480 Higher Engineering Mathematics
3.
d 2 y
dx 2 + 2
d y
dx
+ 5y = 0
[y = e −x (A cos 2x + B sin 2x)]
In Problems 4 to 9, find the particular solution of the
given differential equations for the stated boundary
conditions.
4. 6
d 2 y
dx 2 + 5
d y
dx
− 6y = 0; when x = 0, y = 5 and
d y
dx
=−1.
y = 3e
2
3 x
+ 2e
−
3
2 x
5. 4
d 2 y
dt 2 − 5
d y
dt
+ y = 0; when t = 0, y = 1 and
d y
dt
=−2.
y = 4e
1
4 t
− 3e t
6. (9D 2 + 30D +25)y = 0, where D ≡
d
dx
; when
x = 0, y = 0 and
d y
dx
= 2.
y = 2xe
−
5
3 x
7.
d 2 x
dt 2 − 6
dx
dt
+ 9x = 0; when t = 0, x = 2 and
dx
dt
= 0.
[x = 2(1 − 3t )e 3t ]
8.
d 2 y
dx 2 + 6
d y
dx
+ 13y = 0; when x = 0, y = 4 and
d y
dx
= 0.
[y = 2e −3x (2 cos 2x + 3 sin2x)]
9. (4D 2 + 20D + 125)θ = 0, where D ≡
d
dt
; when
t = 0, θ = 3 and
dθ
dt
= 2.5.
[θ = e −2.5t (3 cos 5t + 2 sin5t )]
50.4 Further worked problems on
practical differential equations
of the form a
d 2 y
dx 2 + b
dy
dx
+ cy = 0
Problem 4. The equation of motion of a body
oscillating on the end of a spring is
d 2 x
dt 2 + 100x = 0,
where x is the displacement in metres of the body
from its equilibrium position after time t seconds.
Determine x in terms of t given that at time t = 0,
x = 2m and
dx
dt
= 0.
An equation of the form
d 2 x
dt 2 + m 2 x = 0 is a differential equation representing simple harmonic motion
(S.H.M.). Using the procedure of Section 50.2:
(a)
d 2 x
dt 2 + 100x = 0 in D-operator form is
(D 2 + 100)x = 0.
(b) The auxiliary equation is m 2 + 100 = 0, i.e.
m 2 =−100 and m =
√
(−100), i.e. m = ± j 10.
(c) Since the roots are complex, the general solution
is x = e 0 (A cos 10t + B sin 10t ),
i.e. x =(A cos 10t +B sin10t) metres
(d) When t = 0, x = 2, thus 2 = A
dx
dt
=−10 A sin 10t + 10B cos 10t
When t = 0,
dx
dt
= 0
thus 0 =−10 A sin 0 + 10B cos 0, i.e. B = 0
Hence the particular solution is
x = 2 cos 10t metres
Problem 5. Given the differential equation
d
2 V
dt 2 = ω 2 V , where ω is a constant, show that its
solution may be expressed as:
V = 7 cosh ωt + 3 sinhωt
given the boundary conditions that when
t = 0, V = 7 and
dV
dt
= 3ω.
Using the procedure of Section 50.2:
(a)
d 2 V
dt 2 = ω
2 V , i.e.
d 2 V
dt 2 − ω
2 V = 0 in D-operator
form is (D 2 − ω 2 )v = 0, where D ≡
d
dx
.
(b) The auxiliary equation is m 2 − ω 2 = 0, from
which, m 2 = ω 2 and m = ±ω.
3.
d 2 y
dx 2 + 2
d y
dx
+ 5y = 0
[y = e −x (A cos 2x + B sin 2x)]
In Problems 4 to 9, find the particular solution of the
given differential equations for the stated boundary
conditions.
4. 6
d 2 y
dx 2 + 5
d y
dx
− 6y = 0; when x = 0, y = 5 and
d y
dx
=−1.
y = 3e
2
3 x
+ 2e
−
3
2 x
5. 4
d 2 y
dt 2 − 5
d y
dt
+ y = 0; when t = 0, y = 1 and
d y
dt
=−2.
y = 4e
1
4 t
− 3e t
6. (9D 2 + 30D +25)y = 0, where D ≡
d
dx
; when
x = 0, y = 0 and
d y
dx
= 2.
y = 2xe
−
5
3 x
7.
d 2 x
dt 2 − 6
dx
dt
+ 9x = 0; when t = 0, x = 2 and
dx
dt
= 0.
[x = 2(1 − 3t )e 3t ]
8.
d 2 y
dx 2 + 6
d y
dx
+ 13y = 0; when x = 0, y = 4 and
d y
dx
= 0.
[y = 2e −3x (2 cos 2x + 3 sin2x)]
9. (4D 2 + 20D + 125)θ = 0, where D ≡
d
dt
; when
t = 0, θ = 3 and
dθ
dt
= 2.5.
[θ = e −2.5t (3 cos 5t + 2 sin5t )]
50.4 Further worked problems on
practical differential equations
of the form a
d 2 y
dx 2 + b
dy
dx
+ cy = 0
Problem 4. The equation of motion of a body
oscillating on the end of a spring is
d 2 x
dt 2 + 100x = 0,
where x is the displacement in metres of the body
from its equilibrium position after time t seconds.
Determine x in terms of t given that at time t = 0,
x = 2m and
dx
dt
= 0.
An equation of the form
d 2 x
dt 2 + m 2 x = 0 is a differential equation representing simple harmonic motion
(S.H.M.). Using the procedure of Section 50.2:
(a)
d 2 x
dt 2 + 100x = 0 in D-operator form is
(D 2 + 100)x = 0.
(b) The auxiliary equation is m 2 + 100 = 0, i.e.
m 2 =−100 and m =
√
(−100), i.e. m = ± j 10.
(c) Since the roots are complex, the general solution
is x = e 0 (A cos 10t + B sin 10t ),
i.e. x =(A cos 10t +B sin10t) metres
(d) When t = 0, x = 2, thus 2 = A
dx
dt
=−10 A sin 10t + 10B cos 10t
When t = 0,
dx
dt
= 0
thus 0 =−10 A sin 0 + 10B cos 0, i.e. B = 0
Hence the particular solution is
x = 2 cos 10t metres
Problem 5. Given the differential equation
d
2 V
dt 2 = ω 2 V , where ω is a constant, show that its
solution may be expressed as:
V = 7 cosh ωt + 3 sinhωt
given the boundary conditions that when
t = 0, V = 7 and
dV
dt
= 3ω.
Using the procedure of Section 50.2:
(a)
d 2 V
dt 2 = ω
2 V , i.e.
d 2 V
dt 2 − ω
2 V = 0 in D-operator
form is (D 2 − ω 2 )v = 0, where D ≡
d
dx
.
(b) The auxiliary equation is m 2 − ω 2 = 0, from
which, m 2 = ω 2 and m = ±ω.