MATH 215/255
Fall 2015
Assignment 4
Due date: October 23, 2015
1. A small object of mass 1 kg is attached to a spring with spring constant 2 N/m. This
spring-mass system is immersed in a viscous medium with damping constant 3 N· s/m.
At time t = 0, the mass is lowered 1/2 m below its equilibrium position, and released.
Show that the mass will creep back to its equilibrium position as t approaches infinity.
Answer.
Let x(t) be the position of the mass away from the equilibrium position
at time t, then our differential equation is:
x00 + 3x0 + 2x = 0,
x(0) = 0.5,
x0 (0) = 0.
The characteristic equation of x00 + 3x0 + 2x = 0 is r2 + 3r + 2 = 0, then r1 = −1 and
r2 = −2, which implies that the general solution to x00 + 3x0 + 2x = 0 is:
x(t) = C1 e−t + C2 e−2t .
Since x(0) = 0.5, then C1 + C2 = 0.5. Since x0 (t) = −C1 e−t − 2C2 e−2t and x0 (0) = 0,
then −C1 − 2C2 = 0. So we get
C1 = 1,
C2 = −0.5.
Hence x(t) = e−t − 0.5e−2t . In particular, we have x(t) → 0 as t → ∞, which means
that the mass will creep back to its equilibrium position as t approaches infinity.
2. A small object of mass 1 kg is attached to a spring with spring-constant 1 N/m and is
immersed in a viscous medium with damping constant 2 N · s/m. At time t = 0, the
mass is lowered 1/4 m and given an initial velocity of 1 m/s in the upward direction.
Show that the mass will overshoot its equilibrium position once, and then creep back
to equilibrium.
Answer.
Let x(t) be the position of the mass away from the equilibrium position
at time t, then our differential becomes:
x00 + 2x0 + x = 0,
x(0) = −1/4,
x0 (0) = 1.
The characteristic equation of x00 + 2x0 + x = 0 is r2 + 2r + 1 = 0, then r1 = r2 = −1,
which implies that the general solution to x00 + 2x0 + x = 0 is:
x(t) = C1 e−t + C2 te−t .
Since x(0) = 1/4, then C1 = −1/4. Since x0 (t) = −C1 e−t + C2 e−t − C2 te−t and
x0 (0) = 1, then −C1 + C2 = 1. So we get C2 = 3/4. Hence
1
3
1
x(t) = − e−t + te−t = e−4 [3t − 1].
4
4
4
So at time t = 1/3, we know that x(1/3) = 0, which means that the mass will
overshoot its equilibrium position once. On the other hand, we know that x(t) → 0
as t → ∞, which means that the mass will creep back to equilibrium as t approaches
infinity.
3. Using the mks units (meters-kilograms-seconds), suppose you have a spring with
spring constant 4 N/m. You want to use it to weigh items. Assume no friction.
You place the mass on the spring and put it in motion. a) You count and find that
the frequency is 0.8 Hz (cycles per second). What is the mass? b) Find a formula for
the mass m given the frequency ω in Hz.
q
q
k
4
Answer. Since the motion is undamped, then the requency is ω0 = m
= m
. a)
q
4
, which implies that m = 1.642 π2 . b) In general,
If ω0 = 0.8π = 1.6π, then 1.6π = m
q
4
we know that ω · 2π = m
, then m = 4ω42 π2 = π21ω2 .
4. Find the general solution to y 00 + 4y = x3 − 1.
Answer. The characterisitc equation of y 00 + 4y = 0 is r2 + 4 = 0, then r1 = 2i and
r2 = −2i, which implies that the general solution to y 00 + 4y = 0 is:
yc (x) = C1 cos(2x) + C2 sin(2x).
Let yp (x) = ax3 + bx2 + cx + d be a particular solution to y 00 + 4y = x3 − 1, then
yp0 = 3ax2 + 2bx + c,
yp00 = 6ax + 2b.
So we get
yp00 + 4yp = 6ax + 2b + 4(ax3 + bx2 + cx + d)
= 4ax3 + 4bx2 + (6a + 4c)x + 4d + 2b
= x3 − 1.
So we have
4a = 1,
4b = 0,
6a + 4c = 0,
4d + 2b = −1.
