differential equation

482 DIFFERENTIAL EQUATIONS
Table 51.1
Form of particular integral for different functions
Type Straightforward cases ‘Snag’ cases See
Try as particular integral: Try as particular integral: problem
(a) f (x) = a constant v = k v = kx (used when C.F. 1, 2
contains a constant)
(b) f (x) = polynomial (i.e. v = a + bx + cx 2 + ··· 3
f (x) = L + Mx + Nx 2 + ···
where any of the coefficients
may be zero)
(c) f (x) = an exponential function v = ke ax (i) v = kxe ax (used when e ax 4, 5
(i.e. f (x) = Ae ax )
appears in the C.F.)
(ii) v = kx 2 e ax (used when e ax 6
and xe ax both appear in
the C.F.)
(d) f (x) = a sine or cosine function v = A sin px + B cos px v = x(A sin px + B cos px) 7, 8
(i.e. f (x) = a sin px + b cos px,
(used when sin px and/or
where a or b may be zero)
cos px appears in the C.F.)
(e) f (x) = a sum e.g. 9
(i) f (x) = 4x 2 − 3 sin 2x (i) v = ax 2 + bx + c
+ d sin 2x + e cos 2x
(ii) f (x) = 2 − x + e 3x
(ii) v = ax + b + ce 3x
(f) f (x) = a product e.g. v = e x (A sin 2x + B cos 2x) 10
f (x) = 2e x cos 2x
51.3 Worked problems on differential
equations of the form
a d2 y
dx 2 + b dy + cy = f (x) where
dx
f (x) is a constant or polynomial
Problem 1. Solve the differential equation
d 2 y
dx 2 + dy − 2y = 4.
dx
Using the procedure of Section 51.2:
(i) d2 y
dx 2 + dy − 2y = 4 in D-operator form is
dx
(D 2 + D − 2)y = 4.
(ii) Substituting m for D gives the auxiliary equation
m 2 + m − 2 = 0. Factorising gives: (m − 1)
(m + 2) = 0, from which m = 1orm =−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 d2 y
dx 2 − 3dy = 9, given the
dx
boundary conditions that when x = 0, y = 0 and
dy
dx = 0.
Using the procedure of Section 51.2:
(i) d2 y
dx 2 − 3 dy = 9 in D-operator form is
dx
(D 2 − 3D)y = 9.