x
x
x
x
2. Find the general solution of:
56
48
sin x +
cos x
17
17
8
( 6 cos x − 7 sin x )
17
dy
d2 y
–3
+ 2y = 2 sin 2θ – 4 cos 2θ
2
dθ
dθ
dy
d2 y
–3
+ 2y = 2 sin 2θ – 4 cos 2θ in D-operator form is:
dθ
dθ 2
( D 2 − 3 D + 2=
)y
2sin 2θ − 4 cos 2θ
m 2 − 3m + 2 =
0
Auxiliary equation is:
i.e.
(m – 2)(m – 1) = 0
from which,
m=2
and
m =1
Hence, C.F., =
u Ae 2θ + Beθ
Let P.I., v = A sin 2θ + B cos 2θ
Hence, ( D 2 − 3D + 2 ) [ A sin 2θ + B cos 2=
θ ] 2sin 2θ − 4 cos 2θ
D(v) = 2A cos 2θ – 2B sin 2θ
Hence,
( D 2 − 3D + 2 ) v
and
D 2 (v) = –4A sin 2θ – 4B cos 2θ
= (– 4A sin 2θ – 4B cos 2θ) – 3(2A cos 2θ – 2B sin 2θ)
+ 2(A sin 2θ + B cos 2θ) = 2 sin 2θ – 4 cos2θ
i.e.
–4A + 6B + 2A = 2
and
–4B – 6A + 2B = –4
i.e.
–2A + 6B = 2
(1)
and
–6A – 2B = –4
(2)
2 × (1) gives:
–4A + 12B = 4
(3)
6 × (2) gives:
–36A – 12B = –24
(4)
(3) + (4) gives:
–40A = –20
Substituting in (1) gives: –1 + 6B = 2
from which, =
A
20 1
=
40 2
from which, 6B = 3 and B =
1279
1
2
© 2014, John Bird
Hence, P.I., v =
1
1
sin 2θ + cos 2θ
2
2
and
y = u + v = A e 2θ + B eθ +
or
y = A e 2θ + B eθ +
3. Find the general solution of:
1
1
sin 2θ + cos 2θ
2
2
1
( sin 2θ + cos 2θ )
2
d2 y
dy
+
– 2y = x 2 + e 2 x
2
dx
dx
dy
d2 y
+
– 2y = x 2 + e 2 x in D-operator form is:
2
dx
dx
( D2 + D − 2) y =
x 2 + e2 x
m2 + m − 2 =
0
Auxiliary equation is:
i.e.
(m – 1)(m + 2) = 0
from which,
m=1
and
m = –2
Hence, C.F., =
u Ae x + Be − 2 x
Let P.I., v = a + bx + cx 2 + d e 2 x
Hence, ( D 2 + D − 2 ) [ a + bx + cx 2 + d e 2 x ] = x 2 + e 2 x
D(v) = b + 2cx + 2d e 2 x
Hence,
( D2 + D − 2) v
and
= (b + 2cx + 2d e 2 x ) + (2c + 4d e 2 x ) – 2(a + bx + cx 2 + d e 2 x ) = x 2 + e 2 x
Equating x 2 coefficients gives:
Equating x coefficients gives:
Equating constants gives:
and
–2c = 1
from which, c = –
2c – 2b = 0
b + 2c – 2a = 0
Equating e 2 x coefficients gives:
Hence, P.I., v = –
D 2 (v) = 2c + 4d e 2 x
1
2
i.e. –1 – 2b = 0
i.e. –
2d + 4d – 2d = 1
and –1 = 2b i.e. b = –
1
2
1
3
3
– 1 – 2a = 0 and 2a = –
i.e. a = –
2
2
4
i.e. 4d = 1 and
d=
1
4
3 1
1
1
– x – x 2 + e2 x
4
2
2
4
y = u + v = A e x + B e− 2 x –
3 1
1
1
– x – x 2 + e2 x
4
2
2
4
1280
© 2014, John Bird
4. Find the general solution of:
dy
d2 y
–2
+ 2y = e t sin t
2
dt
dt
d2 y
dy
et sin t in D-operator form is:
−2
+ 2y =
2
dt
dt
et sin t
( D2 − 2D + 2) y =
m 2 − 2m + 2 =
0
Auxiliary equation is:
i.e.
m=
−(−2) ± (−2) 2 − 4(1)(2) 2 ± −4 2 ± j 2
=
=
= 1± j
2(1)
2
2
Hence, C.F.,
u et ( A cos t + B sin t )
=
Since et sin t occurs in the C.F. and the right-hand side of the differential equation
let P.I., v = t et ( C sin t + D cos t )
then
et sin t
( D 2 − 2 D + 2 ) t et ( C sin t + D cos t ) =
D(v) = ( t et )( C cos t − D sin t ) + ( C sin t + D cos t ) [t et + et ]
D 2 (=
v)
( t et )( −C sin t − D cos t ) + ( C cos t − D sin t ) [t et + et ] + ( C sin t + D cos t ) [t et + et + et ]
+ ( t et + et )( C cos t − D sin t )
Hence,
( t et )( −C sin t − D cos t ) + ( C cos t − D sin t ) [t et + et ] + ( C sin t + D cos t ) [t et + et + et ]
+ ( t et + et )( C cos t − D sin t ) – 2 ( t et )( C cos t − D sin t ) − 2 ( C sin t + D cos t ) [t et + et ]
+ 2 t et ( C sin t + D cos t ) = et sin t
i.e.
–D + 2C – D – 2C = 1
i.e.
–2D = 1
and
C + 2D + C – 2D = 0
i.e.
C=0
and
D =−
1
2
t
 1