Hence
a = 1/4,
c = −3/8,
b = 0,
d = −1/4.
So we get
1
3
1
yp (x) = x3 − x − .
4
8
4
Therefore, the general solution to y 00 + 4y = x3 − 1 is:
1
3
1
y(x) = yc (x) + yp (x) = C1 cos(2x) + C2 sin(2x) + x3 − x − .
4
8
4
5. Solve the initial value problem y 00 + 9y = cos(3x) for y(0) = 0 and y 0 (0) = 1.
Answer. The characteristic equation of y 00 + 9y = 0 is r2 + 9 = 0, then r1 = 3i and
r2 = −3i, which implies that the general solution to y 00 + 9y = 0 is:
yc (x) = C1 cos(3x) + C2 sin(3x).
2
Let yp (x) = x[a cos(3x) + b sin(3x)] be a particular solution to y 00 + 9y = cos(3x), then
yp0 (x) = a cos(3x) + b sin(3x) + x[−3a sin(3x) + 3b cos(3x)]
yp00 (x) = −3a sin(3x) + 3b cos(3x) − 3a sin(3x) + 3b cos(3x) + x[−9a cos(3x) − 9b sin(3x)]
= −6a sin(3x) + 6b cos(3x) + x[−9a cos(3x) − 9b sin(3x)].
So we get
yp00 + 9yp = −6a sin(3x) + 6b cos(3x) + x[−9a cos(3x) − 9b sin(3x)] + 9x[a cos(3x) + b sin(3x)]
= −6a sin(3x) + 6b cos(3x)
= cos(3x).
So −6a = 0 and 6b = 1, that is, a = 0 and b = 1/6. So
1
yp (x) = x sin(3x).
6
Hence the general solution to y 00 + 9y = cos(3x) is:
1
y(x) = yc (x) + yp (x) = C1 cos(3x) + C2 sin(3x) + x sin(3x).
6
Since y(0) = 0, then C1 = 0, which implies that y(x) = C2 sin(3x) + 61 x sin(3x). Since
y 0 (x) = 3C2 cos(3x) + 36 x cos(3x) + 16 sin(3x) and y 0 (0) = 1, then 3C2 = 1, that is,
C2 = 1/3. So we know that
y(x) =
1
1
sin(3x) + x sin(3x).
3
6
6. Let k be a real constant, find a particular solution to y 00 − 2y 0 + y = ekx . (Hint: You
should consider two cases: k = 1 or k 6= 1).
Answer.
The characteristic equation of y 00 − 2y 0 + y = 0 is r2 − 2r + 1 = 0, then
r1 = r2 = 1.
• When k = 1. Let yp (x) = ax2 ex be a particular solution to y 00 − 2y 0 + y = ekx ,
then
yp0 (x) = 2axex +ax2 ex ,
yp00 (x) = 2aex +2axex +2axex +ax2 ex = 2aex +4axee +ax2 ex .
So we have
yp00 − 2yp0 + yp = 2aex + 4axee + ax2 ex − 2[2axex + ax2 ex ] + ax2 ex
= 2aex
= ex .
So 2a = 1, that is, a = 1/2. Hence
1
yp (x) = x2 ex .
2
3
• When k 6= 1, let yp (x) = aekx be a particular solution to y 00 − 2y 0 + y = ekx , then
yp0 (x) = akekx ,
yp00 (x) = ak 2 ekx .
So we get
yp00 − 2yp0 + yp = ak 2 ekx − 2akekx + akekx
= aekx [k 2 − 2k + 1]
= ekx .
So a[k 2 − 2k + 1] = 1, that is a = (k − 1)−2 . So
yp (x) = (k − 1)−2 ekx .
7. Let k be a real constant, find a particular solution to y 00 + y = cos(kx). (Hint: You
should consider two cases: k = ±1 or k 6= ±1).
Answer.
r2 = −i.
The characteristic equation of y 00 + y = 0 is r2 + 1 = 0, then r1 = i and
• When k = ±1, let yp (x) = x[a cos(x) + b sin(x)] be a particular solution to
y 00 + y = cos(kx) = cos(x), then
yp0 = a cos(x) + b sin(x) + x[−a sin(x) + b cos(x)]
yp00 = −a sin(x) + b cos(x) − a sin(x) + b cos(x) + x[−a cos(x) − b sin(x)]
= −2a sin(x) + 2b cos(x) + x[−a cos(x) − b sin(x)].
