Second order differential equations of the form a
d 2 y
dx 2 + b
dy
dx + cy = f (x) 485
(ii) Substituting m for D gives the auxiliary equation m 2 + m − 2 = 0. Factorizing gives: (m − 1)
(m + 2) = 0, from which m = 1 or m =−2.
(iii) Since the roots are real and different, the C.F.,
u = Ae x + Be −2x .
(iv) Since the term on the right hand side of the
given equation is a constant, i.e. f (x) = 4, let
the P.I. also be a constant, say v = k (see
Table 51.1(a)).
(v) Substituting v = k into (D 2 + D − 2)v = 4
gives (D 2 + D − 2)k = 4. Since D(k) = 0 and
D 2 (k) = 0 then −2k = 4, from which, k = −2.
Hence the P.I., v =−2.
(vi) The general solution is given by y = u + v, i.e.
y = Ae
x
+ Be
−2x
− 2.
Problem 2. Determine the particular solution of
the equation
d 2 y
dx 2 − 3
d y
dx
= 9, given the boundary
conditions that when x = 0, y = 0 and
d y
dx
= 0.
Using the procedure of Section 51.2:
(i)
d 2 y
dx 2 − 3
d y
dx
= 9 in D-operator form is
(D 2 − 3D)y = 9.
(ii) Substituting m for D gives the auxiliary equation
m 2 − 3m =0. Factorizing gives: m(m − 3) = 0,
from which, m = 0 or m = 3.
(iii) Since the roots are real and different, the C.F.,
u = Ae 0 + Be 3x , i.e. u = A +Be 3x .
(iv) Since the C.F. contains a constant (i.e. A) then let
the P.I., v = kx (see Table 51.1(a)).
(v) Substituting v = kx into (D 2 − 3D)v = 9 gives
(D 2 − 3D)kx = 9.
D(kx) = k and D 2 (kx) = 0.
Hence (D 2 − 3D)kx = 0 −3k = 9, from which,
k = −3.
Hence the P.I., v = −3x.
(vi) The general solution is given by y = u + v, i.e.
y = A +Be 3x −3x.
(vii) When x = 0, y = 0, thus 0 = A + Be 0 − 0, i.e.
0 = A + B
(1)
d y
dx
= 3Be 3x − 3;
d y
dx
= 0 when x = 0, thus
0 = 3Be 0 − 3 from which, B = 1. From equation (1), A = −1.
Hence the particular solution is
y = −1 + 1e
3x
− 3x,
i.e. y = e
3x
− 3x − 1
Problem 3. Solve the differential equation
2
d 2 y
dx 2 − 11
d y
dx
+ 12y = 3x − 2.
Using the procedure of Section 51.2:
(i) 2
d 2 y
dx 2 − 11
d y
dx
+ 12y = 3x − 2 in D-operator
form is
(2D
2
− 11D + 12)y = 3x − 2.
(ii) Substituting m for D gives the auxiliary
equation 2m 2 − 11m + 12 =0. Factorizing gives:
(2m − 3)(m − 4) = 0, from which, m =
3
2 or
m = 4.
(iii) Since the roots are real and different, the C.F.,
u =Ae
3
2 x + Be 4x
(iv) Since f (x) = 3x − 2 is a polynomial, let the P.I.,
v = ax + b (see Table 51.1(b)).
(v) Substituting v = ax + b into
(2D 2 − 11D +12)v = 3x − 2 gives:
(2D 2 − 11D + 12)(ax + b) = 3x − 2,
i.e. 2D 2 (ax + b) − 11D(ax + b)
+ 12(ax + b) = 3x − 2
i.e.
0 − 11a + 12ax + 12b = 3x − 2
Equating the coefficients of x gives: 12a = 3,
from which, a =
1
4 .
Equating the constant terms gives:
−11a + 12b =−2.
i.e. −11
1
4
+ 12b =−2 from which,
12b =−2 +
11
4
=
3
4
i.e. b =
1
16
Hence the P.I., v = ax + b =
1
4
x +
1
16
(vi) The general solution is given by y = u + v, i.e.
y = Ae
3
2 x + Be
4x
+
1
4
x +
1
16