Math 315 Homework 5 Solutions
√
§ 4.4 #8: Place the equation y = e−t/4 3 cos 4t−sin 4t in the form y = Ae−ct cos(ωt−φ). Then,
on one plot, place the graphs of y = Ae−ct cos(ωt − φ), y = Ae−ct , and y = −Ae−ct . For the last
two, use a different line style and/or color than the first.
√
Solution: Note √a = 3, b = −1, and ω0 = 4.
φ = arctan(−1/ 3) = −π/6. Thus,
Hence, A =
√
a2 + b2 =
√
3 + 1 = 2 and
y(t) = Ae−t/4 cos(ω0 t − φ)
π
.
= 2e−t/4 cos 4t +
6
Using the online program Desmos, we have plotted the curves y1 (t) = 2e−t/4 cos 4t +
2e−t/4 , and y3 (t) = −2e−t/4 , shown below.
π
6
, y2 (t) =
§ 4.4 #12: A 0.1 kg mass is attached to a spring having a spring constant 3.6 kg/s2 . The system is
allowed to come to rest. Then the mass is given a sharp tap, imparting an instantaneous downward
velocity of 0.4 m/s. If there is no damping present, find the amplitude, frequency, and phase of the
resulting motion. Plot the solution.
Solution: We use the equation of motion of a vibrating string
my 00 + µy 0 + ky = F (t),
where m is the mass, µ is the damping constant, k is the spring constant, and F (t) is an external
force. Here m = 0.1 kg, µ = 0, k = 3.6 kg/s2 , and F (t) = 0. Hence, our equation of motion is
0.1y 00 + 3.6y = 0
or, multiplying through by 10,
y 00 + 36y = 0.
Let’s write down our initial conditions. Since the system is allowed to come to rest, y(0) = 0. The
initial velocity is a downward velocity of 0.4 m/s, so y 0 (0) = −0.4. Therefore, the initial value
problem we have to solve is
y 00 + 36y = 0,
y(0) = 0,
y 0 (0) = −0.4.
The characteristic equation is given by
λ2 + 36 = 0.
Thus, λ = ±6i. So the general solution is given by
y(t) = C1 cos(6t) + C2 sin(6t).
Now y(0) = 0 implies
0 = C1 cos(0) + C2 sin(0) = C1 .
Hence C1 = 0. It follows that
y(t) = C2 sin(6t).
Taking the derivative, we get y 0 (t) = 6C2 cos(6t). So y 0 (0) = −0.4 implies
−0.4 = 6C2
⇒
C2 = −
1
0.4
=− .
6
15
Thus, the solution is given by
y(t) = −
which we have plotted below.
1
sin(6t)
15
Now, recall that sin(ωt) = cos(ωt − π/2) (i.e. a sine curve is just a cosine curve shifted π/2 units
to the right). Thus, we can also write the solution as
1
π
y(t) = − cos 6t −
15
2
We get that
A=
1
m,
15
φ=
π
,
2
and ω0 = 6 rad/s
§ 4.5 #20: Find the solution of
y 00 + 2y 0 + 2y = 2 cos 2t
satisfying the initial conditions y(0) = −2 and y 0 (0) = 0.
Solution:
Finding the Homogeneous Solution
Let’s start by finding the general solution to the associated homogeneous equation
y 00 + 2y 0 + 2y = 0.
The characteristic polynomial is given by r2 + 2r + 2 = 0. We use the quadratic formula to get
p
√
−2 ± 4 − 4(1)(2)
−2 ± −4
−2 ± 2i
r=
=
=
= −1 ± i.
2(1)
2
2
Thus, the homogeneous solution is
yh = e−t C1 cos t + C2 sin t .
Finding the Particular Solution
There are two ways to find our particular solution. We can either use the standard method or the
complex method. We’ll go through both methods here.
Method 1: Using the Standard Method
Since the right-hand side of our differential equation is 2 cos 2t, we look for a particular solution in
the form yp (t) = a cos 2t + b sin 2t. Calculating the derivatives of yp (t), we get
yp0 (t) = −2a sin 2t + 2b cos 2t
yp00 (t) = −4a cos 2t − 4b sin 2t
Inserting yp into our equation y 00 + 2y 0 + 2y = 2 cos 2t, we get
2 cos 2t = yp00 (t) + 2yp0 (t) + 2yp
= −4a cos 2t − 4b sin 2t + 2 − 2a sin 2t + 2b cos 2t + 2 a cos 2t + b sin 2t
= −4a cos 2t − 4b sin 2t − 4a sin 2t + 4b cos 2t + 2a cos 2t + 2b sin 2t
= (−2a + 4b) cos 2t + (−2b − 4a) sin 2t.
Equating coefficients gives the system of equations
−2a + 4b = 2
−2b − 4a = 0
From the second equation, we get a = − 21 b. Plugging this into the first equation, we get b + 4b = 2,
2
hence b = 25 . So a = − 12 b = − 10
= − 15 . Therefore, our particular solution is given by
1
2
yp (t) = − cos 2t + sin 2t
5
5
Method 2: Using the Complex Method
One can also use the complex method to find our particular solution. Since the right-hand side
of our differential equation is 2 cos 2t = Re 2e2it , we look for a particular solution to the complex
equation
z 00 + 2z 0 + 2z = 2e2it .
