MTH 264
DELTA COLLEGE
SECTION 4.2 18
(a) Find the general solution of
y 00 + 4y 0 + 20y = 3 + 2 cos 2t
(b) Discuss the long-term behavior of solutions of this equation.
Solution:
The four basic components of this problem are:
(1)
(2)
(3)
(4)
y 00 + 4y 0 + 20y
y 00 + 4y 0 + 20y
y 00 + 4y 0 + 20y
y 00 + 4y 0 + 20y
=
=
=
=
0
the homogeneous problem . . .
3
. . . with a constant driving function . . .
2 cos 2t . . . and a periodic driving function
2e(2i)t the complexification of (***)
The roots of the characteristic equation s2 + 4s + 20 = 0 are −2 ± 4i, so the solution of the homogeneous
problem is yh (t) = k1 e−2t cos 4t + k2 e−2t sin 4t.
Since y2 (t) = 3/20 is clearly a solution to (2), y2 will be one part of the particular solution yp of our problem.
The 2 cos 2t portion of the driving function in (3), is the real part of the complex driving function 2e(2i)t in
(4). That means that the real part of a particular solution y4 for (4) will be a particular solution y3 for (3). We
set our guess for a particular solution of (4) at y4 (t) = Ae(2i)t . Taking derivatives and inserting y4 (t) into (4) we
find that
20 y4 (t)
+4 y40 (t)
+1 y400 (t)
2e(2i)t
Therefore A =
= 20(
= 4(
= 1(
(1 + 0i)Ae(2i)t
(0 + 2i)Ae(2i)t
(−4 + 0i)Ae(2i)t
=
(16 + 8i)Ae(2i)t
)
)
)
1
8 − 4i
1
1
2
=
=
=
− i, and the particular solution for (4) is y4 =
16 + 8i
8 + 4i
80
10 20
1
1
− i e(2i)t .
10 20
Expanding y4 and extracting the real part yields;
1
1
1
1
1
1
1
1
(2i)t
y4 =
− i e
=
− i (cos 2t + i sin 2t) =
cos 2t +
sin 2t + i − cos 2t +
sin 2t
10 20
10 20
10
20
20
10
y3 =
1
1
cos 2t +
sin 2t
10
20
(a) Therefore the general solution of the original
problem is
y(t) = yh (t) + yp (t) = yh (t) + y2 (t) + y3 (t)
y(t) = k1 e−2t cos 4t + k2 e−2t sin 4t +
1
1
3
+
cos 2t +
sin 2t
20 10
20
(b) This is an underdamped harmonic oscillator with
a periodic driving function.
On the top right we have graphed the solution in blue,
the yh in red, and the yp in black for the solution with
initial conditions of y(0) = 0 and y 0 (0) = 0. This graph
illustrates that as t increases, yh → 0 and y → yp .
On the bottom right we have graphed the solution y
in blue and the driving function 3 + 2 cos 2t in black. This
graph illustrates how the solution “lags” the driving
function.