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MAT 275 Test 2 SOLUTIONS, FORM A
For the following multiple-choice problems below, write the letter of the correct answer on the blank to the
right of the problem. Make sure your answer is clearly legible, or it will be counted wrong. There is no
partial credit for these problems. Each is worth 5 points.
1. The solution to a certain differential equation is
y(t) = 4e−3t sin(2t) + 3e−3t cos(2t) + 5 sin(4t) − 2 cos(4t).
The steady-state solution is
D.
5 sin(4t) − 2 cos(4t)
Solution: These are the terms which do not converge to zero when t approaches ∞.
2. The form for a particular solution to the differential equation y 00 − 4y 0 + 4y = te2t is
D.
At3 e2t + Bt2 e2t
Solution: The presence of te2t on the right-hand side suggests a form of Ae2t + Bte2t . However, e2t
and te2t are solutions to the homogeneous equation, so you need to multiply Ae2t + Bte2t by t2 to
get a useable form.
3. The vibration problem 3y 00 + 6y 0 + 8y = 0 is
B.
underdamped.
√
Solution: γ = 6 and
4km =
√
√
72. Since γ <
4km, the problem is underdamped.
4. In the differential equation y 00 + 9y = 5 sin(at), resonance will
A.
occur when a = 3.
Solution: The homogeneous solution is y = C1 sin(3t) + C2 cos(3t). In order for resonance to occur,
the function A sin(at) + B sin(bt) must be a solution to the homogeneous solution; hence a = 3.
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5. [10 points] Find the form of the solution to differential equation
y (v) + 5y 0000 + 15y 000 + 23y 00 − 8y 0 − 36y = 0
(The characteristic equation for y = erx factors into (r + 2)2 (r − 1)(r2 + 2r + 9).)
Solution: The roots of the characteristic equation are −2 (with multiplicity 2), 1, and
−2 ±
√
√
22 − 4 · 9
= −1 ± 8 i,
2
so the form of the solution is
y(x) = C1 e−2x + C2 xe−2x + C3 ex + C4 e−x sin
√
√ 8 x + C5 e−x cos
8x .
(The variable t could have been used instead of x, even though it was stated that y = erx .)
Graded on a 0 − 3 − 5 − 7 − 10-point basis.
6. [15 points] A mass of 25 kilograms is attached to a spring whose constant is 30 N/m. The mass is started
in motion from the equilibrium position with an initial velocity of 250 cm/sec in the upward direction.
Assume there is no air resistance. Set up the initial value problem governing the position y(t) of the
mass, assuming that the position is measured in meters and time in seconds. DO NOT SOLVE THIS
EQUATION.
Solution: The basic equation for vibration is
my 00 + γy + ky = f (t),
where m is the mass, γ is the air resistance factor, k is the spring constant, and f (t) is due to any
force other than gravity, air resistance, or the spring. Here, m = 25, γ = 0, k = 30, and f (t) = 0.
Since the initial position is the equilibrium position, y(0) = 0. The initial velocity must be
converted into meters (by dividing by 100), with downward velocities being positive and negative
velocities upward. Thus y 0 (0) = −2.5, and the complete initial value problem is
25 y 00 + 30 y = 0
y(0) = 0
y 0 (0) = −2.5
Grading: +7 points for the equation, +4 points for each initial condition. Grading for common
mistakes: −2 points for a positive value of y 0 (0), or −0.25 or −250.
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7. [10 points] Determine the value γ for which the differential equation 7y 00 + γy 0 + 8y = 0 is critically
damped.
Solution: The value of γ for which this equation is critically damped is
√
γ=
4km =
√
4·8·7=
√
224 ≈ 14.9666.
√
Grading: +4 points for the
cation.
4km formula, +4 points for substitution, +2 points for simplifi-
8. [15 points] Solve the initial value problem below.
y 00 − 3y 0 − 10y = 0
y(0) = −2
y 0 (0) = 25
Solution: This is a homogeneous differential equation with constant coefficents, so you start with
the auxiliary (characteristic) equation:
r2 − 3r − 10 = 0
(r − 5)(r + 2) = 0
y(t) = C1 e5t + C2 e−2t .
for some constants C1 and C2 . Now use the initial conditions to determine C1 and C2 :
y(0) = −2 :
−2 = C1 + C2
y 0 (t) = 5C1 e5t − 2C2 e−2t
y 0 (0) = 25 :
25 = 5C1 − 2C2
This is a system of linear equations, which can be solved using elimination. Multiplying the first
equation by 2 and adding that (new) equation to the equation for y 0 (0) yields 21 = 7C1 or C1 = 3;
substituting this information into either equation and solving for C2 produces C2 = −5. Thus
y(t) = 3e5t − 5e−2t .
