MTH 264
DELTA COLLEGE
SECTION 2.3 8
Consider any damped harmonic oscillator equation
(1)
m
dy
d2 y
+b
+ ky = 0
dt2
dt
(z) (from the previous exercise) Show that a constant multiple of any solution to (1) is another solution to (1).
(a) Show that the sum of any solutions to (1) is another solution to (1).
(a’) (additionally) Show that any linear combination of solutions to (1) is another solution to (1).
(b) Using the results above and the “guess-and-test” method described in this section, solve the initial-value
problem
(2)
d2 y
dy
+3
+ 2y = 0,
dt2
dt
y(0) = 2,
v(0) = −3
(c) Using the results above and the “guess-and-test” method described in this section, solve the initial-value
problem
(3)
dy
d2 y
+ 2y = 0,
+3
2
dt
dt
y(0) = 3,
v(0) = −5
(d) How many solutions to the equation
d2 y
dy
+3
+ 2y = 0
dt2
dt
(4)
do you get if you use the results above and the “guess-and-test” method described in this section?
Solution:
(z) Assume that y1 is a solution to (1) and k1 is any constant. Since y1 is a solution to (1), the sum of the
d2 y1 dy1
, and ky1 , is 0:
three terms m 2 , b
dt
dt
(5)
m
d 2 y1
dy1
+b
+ ky1 = 0
dt2
dt
Now consider a constant multiple k1 y1 of this solution. Since differentiation is linear,
d
dy1
(k1 y1 ) = k1
and
dt
dt
d2
d 2 y1
d2
d
(k1 y1 ) = k1 2 , so when we sum m 2 (k1 y1 ), b (k1 y1 ), and k(k1 y1 ), we still get 0 thanks to equation (5):
2
dt
dt
dt
dt
d2
d
m 2 (k1 y1 ) + b (k1 y1 ) + k(k1 y1 )
dt
dt
d 2 y1
dy1
= m k1 2 + b k1
+ k (k1 y1 )
dt
dt
d 2 y1
dy1
= k1 m 2 + k1 b
+ k1 (ky1 )
dt
dt
d 2 y1
dy1
= k1 m 2 + b
+ ky1 = k1 (0) = 0
dt
dt
So we have shown that k1 y1 is a solution to (*).
(a) In a similar way we can show that the sum of any two solutions to (1) is also a solution to (1). Assume
that y1 and y2 are solutions to (1). That means that
(6)
m
dy1
d 2 y1
+b
+ ky1 = 0
2
dt
dt
(7)
m
dy2
d 2 y2
+b
+ ky2 = 0
2
dt
dt
Now consider the sum y1 + y2 of these two solutions. Since differentiation is linear,
and
d
dy1
dy2
(y1 + y2 ) =
+
dt
dt
dt
d2 y1
d2 y2
d2
(y1 + y2 ) =
+
, so y1 + y2 is also a solution thanks to equations (6) and (7):
2
2
dt
dt
dt2
d
d2
m 2 (y1 + y2 ) + b (y1 + y2 ) + k(y1 + y2 )
dt
dt
d2 y2
dy1
d2 y1
dy2
=
m 2 +m 2 + b
+b
+ (ky1 + ky2 )
dt
dt
dt
dt
dy1
dy2
d2 y1
d 2 y2
=
m 2 +b
+ ky1 + m 2 + b
+ ky2
dt
dt
dt
dt
= 0+0=0
So we have shown that y1 + y2 is a solution to (*).
(a’) Putting (z) and (a) together we actually have something very strong.
If y1 and y2 are two solutions to equation (1), and k1 and k2 are any two constants, then the linear combination k1 y1 + k2 y2 is also a solution to equation (1).
This is true because the constant multiples k1 y1 and k2 y2 are solutions to (1) as we showed in (z), and then
their sum is a solution to (1) as we showed in (a).
(b) Now to use the “guess-and-test” method described in this section, we will assume that y = est is a solution
to (2) and determine the conditions on s for which this is true. If y = est , then
d
d2 st
(e ) + 3 (est ) + 2(est )
dt2
dt
= s2 est + 3sest + 2est
=
(s2 + 3s + 2)est
=
(s + 2)(s + 1)est
Since est is never equal to 0, no matter what the values of s or t, the only way for this sum to be 0 and for
y = est to satisfy the differential equation is that (s + 2)(s + 1) = 0, which happens if s = −2 or s = −1. So
y1 = e−2t and y2 = e−t are solutions to the differential equation in (2).
By our observation in (a’), this means that y(t) = k1 e−2t + k2 e−t solves the differential equation in (2) for any
constants k1 and k2 . We only need to find the values of k1 and k2 that will create the initial conditions required
in the initial-value problem (2).
Differentiating y(t) = k1 e−2t + k2 e−t yields y 0 (t) = −2k1 e−2t − k2 e−t . Substituting t = 0 into these last two
equations yields:
y(0)
= k1 e−2(0) + k2 e−(0) = k1 + k2
y 0 (0)
= −2k1 e−2(0) − k2 e−(0) = −2k1 − k2
This gives us two equations in two unknowns according to the initial conditions in (2):
2
=
k1 + k2
−3
=
−2k1 − k2
Solve this by the method of your choice, say the elimination method, and you have k1 = 1 and k2 = 1. So
the function that solves the differential equation in (2) and meets the initial conditions in (2), i.e. the function
that solves the initial-value problem (2), is:
y(t) = e−2t + e−t
(c) Now we see that this technique is quite mechanical and can be repeated fairly effortlessly. To solve the
initial-value problem (3) we need only change the initial values from y(0) = 2 and y 0 (0) = −3 to y(0) = 3 and
y 0 (0) = −5.
3
=
k1 + k2
−5
=
−2k1 − k2
The solution to this system is k1 = 2 and k2 = 1. So the solution to the initial-value problem (3) is:
y(t) = 2e−2t + e−t
(d) This is the exciting and efficient answer:
Infinitely many solutions. For each pair of initial conditions y(0) = y0 and v(0) = v0 , we can generate a
solution to (4) that satisfies the specified initial conditions.
We will learn soon that this is the case for every equation that has the form of (4).