− et cos t
Hence, P.I., v = t et  − cos t  =
2
 2

and
t
y = u + v = et ( A cos t + B sin t ) − et cos t
2
1281
© 2014, John Bird
dy
dy
1
d2 y
–7
+ 10y = e 2 x + 20; when x = 0, y = 0 and
=–
2
3
dx
dx
dx
5. Find the particular solution of:
d2 y
dy
−7
+ 10 y =e 2 x + 20 in D-operator form is:
2
dx
dx
( D 2 − 7 D + 10 ) y = e2 x + 20
m 2 − 7 m + 10 =
0
Auxiliary equation is:
i.e.
(m – 5)(m – 2) = 0
from which,
m=5
and
m=2
Hence, C.F., =
u A e5 x + B e 2 x
Let P.I., v = kx e 2 x + a
Thus,
( D 2 − 7 D + 10 ) [ kx e2 x + a ] =
D(v) = ( kx )( 2 e 2 x ) + ( k e 2 x )
Hence,
e 2 x + 20
D 2 (v ) =
and
( kx )( 4 e2 x ) + ( 2 e2 x )( k ) + 2k e2 x
( kx )( 4 e2 x ) + ( 2 e2 x )( k ) + 2k e2 x − 7 ( kx )( 2 e2 x ) − 7 ( k e2 x ) +10k x e2 x + 10a =e2 x + 20
from which,
10a = 20
Also,
2k + 2k – 7k = 1
i.e.
and
a=2
–3k = 1
from which,
k= −
1
3
1
Hence, P.I., v = − x e 2 x + 2
3
1
y = u + v = A e5 x + B e 2 x − x e 2 x + 2
3
and
When x = 0, y = 0, hence,
0=A+B+2
i.e.
A + B = –2
(1)
 1 
dy
 1 
= 5 A e5 x + 2 B e 2 x −  x  ( 2 e 2 x ) + ( e 2 x )   
dx
 3 
 3 
When x = 0,
dy
1
= − , hence,
dx
3
−
1
1
= 5A + 2B −
3
3
5A + 2B = 0
(2)
2 × (1) gives:
2A + 2B = –4
(3)
(2) – (3) gives:
3A
1282
i.e.
=4
from which, A =
4
3
© 2014, John Bird
4
+ B = –2
3
Substituting in (1) gives:
Hence,
y=u+v=
6. Find the particular solution of:
from which,
B = –2 –
4
10
=−
3
3
4 5 x 10 2 x 1 2 x
e − e − xe + 2
3
3
3
2
d2 y d y
21
–
– 6y = 6e x cos x; when x = 0, y = –
and
d x2
dx
29
dy
12
= –7
dx
29
2
d2 y d y
–
– 6y = 6e x cos x in D-operator form is (2D2 – D – 6)y = 6e x cos x
2
dx
dx
The auxiliary equation is 2m 2 – m – 6 = 0
Using the quadratic formula, m =
1 ± [(−1) 2 − 4(2)(−6)]
1 ± 49 1 ± 7
3
=
= 2 or –
=
2(2)
4
2
2
Since the roots are complex, the C.F., u = A e 2 x + B e
−
3
x
2
Since the right-hand side of the given differential equation is a product of an exponential and a
cosine function, let the P.I., v = e x (C sin x + D cos x)
Substituting v into (2D2 – D – 6)v = 6e x cos x gives:
(2D2 – D – 6)[e x (C sin x + D cos x)] = 6e x cos x
D(v) = e x (C cos x – D sin x) + (C sin x + D cos x)(e x ) ≡ e x {(C + D)cos x + (C – D)sin x}
D2(v) = e x (–C sin x – D cos x) + e x (C cos x – D sin x)
+ e x (C cos x – D sin x) + e x (C sin x + D cos x)
≡ e x {–2D sin x + 2C cos x}
Hence (2D2 – D – 6)v = 2e x {–2D sin x + 2C cos x } – e x {(C + D)cos x + (C – D)sin x}
– 6 e x (C sin x + D cos x) = 6e x cos x
Equating coefficients of e x sin x gives: – 4D – C + D – 6C = 0
i.e.
Equating coefficients of e x cos x gives:
–7C – 3D = 0
(1)
4C – C – D – 6D = 6
1283
© 2014, John Bird
i.e.
3C – 7D = 6
(2)
3 × (1) gives:
–21C – 9D = 0
(3)
7 × (2) gives:
21C – 49D = 42
(4)
(3) + (4) gives:
– 58D = 42
Substituting in (1) gives:
Hence the P.I., v = e x (
–7C +
63
=0
29
9
21
sin x − cos x)
29
29
from which, D = –
from which, 7C =
42
21
= −
58
29
63
9
and C =
29
29
3e x
( 3sin x − 7 cos x )
29
or v =
The general solution, y = u + v
y = A e2 x + B e
i.e.
When x = 0, y = –
21
, hence,
29
–
−
3
x
2
+
3e x
( 3sin x − 7 cos x )
29
21
21
=A+B–
from which,
29
29
A+B=0
(1)
dy
3 − 3 x  3e x 
 3e x 
=2 A e 2 x − B e 2 + 
+
+
−
x
x
x
x
3cos
7
sin
3sin
7
cos
(
)
(
)



dx
2
 29 
 29 
When x = 0,
dy
12
12
9
21
3
= –7 , hence, –7
= 2A – B +
–
dx
29
29
29
29
2
i.e.
Solving equations (1) and (2) gives:
2A –
3
B = –7
2
A = –2 and B = 2
Hence the particular solution is: y = −2 e 2 x + 2 e
i.e.
(2)
y = 2e
−
3
x
2
−
3
x
2
+
− 2 e2 x +
1284
3e x
( 3sin x − 7 cos x )
29
3e x
( 3sin x − 7 cos x )
29
© 2014, John Bird