So
yp00 + yp = −2a sin(x) + 2b cos(x) + x[−a cos(x) − b sin(x)] + x[a cos(x) + b sin(x)]
= −2a sin(x) + 2b cos(x)
= cos(x).
Then −2a = 0 and 2b = 1, then a = 0 and b = 1/2. Hence
1
yp (x) = x sin(x).
2
• When k 6= ±1, let yp = a cos(kx) + b sin(kx) be a particular solution to y 00 + ky =
cos(kx), then
yp0 = −ak sin(kx) + bk cos(kx),
yp00 = −ak 2 cos(kx) − bk 2 sin(kx).
So we get
yp00 + yp = −ak 2 cos(kx) + bk 2 sin(kx) + a cos(kx) + b sin(kx)
= a(1 − k 2 ) cos(kx) + b(1 − k 2 ) sin(kx)
= cos(kx).
Then a(1 − k 2 ) = 1 and b(1 − k 2 ) = 0, then a =
yp (x) =
4
1
1−k2
1
cos(kx).
1 − k2
and b = 0. Hence
8. Find a form of a particular solution to y 00 + 4y = x sin(2x) (You do not need to find
the coefficients).
Answer. The characteristic equation of y 00 + 4y = 0 is r2 + 4 = 0, then r1 = 2i and
r2 = −2i. Let yp (x) = x · [(ax + b) cos(2x) + (cx + d) sin(2x)] be a particular solution
to y 00 + 4y = x sin(2x). Indeed, one can use wolfram to get yp (x) = − 81 x2 cos(2x) +
1
16 x sin(2x), that is, a = −1/8, b = c = 0 and d = 1/16.
9. Find the general solution to y 00 + 2y 0 = 1 + e−2x .
Answer. The characterisitc equation of y 00 + 2y 0 = 0 is r2 + 2r = 0, then r1 = 0 and
r2 = −2, which implies that the general solution to y 00 + 2y 0 = 0 is:
yc (x) = C1 + C2 e−2x .
Let yp (x) = ax + bxe−2x be a particular solution to y 00 + 2y 0 = 1 + e−2x , then
yp0 = a + be−2x − 2bxe−2x ,
yp00 = −2be−2x − 2be−2x + 4bxe−2x = −4be−2x + 4bxe−2x .
Then
yp00 + 2yp0 = −4be−2x + 4bxe−2x + 2[a + be−2x − 2bxe−2x ]
= −2be−2x + 2a
= 1 + e−2x
Then −2b = 1 and 2a = 1, that is, a = 1/2 and b = −1/2. Hence
x 1
yp (x) = − xe−2x .
2 2
So the general solution to y 00 + 2y 0 = 1 + e−2x is:
y(x) = yc (x) + yp (x) = C1 + C2 e−2x +
x 1 −2x
− xe .
2 2
10. Find a particular solution to y 00 − 4y = xe2x .
Answer.
The characteristic equation of y 00 − 4y = 0 is r2 − 4 = 0, then r1 = 2
and r2 = −2. Let yp (x) = x(ax + b)e2x = (ax2 + bx)e2x be a particular solution to
y 00 − 4y = xe2x , then
yp0 = (2ax + b)e2x + 2(ax2 + bx)e2x = e2x [2ax2 + (2a + 2b)x + b]
yp00 = 2ae2x + 2(2ax + b)e2x + 2(2ax + b)e2x + 4(ax2 + bx)e2x
= e2x [2a + 4ax + 2b + 4ax + 2b + 4ax2 + 4bx)
= e2x [4ax2 + (8a + 4b)x + 2a + 4b].
So
yp00 − 4yp = e2x [4ax2 + (8a + 4b)x + 2a + 4b] − 4(ax2 + bx)e2x
= e2x [8x + 2a + 4b]
= xe2x .
So 8a = 1 and 2a + 4b = 0, which implies that a = 1/8 and b = −1/16. Hence
1
yp (x) = 18 x2 e2x − 16
xe2x .
5