488 Higher Engineering Mathematics
from which, 2ke
2x
= 3e
2x and k =
3
2
Hence the P.I., v = kx
2 e 2x =
3
2 x 2 e 2x .
(vi) The general solution, y = u + v, i.e.
y = (Ax + B)e
2x
+
3
2 x
2 e
2x
Now try the following exercise
Exercise 190 Further problems on
differential equations of the form
a
d 2 y
dx 2 + b
dy
dx
+cy = f (x) where f (x) is an
exponential function
In Problems 1 to 4, find the general solutions of the
given differential equations.
1.
d 2 y
dx 2 −
d y
dx
− 6y = 2e x
y = Ae 3x + Be −2x −
1
3 e x
2.
d 2 y
dx 2 − 3
d y
dx
− 4y = 3e −x
y = Ae 4x + Be −x −
3
5 xe −x
3.
d 2 y
dx 2 + 9y = 26e 2x
[ y = A cos 3x + B sin 3x + 2e 2x ]
4. 9
d 2 y
dt 2 − 6
d y
dt
+ y = 12e
t
3
y = (At + B)e
1
3 t +
2
3 t 2 e
1
3 t
In problems 5 and 6 find the particular solutions of
the given differential equations.
5. 5
d 2 y
dx 2 + 9
d y
dx
− 2y = 3e x ; when x = 0, y =
1
4
and
d y
dx
= 0.
y =
5
44
e −2x − e
1
5 x
+
1
4
e x
6.
d 2 y
dt 2 − 6
d y
dt
+ 9y = 4e 3t ; when t = 0, y = 2
and
d y
dt
= 0
[y = 2e 3t (1 − 3t + t 2 )]
51.5 Worked 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
Problem 7. Solve the differential equation
2
d 2 y
dx 2 + 3
d y
dx
− 5y = 6 sin2x.
Using the procedure of Section 51.2:
(i) 2
d 2 y
dx 2 + 3
d y
dx
− 5y = 6 sin2x in D-operator form
is (2D 2 + 3D − 5)y = 6 sin2x
(ii) The auxiliary equation is 2m 2 + 3m −5 = 0, from
which,
(m − 1)(2m + 5) = 0,
i.e. m = 1 or m = −
5
2
(iii) Since the roots are real and different the C.F.,
u = Ae
x
+ Be
−
5
2 x .
(iv) Let the P.I., v = A sin 2x + B cos 2x (see
Table 51.1(d)).
(v) Substituting v = A sin 2x + B cos 2x into
(2D 2 + 3D −5)v = 6 sin2x gives:
(2D 2 + 3D−5)(A sin 2x + B cos 2x) = 6 sin2x.
D(A sin 2x + B cos 2x)
= 2 A cos 2x − 2B sin 2x
D 2 (A sin 2x + B cos 2x)
= D(2 A cos 2x − 2B sin 2x)
= −4 A sin 2x − 4B cos 2x
Hence (2D 2 + 3D −5)(A sin 2x + B cos 2x)
=−8 A sin 2x − 8B cos 2x + 6 A cos 2x
− 6B sin 2x − 5 A sin 2x − 5B cos 2x
= 6 sin 2x
Equating coefficient of sin 2x gives:
−13 A − 6B = 6
( 1 )
from which, 2ke
2x
= 3e
2x and k =
3
2
Hence the P.I., v = kx
2 e 2x =
3
2 x 2 e 2x .
(vi) The general solution, y = u + v, i.e.
y = (Ax + B)e
2x
+
3
2 x
2 e
2x
Now try the following exercise
Exercise 190 Further problems on
differential equations of the form
a
d 2 y
dx 2 + b
dy
dx
+cy = f (x) where f (x) is an
exponential function
In Problems 1 to 4, find the general solutions of the
given differential equations.
1.
d 2 y
dx 2 −
d y
dx
− 6y = 2e x
y = Ae 3x + Be −2x −
1
3 e x
2.
d 2 y
dx 2 − 3
d y
dx
− 4y = 3e −x
y = Ae 4x + Be −x −
3
5 xe −x
3.
d 2 y
dx 2 + 9y = 26e 2x
[ y = A cos 3x + B sin 3x + 2e 2x ]
4. 9
d 2 y
dt 2 − 6
d y
dt
+ y = 12e
t
3
y = (At + B)e
1
3 t +
2
3 t 2 e
1
3 t
In problems 5 and 6 find the particular solutions of
the given differential equations.
5. 5
d 2 y
dx 2 + 9
d y
dx
− 2y = 3e x ; when x = 0, y =
1
4
and
d y
dx
= 0.
y =
5
44
e −2x − e
1
5 x
+
1
4
e x
6.
d 2 y
dt 2 − 6
d y
dt
+ 9y = 4e 3t ; when t = 0, y = 2
and
d y
dt
= 0
[y = 2e 3t (1 − 3t + t 2 )]
51.5 Worked 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
Problem 7. Solve the differential equation
2
d 2 y
dx 2 + 3
d y
dx
− 5y = 6 sin2x.
Using the procedure of Section 51.2:
(i) 2
d 2 y
dx 2 + 3
d y
dx
− 5y = 6 sin2x in D-operator form
is (2D 2 + 3D − 5)y = 6 sin2x
(ii) The auxiliary equation is 2m 2 + 3m −5 = 0, from
which,
(m − 1)(2m + 5) = 0,
i.e. m = 1 or m = −
5
2
(iii) Since the roots are real and different the C.F.,
u = Ae
x
+ Be
−
5
2 x .
(iv) Let the P.I., v = A sin 2x + B cos 2x (see
Table 51.1(d)).
(v) Substituting v = A sin 2x + B cos 2x into
(2D 2 + 3D −5)v = 6 sin2x gives:
(2D 2 + 3D−5)(A sin 2x + B cos 2x) = 6 sin2x.
D(A sin 2x + B cos 2x)
= 2 A cos 2x − 2B sin 2x
D 2 (A sin 2x + B cos 2x)
= D(2 A cos 2x − 2B sin 2x)
= −4 A sin 2x − 4B cos 2x
Hence (2D 2 + 3D −5)(A sin 2x + B cos 2x)
=−8 A sin 2x − 8B cos 2x + 6 A cos 2x
− 6B sin 2x − 5 A sin 2x − 5B cos 2x
= 6 sin 2x
Equating coefficient of sin 2x gives:
−13 A − 6B = 6
( 1 )