490 Higher Engineering Mathematics
Now try the following exercise
Exercise 191 Further problems on
differential equations of the form
a
d 2 y
dx 2 + b
dy
dx
+ cy = f (x) where f (x) is a sine or
cosine function
In Problems 1 to 3, find the general solutions of the
given differential equations.
1. 2
d 2 y
dx 2 −
d y
dx
− 3y = 25 sin 2x
y = Ae
3
2 x
+ Be −x
−
1
5 (11 sin 2x − 2 cos 2x)
2.
d 2 y
dx 2 − 4
d y
dx
+ 4y = 5 cos x
y = (Ax + B)e 2x −
4
5 sin x +
3
5 cos x
3.
d 2 y
dx 2 + y = 4 cos x
[ y = A cos x + B sin x + 2x sin x]
4. Find the particular solution of the differential equation
d 2 y
dx 2 − 3
d y
dx
− 4y = 3 sin x; when
x = 0, y = 0 and
d y
dx
= 0.
⎡
⎢
⎢
⎣
y =
1
170
(6e 4x − 51e −x )
−
1
34
(15 sin x − 9 cos x)
⎤
⎥
⎥
⎦
5. A differential equation representing the
motion of a body is
d 2 y
dt 2 + n 2 y = k sin pt ,
where k, n and p are constants. Solve the equation (given n =0 and p 2 = n 2 ) given that when
t = 0, y =
d y
dt
= 0.
y =
k
n 2 − p 2
sin pt −
p
n
sin nt
6. The motion of a vibrating mass is given by
d 2 y
dt 2 + 8
d y
dt
+ 20y = 300 sin4t . Show that the
general solution of the differential equation is
given by:
y = e
−4t
(A cos 2t + B sin 2t )
+
15
13
(sin 4t − 8 cos4t )
7. L
d 2 q
dt 2 + R
dq
dt
+
1
C
q = V 0 sin ωt represents the
variation of capacitor charge in an electric circuit. Determine an expression for
q at time t seconds given that R = 40 ,
L =0.02 H, C = 50 × 10 −6 F, V 0 = 540.8 V
and ω = 200 rad/s and given the boundary
conditions that when t = 0, q = 0 and
dq
dt
= 4.8
q = (10t + 0.01)e −1000t
+ 0.024 sin 200t − 0.010 cos 200t
51.6 Worked problems on
differential equations of the
form a
d 2 y
d x 2 + b
dy
d x
+ cy = f (x)
where f (x) is a sum or a product
Problem 9. Solve
d
2 y
dx 2 +
d y
dx
− 6y = 12x − 50 sin x.
Using the procedure of Section 51.2:
(i)
d 2 y
dx 2 +
d y
dx
− 6y = 12x − 50 sin x in D-operator
form is
(D
2
+ D − 6)y = 12x − 50 sin x
(ii) The auxiliary equation is (m 2 + m − 6) = 0, from
which,
(m − 2)(m + 3) = 0,
i.e. m = 2 or m = −3
(iii) Since the roots are real and different, the C.F.,
u = Ae
2x
+ Be
−3x .
(iv) Since the right hand side of the given differential
equation is the sum of a polynomial and a sine
function let the P.I. v = ax + b + c sin x + d cos x
(see Table 51.1(e)).
Now try the following exercise
Exercise 191 Further problems on
differential equations of the form
a
d 2 y
dx 2 + b
dy
dx
+ cy = f (x) where f (x) is a sine or
cosine function
In Problems 1 to 3, find the general solutions of the
given differential equations.
1. 2
d 2 y
dx 2 −
d y
dx
− 3y = 25 sin 2x
y = Ae
3
2 x
+ Be −x
−
1
5 (11 sin 2x − 2 cos 2x)
2.
d 2 y
dx 2 − 4
d y
dx
+ 4y = 5 cos x
y = (Ax + B)e 2x −
4
5 sin x +
3
5 cos x
3.
d 2 y
dx 2 + y = 4 cos x
[ y = A cos x + B sin x + 2x sin x]
4. Find the particular solution of the differential equation
d 2 y
dx 2 − 3
d y
dx
− 4y = 3 sin x; when
x = 0, y = 0 and
d y
dx
= 0.
⎡
⎢
⎢
⎣
y =
1
170
(6e 4x − 51e −x )
−
1
34
(15 sin x − 9 cos x)
⎤
⎥
⎥
⎦
5. A differential equation representing the
motion of a body is
d 2 y
dt 2 + n 2 y = k sin pt ,
where k, n and p are constants. Solve the equation (given n =0 and p 2 = n 2 ) given that when
t = 0, y =
d y
dt
= 0.
y =
k
n 2 − p 2
sin pt −
p
n
sin nt
6. The motion of a vibrating mass is given by
d 2 y
dt 2 + 8
d y
dt
+ 20y = 300 sin4t . Show that the
general solution of the differential equation is
given by:
y = e
−4t
(A cos 2t + B sin 2t )
+
15
13
(sin 4t − 8 cos4t )
7. L
d 2 q
dt 2 + R
dq
dt
+
1
C
q = V 0 sin ωt represents the
variation of capacitor charge in an electric circuit. Determine an expression for
q at time t seconds given that R = 40 ,
L =0.02 H, C = 50 × 10 −6 F, V 0 = 540.8 V
and ω = 200 rad/s and given the boundary
conditions that when t = 0, q = 0 and
dq
dt
= 4.8
q = (10t + 0.01)e −1000t
+ 0.024 sin 200t − 0.010 cos 200t
51.6 Worked problems on
differential equations of the
form a
d 2 y
d x 2 + b
dy
d x
+ cy = f (x)
where f (x) is a sum or a product
Problem 9. Solve
d
2 y
dx 2 +
d y
dx
− 6y = 12x − 50 sin x.
Using the procedure of Section 51.2:
(i)
d 2 y
dx 2 +
d y
dx
− 6y = 12x − 50 sin x in D-operator
form is
(D
2
+ D − 6)y = 12x − 50 sin x
(ii) The auxiliary equation is (m 2 + m − 6) = 0, from
which,
(m − 2)(m + 3) = 0,
i.e. m = 2 or m = −3
(iii) Since the roots are real and different, the C.F.,
u = Ae
2x
+ Be
−3x .
(iv) Since the right hand side of the given differential
equation is the sum of a polynomial and a sine
function let the P.I. v = ax + b + c sin x + d cos x
(see Table 51.1(e)).