differential equation

476 DIFFERENTIAL EQUATIONS
(c) If the roots of the auxiliary equation are:
(i) real and different, say m = α and m = β,
then the general solution is
y = Ae αx + Be βx
(ii) real and equal, say m = α twice, then the
general solution is
y = (Ax + B)e αx
(iii) complex, say m = α ± jβ, then the general
solution is
y = e αx {A cos βx + B sin βx}
(d) Given boundary conditions, constants A and B,
may be determined and the particular solution
of the differential equation obtained.
The particular solutions obtained in the worked
problems of Section 50.3 may each be verified by
substituting expressions for y, dy
dx and d2 y
into the
dx2 original equation.
50.3 Worked problems on differential
equations of the form
a d2 y
dx 2 + bdy dx + cy = 0
Problem 1. Determine the general solution
of 2 d2 y
dx 2 + 5dy − 3y = 0. Find also the particular
solution given that when x = 0, y = 4 and
dx
dy
dx = 9.
Using the above procedure:
(a) 2 d2 y
dx 2 + 5dy − 3y = 0 in D-operator form is
dx
(2D 2 + 5D − 3)y = 0, where D ≡ d dx
(b) Substituting m for D gives the auxiliary equation
2m 2 + 5m − 3 = 0.
Factorising gives: (2m − 1)(m + 3) = 0, from
which, m =
2 1 or m =−3.
(c) Since the roots are real and different the general
solution is y = Ae 2 1 x + Be −3x .
(d) When x = 0, y = 4,
hence 4 = A + B (1)
Since
then
When
y = Ae 1 2 x + Be −3x
dy
dx = 1 2 Ae 1 2 x − 3Be −3x
x = 0, dy
dx = 9
thus 9 = 1 A − 3B (2)
2
Solving the simultaneous equations (1) and (2)
gives A = 6 and B =−2.
Hence the particular solution is
y = 6e 1 2 x − 2e −3x
Problem 2. Find the general solution of
9 d2 y
dt 2 − 24dy + 16y = 0 and also the particular
dt
solution given the boundary conditions that
when t = 0, y = dy
dt = 3.
Using the procedure of Section 50.2:
(a) 9 d2 y
dt 2 − 24dy + 16y = 0 in D-operator form is
dt
(9D 2 − 24D + 16)y = 0 where D ≡ d dt
(b) Substituting m for D gives the auxiliary equation
9m 2 − 24m + 16 = 0.
Factorising gives: (3m − 4)(3m − 4) = 0, i.e.
m = 4 3 twice.
(c) Since the roots are real and equal, the general
solution is y = (At + B)e 4 3 t .
(d) When t = 0, y = 3 hence 3 = (0 + B)e 0 , i.e. B = 3.
Since y = (At + B)e 4 3 t
then dy ( ) 4
dt = (At + B) 3 e 4 3 t + Ae 4 3 t , by the
product rule.