Then the particular solution yp to our equation will be given by yp = Re z. So, let z(t) = ae2it be the
particular solution to the complex equation. Then z 0 (t) = 2iae2it and z 00 (t) = 2i(2ia)e2it = −4ae2it .
We plug z into our complex equation to get
2e2it = z 00 + 2z 0 + 2z
= −4ae2it + 4iae2it + 2ae2it
= (−2a + 4ia)e2it .
Hence, we get that 2 = −2a + 4ia. It follows that
a=
2
1
=
.
−2 + 4i
−1 + 2i
Now we multiply and divide by the conjugate of the denominator to get
a=
1
−1 − 2i
−1 − 2i
1 2
·
=
= − − i.
−1 + 2i −1 − 2i
1+4
5 5
It follows that
z(t) =
−
1 2 2it
− i e .
5 5
Now we need to take the real part of z(t) to find our particular solution yp (t). To do this, we rewrite
z(t) using the identity eiθ = cos(θ) + i sin(θ), so that
1 2 − i cos 2t + i sin 2t
5 5
1
1
2
2
= − cos 2t − i sin 2t − i cos 2t + sin 2t.
5
5
5
5
z(t) =
−
Hence,
1
2
yp (t) = Re z(t) = − cos 2t + sin 2t
5
5
Finding the Solution
Now we combine the homogeneous solution and the particular solution to form the general solution
y(t) = yp (t) + yh (t)
1
2
= − cos 2t + sin 2t + e−t C1 cos t + C2 sin t .
5
5
We use our initial conditions to find C1 and C2 . Note
y(0) = −2
Hence, C1 = −2 +
y 0 (t) =
1
5
⇒
1
−2 = − + C1 .
5
= − 95 . To find C2 , we need to take the derivative of y(t):
2
4
sin 2t + cos 2t − e−t C1 cos t + C2 sin t + e−t − C1 sin t + C2 cos t .
5
5
Hence
y 0 (0) = 0
⇒
⇒
4
− C1 + C2
5
4
4 9
13
C2 = − + C1 = − − = − .
5
5 5
5
0=
Thus, our solution is given by
9
1
2
13
y(t) = − cos 2t + sin 2t + e−t − cos t −
sin t .
5
5
5
5
§ 4.5 #33: Find the particular solution for the differential equation
y 00 + 25y = 2 + 3t + cos 5t.
Solution: First, we find a particular solution y1 (t) to the equation
y 00 + 25y = 2 + 3t.
Since the right-hand side of this equation is 2 + 3t, we let our particular solution be of the form
y1 (t) = a + bt.
Then y10 (t) = b and y100 (t) = 0. Plugging y1 (t) into our equation gives
2 + 3t = y100 (t) + 25y1 (t) = 0 + 25(a + bt) = 25a + 25bt
Equation coefficients gives 25a = 2 and 25b = 3. Hence, we get that a =
y1 (t) =
2
25
and b =
3
25 .
Hence,
3
2
+ t.
25 25
Now, we find a particular solution y2 (t) to the equation
y 00 + 25y = cos 5t.
We use the complex method (although the standard method can be used as well). Since cos 5t =
Re e5it , we consider the complex equation
z 00 + 25z = e5it .
We consider a solution of the form z(t) = ae5it . Then z 0 (t) = 5iae5it and z 00 (t) = −25ae5it .
Plugging these in to the complex equation, we get
e5it = z 00 (t) + 25z(t)
= −25ae5it + 25ae5it = 0
which is inconsistent. Therefore, we must try a new solution of the form z(t) = ate5it (we multiply
the original solution we tried by t). Then z 0 (t) = ae5it + 5iate5it and
z 00 (t) = 5iae5it + 5iae5it − 25ate5it = 10iae5it − 25ate5it .
Plugging these values into our complex equation gives
e5it = z 00 (t) + 25z(t)
= 10iae5it − 25ate5it + 25ate5it
= 10iae5it .
Hence, we get that 1 = 10ia. So
a=
1
1
=− i
10i
10
(which you can get by multiplying and dividing by i, or using the fact that 1/i = −i). Hence, the
solution to our complex equation is
1
ite5it
10
1 = − it cos 5t + i sin 5t
10
1
1
= − it cos 5t + t sin 5t.
10
10
z(t) = −
(Note we used the identity eiθ = cos θ + i sin θ.) Therefore, the particular solution y2 (t) to the
equation
y 00 + 25y = cos 5t
is given by
y2 (t) = Re z(t) =
1
t sin 5t.
10
Note we took the real part of z(t) since cos 5t = Re e5it . Now, the particular solution yp (t) to our
original differential equation
y 00 + 25y = 2 + 3t + cos 5t.
is given by
yp (t) = y1 (t) + y2 (t) =
(If this is unclear, see Theorem 5.22 on p. 170.)
2
3
1
+ t + t sin 5t.
25 25
10