Second order differential equations of the forma
d^2 y
dx^2 +b
dy
dx+cy=f(x)^487
−^23 =( 0 +B)e^0 +^13 e^0 , from which,B=−1.
dy
dx
=(Ax+B)ex+ex(A)+^43 e^4 x.
Whenx=0,
dy
dx
= 4
1
3
, thus
13
3
=B+A+
4
3
from which,A=4, sinceB=−1.
Hence the particular solution is:
y=( 4 x− 1 )ex+^13 e^4 x
Problem 5. Solve the differential equation
2
d^2 y
dx^2
−
dy
dx
− 3 y=5e
3
2 x.
Using the procedure of Section 51.2:
(i) 2
d^2 y
dx^2
−
dy
dx
− 3 y=5e
3
2 x in D-operator form is
(2D^2 −D− 3 )y=5e
3
2 x.
(ii) Substituting m for D gives the auxiliary
equation 2m^2 −m− 3 =0. Factorizing gives:
( 2 m− 3 )(m+ 1 )=0, from which, m=^32 or
m=−1. Since the roots are real and different then
the C.F.,u=Ae
3
2 x+Be−x.
(iii) Since e
3
2 x appears in the C.F. and in the
right hand side of the differential equation, let the
P.I.,v=kxe
3
2 x(see Table 51.1(c), snag case (i)).
(iv) Substitutingv=kxe
3
2 x into(2D^2 −D− 3 )v=
5e
3
2 xgives:(2D^2 −D− 3 )kxe
3
2 x=5e
3
2 x.
D
(
kxe
3
2 x
)
=(kx)
(
3
2 e
3
2 x
)
+
(
e
3
2 x
)
(k),
by the product rule,
=ke
3
2 x
( 3
2 x+^1
)
D^2
(
kxe
3
2 x
)
=D
[
ke
3
2 x
( 3
2 x+^1
)
]
=
(
ke
3
2 x
)
( 3
2
)
+
( 3
2 x+^1
)
(
3
2 ke
3
2 x
)
=ke
3
2 x
( 9
4 x+^3
)
Hence(2D^2 −D− 3 )
(
kxe
3
2 x
)
= 2
[
ke
3
2 x
( 9
4 x+^3
)
]
−
[
ke
3
2 x
( 3
2 x+^1
)
]
− 3
[
kxe
3
2 x
]
=5e
3
2 x
i.e.^92 kxe
3
2 x+ 6 ke
3
2 x−^32 xke
3
2 x−ke
3
2 x
− 3 kxe
3
2 x=5e
3
2 x
Equating coefficients of e
3
2 xgives: 5k=5, from
which,k=1.
Hence the P.I.,v=kxe
3
2 x=xe
3
2 x.
(v) The general solution is y=u+v,i.e.
y=Ae
3
2 x+Be−x+xe
3
2 x.
Problem 6. Solve
d^2 y
dx^2
− 4
dy
dx
+ 4 y=3e^2 x.
Using the procedure of Section 51.2:
(i)
d^2 y
dx^2
− 4
dy
dx
+ 4 y=3e^2 x in D-operator form is
(D^2 −4D+ 4 )y=3e^2 x.
(ii) Substituting m for D gives the auxiliary
equation m^2 − 4 m+ 4 =0. Factorizing gives:
(m− 2 )(m− 2 )=0, from which,m=2 twice.
(iii) Since the roots are real and equal, the C.F.,
u=(Ax+B)e^2 x.
(iv) Since e^2 xandxe^2 xboth appear in the C.F. let the
P.I.,v=kx^2 e^2 x(see Table 51.1(c), snag case (ii)).
(v) Substitutingv=kx^2 e^2 x into(D^2 −4D+ 4 )v=
3e^2 xgives:(D^2 −4D+ 4 )(kx^2 e^2 x)=3e^2 x
D(kx^2 e^2 x)=(kx^2 )(2e^2 x)+(e^2 x)( 2 kx)
= 2 ke^2 x(x^2 +x)
D^2 (kx^2 e^2 x)=D[2ke^2 x(x^2 +x)]
=( 2 ke^2 x)( 2 x+ 1 )+(x^2 +x)( 4 ke^2 x)
= 2 ke^2 x( 4 x+ 1 + 2 x^2 )
Hence(D^2 −4D+ 4 )(kx^2 e^2 x)
=[2ke^2 x( 4 x+ 1 + 2 x^2 )]
−4[2ke^2 x(x^2 +x)]+4[kx^2 e^2 x]
=3e^2 x