488 Higher Engineering Mathematics
from which, 2ke^2 x=3e^2 xandk=^32
Hence the P.I.,v=kx^2 e^2 x=^32 x^2 e^2 x.
(vi) The general solution,y=u+v,i.e.
y=(Ax+B)e^2 x+ 23 x^2 e^2 x
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)wheref(x)is an
exponential function
In Problems 1 to 4, find the general solutionsof the
given differential equations.
1.
d^2 y
dx^2
−
dy
dx
− 6 y=2ex
[
y=Ae^3 x+Be−^2 x−^13 ex
]
2.
d^2 y
dx^2
− 3
dy
dx
− 4 y=3e−x
[
y=Ae^4 x+Be−x−^35 xe−x
]
3.
d^2 y
dx^2
+ 9 y=26e^2 x
[y=Acos 3x+Bsin3x+2e^2 x]
- 9
d^2 y
dt^2
− 6
dy
dt
+y=12e
t
3
[
y=(At+B)e
1
3 t+^23 t^2 e
1
3 t
]
In problems 5 and 6 find the particular solutions of
the given differential equations.
- 5
d^2 y
dx^2
+ 9
dy
dx
− 2 y=3ex;whenx=0,y=
1
4
and
dy
dx
=0.
[
y=
5
44
(
e−^2 x−e
1
5 x
)
+
1
4
ex
]
6.
d^2 y
dt^2
− 6
dy
dt
+ 9 y=4e^3 t;whent=0,y= 2
and
dy
dt
=0[y=2e^3 t( 1 − 3 t+t^2 )]
51.5 Worked problemson
differential equations of the
forma
d^2 y
dx^2
+b
dy
dx
+cy=f(x)
wheref(x)is a sine or cosine
function
Problem 7. Solve the differential equation
2
d^2 y
dx^2
+ 3
dy
dx
− 5 y=6sin2x.
Using the procedure of Section 51.2:
(i) 2
d^2 y
dx^2
+ 3
dy
dx
− 5 y=6sin2xin D-operator form
is(2D^2 +3D− 5 )y=6sin2x
(ii) The auxiliary equation is 2m^2 + 3 m− 5 =0, from
which,
(m− 1 )( 2 m+ 5 )= 0 ,
i.e. m=1orm=−^52
(iii) Since the roots are real and different the C.F.,
u=Aex+Be−
5
2 x.
(iv) Let the P.I., v=Asin 2x+Bcos2x (see
Table 51.1(d)).
(v) Substitutingv=Asin2x+Bcos2xinto
(2D^2 +3D− 5 )v=6sin2xgives:
(2D^2 +3D− 5 )(Asin 2x+Bcos2x)=6sin2x.
D(Asin 2x+Bcos2x)
= 2 Acos 2x− 2 Bsin2x
D^2 (Asin 2x+Bcos2x)
=D( 2 Acos 2x− 2 Bsin2x)
=− 4 Asin 2x− 4 Bcos2x
Hence (2D^2 +3D− 5 )(Asin 2x+Bcos2x)
=− 8 Asin 2x− 8 Bcos2x+ 6 Acos 2x
− 6 Bsin2x− 5 Asin2x− 5 Bcos2x
=6sin2x
Equating coefficient of sin2xgives:
− 13 A− 6 B=6(1)