486 Higher Engineering Mathematics
Now try the following exercise
Exercise 189 Further 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
constant or polynomial.
In Problems 1 and 2, find the general solutions of
the given differential equations.
1. 2
d 2 y
dx 2 + 5
d y
dx
− 3y = 6
y = Ae
1
2 x + Be −3x − 2
2. 6
d 2 y
dx 2 + 4
d y
dx
− 2y = 3x − 2
y = Ae
1
3 x + Be −x − 2 −
3
2 x
In Problems 3 and 4 find the particular solutions of
the given differential equations.
3. 3
d 2 y
dx 2 +
d y
dx
− 4y = 8; when x = 0, y = 0 and
d y
dx
= 0.
y =
2
7 (3e
−
4
3 x + 4e x ) − 2
4. 9
d 2 y
dx 2 − 12
d y
dx
+ 4y = 3x − 1; when x = 0,
y = 0 and
d y
dx
= −
4
3
y = −
2 +
3
4 x
e
2
3 x + 2 +
3
4 x
5. The charge q in an electric circuit at time t satisfies the equation L
d 2 q
dt 2 + R
dq
dt
+
1
C
q = E,
where L, R, C and E are constants. Solve the
equation given L = 2H , C = 200 ×10 −6 F and
E = 250 V, when (a) R = 200 and (b) R is
negligible. Assume that when t = 0, q = 0 and
dq
dt
= 0
⎡
⎢
⎢
⎣
(a) q =
1
20
−
5
2
t +
1
20
e −50t
(b) q =
1
20
(1 − cos 50t )
⎤
⎥
⎥
⎦
6. In a galvanometer the deflection θ satisfies
the differential equation
d 2 θ
dt 2 + 4
dθ
dt
+ 4 θ = 8.
Solve the equation for θ given that when t = 0,
θ =
dθ
dt
= 2.
[θ = 2(t e −2t + 1)]
51.4 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 an exponential function
Problem 4. Solve the equation
d 2 y
dx 2 − 2
d y
dx
+ y = 3e 4x given the boundary
conditions that when x = 0, y = −
2
3 and
d y
dx
= 4
1
3
Using the procedure of Section 51.2:
(i)
d 2 y
dx 2 − 2
d y
dx
+ y = 3e 4x in D-operator form is
(D 2 − 2D + 1)y = 3e 4x .
(ii) Substituting m for D gives the auxiliary
equation m 2 − 2m + 1 =0. Factorizing gives:
(m − 1)(m − 1) = 0, from which, m = 1 twice.
(iii) Since the roots are real and equal the C.F.,
u = (Ax + B)e x .
(iv) Let the particular integral, v = ke
4x (see
Table 51.1(c)).
(v) Substituting v = ke 4x into
(D
2
− 2D + 1)v = 3e
4x gives:
(D
2
− 2D + 1)ke
4x
= 3e
4x
i.e. D 2 (ke 4x ) − 2D(ke 4x ) + 1(ke 4x ) = 3e 4x
i.e.
16ke 4x − 8ke 4x + ke 4x = 3e 4x
Hence 9ke 4x = 3e 4x , from which, k =
1
3
Hence the P.I., v = ke 4x =
1
3 e 4x .
(vi) The general solution is given by y = u + v, i.e.
y = (Ax + B)e x +
1
3 e 4x .
(vii) When x = 0, y = −
2
3 thus
Now try the following exercise
Exercise 189 Further 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
constant or polynomial.
In Problems 1 and 2, find the general solutions of
the given differential equations.
1. 2
d 2 y
dx 2 + 5
d y
dx
− 3y = 6
y = Ae
1
2 x + Be −3x − 2
2. 6
d 2 y
dx 2 + 4
d y
dx
− 2y = 3x − 2
y = Ae
1
3 x + Be −x − 2 −
3
2 x
In Problems 3 and 4 find the particular solutions of
the given differential equations.
3. 3
d 2 y
dx 2 +
d y
dx
− 4y = 8; when x = 0, y = 0 and
d y
dx
= 0.
y =
2
7 (3e
−
4
3 x + 4e x ) − 2
4. 9
d 2 y
dx 2 − 12
d y
dx
+ 4y = 3x − 1; when x = 0,
y = 0 and
d y
dx
= −
4
3
y = −
2 +
3
4 x
e
2
3 x + 2 +
3
4 x
5. The charge q in an electric circuit at time t satisfies the equation L
d 2 q
dt 2 + R
dq
dt
+
1
C
q = E,
where L, R, C and E are constants. Solve the
equation given L = 2H , C = 200 ×10 −6 F and
E = 250 V, when (a) R = 200 and (b) R is
negligible. Assume that when t = 0, q = 0 and
dq
dt
= 0
⎡
⎢
⎢
⎣
(a) q =
1
20
−
5
2
t +
1
20
e −50t
(b) q =
1
20
(1 − cos 50t )
⎤
⎥
⎥
⎦
6. In a galvanometer the deflection θ satisfies
the differential equation
d 2 θ
dt 2 + 4
dθ
dt
+ 4 θ = 8.
Solve the equation for θ given that when t = 0,
θ =
dθ
dt
= 2.
[θ = 2(t e −2t + 1)]
51.4 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 an exponential function
Problem 4. Solve the equation
d 2 y
dx 2 − 2
d y
dx
+ y = 3e 4x given the boundary
conditions that when x = 0, y = −
2
3 and
d y
dx
= 4
1
3
Using the procedure of Section 51.2:
(i)
d 2 y
dx 2 − 2
d y
dx
+ y = 3e 4x in D-operator form is
(D 2 − 2D + 1)y = 3e 4x .
(ii) Substituting m for D gives the auxiliary
equation m 2 − 2m + 1 =0. Factorizing gives:
(m − 1)(m − 1) = 0, from which, m = 1 twice.
(iii) Since the roots are real and equal the C.F.,
u = (Ax + B)e x .
(iv) Let the particular integral, v = ke
4x (see
Table 51.1(c)).
(v) Substituting v = ke 4x into
(D
2
− 2D + 1)v = 3e
4x gives:
(D
2
− 2D + 1)ke
4x
= 3e
4x
i.e. D 2 (ke 4x ) − 2D(ke 4x ) + 1(ke 4x ) = 3e 4x
i.e.
16ke 4x − 8ke 4x + ke 4x = 3e 4x
Hence 9ke 4x = 3e 4x , from which, k =
1
3
Hence the P.I., v = ke 4x =
1
3 e 4x .
(vi) The general solution is given by y = u + v, i.e.
y = (Ax + B)e x +
1
3 e 4x .
(vii) When x = 0, y = −
2
3 thus