4.5 The Superposition Principle and Undetermined Coefficients Revisited
Theorem 3 (Superposition Principle). If y
1
is a solution to the equation
ay
00
+ by
0
+ cy = f
1
(t),
and y
2
is a solution to
ay
00
+ by
0
+ cy = f
2
(t),
then for any constants k
1
and k
2
, the function k
1
y
1
+ k
2
y
2
is a solution to the differential equation
ay
00
+ by
0
+ cy = k
1
f
1
(t) + k
2
f
2
(t).
Proof.
ay
00
+ by
0
+ cy = a(k
1
y
1
+ k
2
y
2
)
00
+ b(k
1
y
1
+ k
2
y
2
)
0
+ c(k
1
y
1
+ k
2
y
2
)
= k
1
(ay
00
1
+ by
0
1
+ cy
1
) + k
2
(ay
00
2
+ by
0
2
+ cy
2
)
= k
1
f
1
(t) + k
2
f
2
(t).
Example 1. Given that y
1
(t) = (1/4) sin 2t is a solution to y
00
+ 2y
0
+ 4y = cos 2t and that y
2
(t) = t/4 − 1/8
is a solution to y
00
+ 2y
0
+ 4y = t, find solutions to the following:
a) y
00
+ 2y
0
+ 4y = t + cos 2t.
b) y
00
+ 2y
0
+ 4y = 2t − 3 cos 2t.
c) y
00
+ 2y
0
+ 4y = 11t − 12 cos 2t.
Thus, by superposition principle, the general solution to a nonhomogeneous equation is the sum of the
general solution to the homogeneous equation and one particular solution. That is, if the general solution
to ay
00
+ by
0
+ cy = 0 is c
1
y
1
(t) + c
2
y
2
(t), and if a particular solution to ay
00
+ by
0
+ cy = f(t) is y
p
(t), then
the general solution to ay
00
+ by
0
+ cy = f(t) is c
1
y
1
(t) + c
2
y
2
(t) + y
p
(t).
Theorem 4 (Existence and Uniqueness: Nonhomogeneous Case). For any real numbers a, b, c, t
0
, y
0
, y
1
,
with a 6= 0, suppose y
p
(t) is a particular solution to
ay
00
+ by
0
+ cy = f(t) (1)
in an interval I containing t
0
, and that y
1
(t), y
2
(t) are linearly independent solutions to the associated
homogeneous equation
ay
00
+ by
0
+ cy = 0 (2)
in I. Then there exists a unique solution in I to the initial value problem
ay
00
+ by
0
+ cy = f(t), y(t
0
) = y
0
, y
0
(t
0
) = y
1
, (3)
and it is given by
y(t) = y
p
(t) + c
1
y
1
(t) + c
2
y
2
(t), (4)
for the appropriate choice of the constants c
1
, c
2
.
Example 2. Find a general solution to y
00
= 2y
0
− y + 2e
x
, if a particular solution is y
p
(x) = x
2
e
x
.
1
To find a particular solution to the differential equation
ay
00
+ by
0
+ cy = P
m
(t)e
rt
,
where P
m
(T ) is a polynomial of degree m, use the form
y
p
(t) = t
s
(A
m
t
m
+ · · · + A
1
t + A
0
)e
rt
; (5)
if r is not a root of the associated auxiliary equation, take s = 0; if r is a simple root of the associated
auxiliary equation, take s = 1; and if r is a double root of the associated auxiliary equation, take s = 2.
To find a particular solution to the differential equation
ay
00
+ by
0
+ cy = P
m
(t)e
αt
cos βt + Q
n
(t)e
αt
sin βt, β 6= 0,
where P
m
(t) is a polynomial of degree m and Q
n
(t) is a polynomial of degree n, use the form
y
p
(t) = t
s
(A
k
t
k
+ · · · + A
1
t + A
0
)e
αt
cos βt + t
s
(B
k
t
k
+ · · · + B
1
t + B
0
)e
αt
sin βt, (6)
where k is the larger of m and n. If α + iβ is not a root of the associated auxiliary equation, take s = 0; if
α + iβ is a root of the associated auxiliary equation, take s = 1.
Example 3. Decide whether the method of undetermined coefficients together with superposition principle
can be applied to find a particular solution of the following equation. Do not solve the equation.
2y
00
− y
0
+ 6y = t
2
e
−t
sin t − 8t cos 3t + 10
t
.
Example 4. Find a general solution to y
00
(x) + 6y
0
(x) + 10y(x) = 10x
4
+ 24x
3
+ 2x
2
− 12x + 18.
Example 5. Find the solution to the initial value problem:
y
00
+ y
0
− 12y = e
t
+ e
2t
− 1; y(0) = 1, y
0
(0) = 3.
Example 6. Determine the form of a particular solution for the differential equation. Do not solve.
y
00
+ 5y
0
+ 6y = sin t − cos 2t.
2