4First-order differential equations
IA Differential Equations
4.3 Forced (inhomogeneous) equations
Recall that a homogeneous equation is an equation like 5
y
0
−
3
y
= 0, with no
x
or constant terms floating around. A forced, or inhomogeneous, equation is
one that is not homogeneous. For example, 5
y
0
−
3
y
= 10 or 5
y
0
−
3
y
=
x
2
are
forced equations. We call the “extra” terms 10 and x
2
the forcing terms.
4.3.1 Constant forcing
Example. Consider 5
y
0
−
3
y
= 10. We can spot that there is a equilibrium
(constant) solution y = y
p
= −
10
3
with y
0
p
= 0.
The particular solution
y
p
is a solution of the ODE. Now suppose the general
solution is
y
=
y
p
+
y
c
. We see that 5
y
0
c
−
3
y
c
= 0. So
y
c
satisfies the homogeneous
equation we already solved, and
y = −
10
3
+ Ae
3x/5
.
Note that any boundary conditions to determine
A
must be applied to the full
solution y and not the complementary function y
c
.
This is the general method of solving forced equations. We first find one
particular solution to the problem, often via educated guesses. Then we solve
the homogeneous equation to get a general complementary solution. Then the
general solution to the full equation is the sum of the particular solution and
the general complementary solution.
4.3.2 Eigenfunction forcing
This is the case when the forcing term is an eigenfunction of the differential
operator.
Example. In a radioactive rock, isotope A decays into isotope B at a rate
proportional to the number
a
of remaining nuclei A, and B also decays at a rate
proportional to the number b of remaining nuclei B. Determine b(t).
We have
da
dt
= −k
a
a
db
dt
= k
a
a − k
b
b.
Solving the first equation, we obtain a = a
0
e
−k
a
t
. Then we have
db
dt
+ k
b
b = k
a
a
0
e
−k
a
t
.
We usually put the variables involving
b
on the left hand side and the others on
the right. The right-hand term k
a
a
0
e
−k
a
t
is the forcing term.
Note that the forcing term is an eigenfunction of the differential operator
on the LHS. So that suggests that we can try a particular integral
b
p
=
Ce
−k
a
t
.
Substituting it in, we obtain
−k
a
C + k
b
C = k
a
a
0
C =
k
a
k
b
− k
a
a
0
.
Then write
b
=
b
p
+
b
c
. We get
b
0
c
+
k
b
b
c
= 0 and
b
c
=
De
−k
b
t
. All together, we
have the general solution
b =
k
a
k
b
− k
a
a
0
e
−k
a
t
+ De
−k
b
t
.
Assume the following boundary condition:
b
= 0 when
t
= 0, in which case we
can find
b =
k
a
k
b
− k
a
a
0
e
−k
a
t
− e
−k
b
t
.
x
y
O
a
b
The isotope ratio is
b
a
=
k
a
k
b
− k
a
h
1 − e
(k
a
−k
b
)t
i
.
So given the current ratio
b/a
, with laboratory determined rates
k
a
and
k
b
, we
can determine the value of t, i.e. the age of the rock.