(Writing y as a function of x was also acceptable.)
Grading: +4 points for the auxiliary equation, +3 points for the form of y(t), +4 points for
substituting the initial conditions, +4 points for finding C1 and C2 .
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9. [15 points] Calculate the Wronskian W (et , t2 , et−1 ). Are the functions et , t2 , and et−1 linearly independent or linearly dependent?
Solution:
t
e
t 2 t−1
W (e , t , e ) = et
et
t2 et−1 2t et−1 2 et−1 = (et · 2t · et−1 ) + (t2 · et−1 · et ) + (et−1 · et · 2) − (et−1 · 2t · et ) − (et · et−1 · 2) − (t2 · et · et−1 )
= 0,
1 t
· e .)
e
Grading: +4 points for setting up the determinant, +4 points for evaluating it (usually done
with Sarrus’s Method), +2 points for simplifying, +5 points for “linearly dependent.”
which means the functions are linearly dependent. (Indeed, et−1 =
10. [15 points] The form of a particular solution to the differential equation y 00 + 9y = −7 sin(3t) is
yp (t) = At sin(3t) + Bt cos(3t). Find the actual particular solution.
Solution: Starting off with
y = At sin(3t) + Bt cos(3t),
taking derivatives twice yields
y 0 = 3At cos(3t) + A sin(3t) − 3Bt sin(3t) + B cos(3t),
y 00 = 3A cos(3t) − 9At sin(3t) + 3A cos(3t) − 3B sin(3t) − 9Bt cos(3t) − 3B sin(3t).
Now you can either substitute these formulas into the differential equation, or just compare coefficients. Below, the latter method is shown:
(−9A) + 9(A) = 0
or
0=0
(no information)
[t sin(3t)]
(−9B) + 9(B) = 0
or
0=0
(no information)
[t cos(3t)]
(−6B) + 9(0) = −7
(6A) + 9(0) = 0
so
so
7
B=
6
A=0
[sin(3t)]
[cos(3t)]
Thus
yp (t) = At sin(3t) + Bt cos(3t) =
7
t cos(3t).
6
Grading: +5 points for finding the formulas for y 0 and y 00 , +5 points for setting the coefficients
equal to each other, +3 points for solving for A and B, +2 points for the final answer.
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MAT 275 Test 2 SOLUTIONS, FORM B
For the following multiple-choice problems below, write the letter of the correct answer on the blank to the
right of the problem. Make sure your answer is clearly legible, or it will be counted wrong. There is no
partial credit for these problems. Each is worth 5 points.
1. The solution to a certain differential equation is
y(t) = 3e−2t sin(4t) + 5e−2t cos(4t) − 7 sin(3t) + 2 cos(3t).
The transient solution is
A.
3e−2t sin(4t) + 5e−2t cos(4t)
Solution: These are the terms which converge to zero when t approaches ∞.
2. The form for a particular solution to the differential equation y 00 − 5y 0 + 6y = te3t is
C.
At2 e3t + Bte3t
Solution: The presence of te3t on the right-hand side suggests a form of Ae3t + Bte3t . However,
e3t is a solution to the homogeneous equation, so you need to multiply Ae3t + Bte3t by t to get a
useable form. (te3t is not a solution to the homogeneous equation.
3. The vibration problem 3y 00 + 11y 0 + 8y = 0 is
D.
overdamped.
√
Solution: γ = 11 and
4km =
√
√
96. Since γ >
4km, the problem is underdamped.
4. In the differential equation y 00 + 16y = −2 sin(at), resonance will
C.
occur when a = 4.
Solution: The homogeneous solution is y = C1 sin(4t) + C2 cos(4t). In order for resonance to occur,
the function A sin(at) + B sin(bt) must be a solution to the homogeneous solution; hence a = 4.
1
5. [10 points] Find the form of the solution to differential equation
y (v) − 7y 0000 + 2y 000 + 34y 00 + 93y 0 − 315y = 0
(The characteristic equation for y = erx factors into (r − 3)2 (r − 5)(r2 + 4r + 7).)
Solution: The roots of the characteristic equation are 3 (with multiplicity 2), 5, and
−4 ±
√
√
42 − 4 · 7
= −2 ± 3 i,
2
so the form of the solution is
y(x) = C1 e3x + C2 xe3x + C3 e5x + C4 e−2x sin
√
√ 3 x + C5 e−2x cos
3x .
(The variable t could have been used instead of x, even though it was stated that y = erx .)
Graded on a 0 − 3 − 5 − 7 − 10-point basis.
6. [15 points] A mass of 35 kilograms is attached to a spring whose constant is 10 N/m. The mass is
started in motion from the equilibrium position with an initial velocity of 50 cm/sec in the downward
direction. Assume there is no air resistance. Set up the initial value problem governing the position y(t)
of the mass, assuming that the position is measured in meters and time in seconds. DO NOT SOLVE
THIS EQUATION.
