Second order differential equations of the form a
d 2 y
dx 2 + b
dy
dx + cy = f (x) 487
−
2
3 = (0 + B)e
0
+
1
3 e
0 , from which, B =−1.
d y
dx
= (Ax + B)e x + e x (A) +
4
3 e 4x .
When x = 0,
d y
dx
= 4
1
3
, thus
13
3
= B + A +
4
3
from which, A = 4, since B =−1.
Hence the particular solution is:
y = (4x − 1)e
x
+
1
3 e
4x
Problem 5. Solve the differential equation
2
d 2 y
dx 2 −
d y
dx
− 3y = 5e
3
2 x .
Using the procedure of Section 51.2:
(i) 2
d 2 y
dx 2 −
d y
dx
− 3y = 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:
(2m − 3)(m + 1) = 0, from which, m =
3
2 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) Substituting v = kxe
3
2 x into (2D 2 − D − 3)v =
5e
3
2 x gives: (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.
9
2 kxe
3
2 x + 6ke
3
2 x −
3
2 xke
3
2 x − ke
3
2 x
− 3kxe
3
2 x = 5e
3
2 x
Equating coefficients of e
3
2 x gives: 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
d y
dx
+ 4y = 3e 2x .
Using the procedure of Section 51.2:
(i)
d
2 y
dx 2 − 4
d y
dx
+ 4y = 3e 2x in D-operator form is
(D 2 − 4D +4)y = 3e 2x .
(ii) Substituting m for D gives the auxiliary
equation m 2 − 4m + 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 2x .
(iv) Since e 2x and xe 2x both appear in the C.F. let the
P.I., v = kx 2 e 2x (see Table 51.1(c), snag case (ii)).
(v) Substituting v = kx 2 e 2x into (D 2 − 4D + 4)v =
3e 2x gives: (D 2 − 4D + 4)(kx 2 e 2x ) = 3e 2x
D(kx
2 e
2x
) = (kx
2
)(2e
2x
) + (e
2x
)(2kx)
= 2ke
2x
(x
2
+ x)
D
2
(kx
2 e
2x
) = D[2ke
2x
(x
2
+ x)]
= (2ke
2x
)(2x + 1) + (x
2
+ x)(4ke
2x
)
= 2ke
2x
(4x + 1 + 2x
2
)
Hence (D
2
− 4D + 4)(kx
2 e
2x
)
= [2ke
2x
(4x + 1 + 2x
2
)]
− 4[2ke
2x
(x
2
+ x)] + 4[kx
2 e
2x ]
= 3e
2x
d 2 y
dx 2 + b
dy
dx + cy = f (x) 487
−
2
3 = (0 + B)e
0
+
1
3 e
0 , from which, B =−1.
d y
dx
= (Ax + B)e x + e x (A) +
4
3 e 4x .
When x = 0,
d y
dx
= 4
1
3
, thus
13
3
= B + A +
4
3
from which, A = 4, since B =−1.
Hence the particular solution is:
y = (4x − 1)e
x
+
1
3 e
4x
Problem 5. Solve the differential equation
2
d 2 y
dx 2 −
d y
dx
− 3y = 5e
3
2 x .
Using the procedure of Section 51.2:
(i) 2
d 2 y
dx 2 −
d y
dx
− 3y = 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:
(2m − 3)(m + 1) = 0, from which, m =
3
2 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) Substituting v = kxe
3
2 x into (2D 2 − D − 3)v =
5e
3
2 x gives: (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.
9
2 kxe
3
2 x + 6ke
3
2 x −
3
2 xke
3
2 x − ke
3
2 x
− 3kxe
3
2 x = 5e
3
2 x
Equating coefficients of e
3
2 x gives: 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
d y
dx
+ 4y = 3e 2x .
Using the procedure of Section 51.2:
(i)
d
2 y
dx 2 − 4
d y
dx
+ 4y = 3e 2x in D-operator form is
(D 2 − 4D +4)y = 3e 2x .
(ii) Substituting m for D gives the auxiliary
equation m 2 − 4m + 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 2x .
(iv) Since e 2x and xe 2x both appear in the C.F. let the
P.I., v = kx 2 e 2x (see Table 51.1(c), snag case (ii)).
(v) Substituting v = kx 2 e 2x into (D 2 − 4D + 4)v =
3e 2x gives: (D 2 − 4D + 4)(kx 2 e 2x ) = 3e 2x
D(kx
2 e
2x
) = (kx
2
)(2e
2x
) + (e
2x
)(2kx)
= 2ke
2x
(x
2
+ x)
D
2
(kx
2 e
2x
) = D[2ke
2x
(x
2
+ x)]
= (2ke
2x
)(2x + 1) + (x
2
+ x)(4ke
2x
)
= 2ke
2x
(4x + 1 + 2x
2
)
Hence (D
2
− 4D + 4)(kx
2 e
2x
)
= [2ke
2x
(4x + 1 + 2x
2
)]
− 4[2ke
2x
(x
2
+ x)] + 4[kx
2 e
2x ]
= 3e
2x