Solution: The basic equation for vibration is
my 00 + γy + ky = f (t),
where m is the mass, γ is the air resistance factor, k is the spring constant, and f (t) is due to any
force other than gravity, air resistance, or the spring. Here, m = 35, γ = 0, k = 10, and f (t) = 0.
Since the initial position is the equilibrium position, y(0) = 0. The initial velocity must be
converted into meters (by dividing by 100), with downward velocities being positive and negative
velocities upward. Thus y 0 (0) = 0.5, and the complete initial value problem is
35 y 00 + 10 y = 0
y(0) = 0
y 0 (0) = 0.5
Grading: +7 points for the equation, +4 points for each initial condition. Grading for common
mistakes: −2 points for a negative value of y 0 (0), or 50 or 0.05.
2
7. [10 points] Determine the value γ for which the differential equation 3y 00 + γy 0 + 4y = 0 is critically
damped.
Solution: The value of γ for which this equation is critically damped is
√
4km =
γ=
√
4·4·3=
√
48 ≈ 6.9282.
√
Grading: +4 points for the
cation.
4km formula, +4 points for substitution, +2 points for simplifi-
8. [15 points] Solve the initial value problem below.
y 00 + y 0 − 12y = 0
y(0) = −2
y 0 (0) = 29
Solution: This is a homogeneous differential equation with constant coefficents, so you start with
the auxiliary (characteristic) equation:
r2 + r − 12 = 0
(r + 4)(r − 3) = 0
y(t) = C1 e−4t + C2 e3t .
for some constants C1 and C2 . Now use the initial conditions to determine C1 and C2 :
y(0) = −2 :
−2 = C1 + C2
y 0 (t) = −4C1 e−4t + 3C2 e3t
y 0 (0) = 29 :
29 = −4C1 + 3C2
This is a system of linear equations, which can be solved using elimination. Multiplying the first
equation by 4 and adding that (new) equation to the equation for y 0 (0) yields 21 = 7C2 or C2 = 3;
substituting this information into either equation and solving for C1 produces C1 = −5. Thus
y(t) = −5e−4t + 3e3t .
(Writing y as a function of x was also acceptable.)
Grading: +4 points for the auxiliary equation, +3 points for the form of y(t), +4 points for
substituting the initial conditions, +4 points for finding C1 and C2 .
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9. [15 points] Calculate the Wronskian W (sin x, cos x, sin(x + 1)). Are the functions sin x, cos x, and
sin(x + 1) linearly independent or linearly dependent?
Solution:
sin x
cos x
sin(x + 1) W (sin x, cos x, sin(x + 1)) = cos x − sin x cos(x + 1) − sin x − cos x − sin(x + 1) = (sin x · − sin x · − sin(x + 1)) + (cos x · cos(x + 1) · − sin x) + (sin(x + 1) · cos x · − cos x)
− (sin(x + 1) · − sin x · − sin x) − (sin x · cos(x + 1) · − cos x) − (cos x · cos x · − sin(x + 1))
= 0,
which means the functions are linearly dependent. (Indeed, sin(x+1) = (cos 1) sin x+(sin 1) cos x.)
Grading: +4 points for setting up the determinant, +4 points for evaluating it (usually done
with Sarrus’s Method), +2 points for simplifying, +5 points for “linearly dependent.”
10. [15 points] The form of a particular solution to the differential equation y 00 + 4y = 3 cos(2t) is
yp (t) = At sin(2t) + Bt cos(2t). Find the actual particular solution.
Solution: Starting off with
y = At sin(2t) + Bt cos(2t),
taking derivatives twice yields
y 0 = 2At cos(2t) + A sin(2t) − 2Bt sin(2t) + B cos(2t),
y 00 = 2A cos(2t) − 4At sin(2t) + 2A cos(2t) − 2B sin(2t) − 4Bt cos(2t) − 2B sin(2t).
Now you can either substitute these formulas into the differential equation, or just compare coefficients. Below, the latter method is shown:
(−4A) + 4(A) = 0
or
0=0
(no information)
[t sin(2t)]
(−4B) + 4(B) = 0
or
0=0
(no information)
[t cos(2t)]
(−4B) + 4(0) = 0
so
(4A) + 4(0) = 3
so
B=0
3
A=
4
[sin(2t)]
[cos(2t)]
Thus
yp (t) = At sin(2t) + Bt cos(2t) =
3
t cos(2t).
4
Grading: +5 points for finding the formulas for y 0 and y 00 , +5 points for setting the coefficients
equal to each other, +3 points for solving for A and B, +2 points for the final answer.
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