Exam II Review Solutions
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Qualitative Behavior
Consider the following 2nd order ODE’s
(a) y 00 + 2y 0 + y = 0
Solution: The characteristic polynomial is r2 + 2r + 1 = 0, which has
the root r = −1 with multiplicity 2. Thus, this is an example of free
(since the equation is unforced) critically damped motion.
Figure 1: A typical critically damped solution.
(b) 3y 00 +
√
2y 0 + y = 0
√
2 + 2r + 1 = 0, which
Solution: The characteristic
polynomial
is
3r
√
√
has the roots r = −6 2 ± 610 i. Thus, this is an example of free (since
the equation is unforced) under-damped motion.
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Figure 2: A typical under-damped solution.
(c) y 00 + 1.1y = cos(t)
Solution: Since the frequency of the homogeneous solution is approximately equal to the frequency of the forcing term, the solution to the
equation will exhibit beats.
Figure 3: A typical beats solution.
(d) y 00 + 4y = sin(2t + π/2)
Solution: Since the frequency of the homogeneous solution is equal to
the frequency of the forcing term, the solution to the equation will
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exhibit resonance.
Figure 4: A typical resonance solution.
Without solving, do the following for each equation given above: name the
phenomena exhibited by the equation and give justification. Finally, sketch
the qualitative behavior of the solution.
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Solution Techniques
For each of the equations listed below, state which method is the most
appropriate solution technique (give justification). Then, solve the equation.
(a) y 00 − y 0 − 2y = et ,
y(0) = y 0 (0) = 1
This is a linear second-order equation with constant coefficients for
which there is a standard guess for the particular solution (see table
4.4.1 in the text). Thus, we may use Undetermined Coefficients.
Solution: y(t) =
e−t et
− + e2t
2
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(b) y 00 + y = sec2 (x) Hint:
R
sec(x)dx = ln | sec(x) + tan(x)| + c
This is a linear second-order equation with constant coefficients, but
with a forcing term for which we do not have a standard guess for the
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particular solution (see table 4.4.1 in the text). Thus, we should try
Variation of Parameters.
Solution: y(t) = c1 cos(x) + c2 sin(x) − 1 + sin(x) ln | sec(x) + tan(x)|
(c) y 00 + y = 1 + U t −
π
2
[sin(t) − 1] ,
y(0) = 1, y 0 (0) = 0
This is a linear second-order equation with constant coefficients, but
with a discontinuous forcing term. Thus, we should try Laplace Transform.
Solution: y(t) = 1 − U(t − π2 ) 1 − sin(t) −
3
1
2
t−
π
2
cos(t)
Spring-Mass System
A force of 400 Newtons stretches a spring 2 meters. A mass of 50 kilograms
is attached to the end of the spring and it is initially released from the
equilibrium position with an upward velocity of 10 m/s. Additionally, the
mass is driven by an external force equal to f (t) = sin(2t).
(a) Write the corresponding IVP.
First, we must find the spring constant. From Newton’s laws we have,
F = kl, thus k = 200N/m. This yields thw following IVP
50y 00 + 200y = sin(2t);
y(0) = 0, y 0 (0) = −10
(b) Solve the IVP using any appropriate method. Name the behavior of
the solution.
Solution: y(t) =
4
−1
400
(1999 sin(2t) + 2t cos(2t)) ;
Resonance
Laplace Transform
(a) For the following equation, find the Laplace transform using only the
definition:
f (t) = et cos(t)
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Solution: F (s)
s−1
(s − 1)2 − 1
(b) Find the Laplace transforms of the following functions using the table:
(i) g(t) = (t + 1)3
Solution: G(s) =
(ii) h(t) =
0
t < π/2
cos(3(t − π/2)) t ≥ π/2
Solution: H(s) =
5
6 + 6s + 3s2 + s3
s4
se−πs/2
s2 + 9
Inverse Laplace Transform
Find the inverse Laplace transform of the following functions:
(a) Y (s) =
2s+5
s2 +6s+34
Solution: y(t) = 15 e−3t (10 cos(5t) − sin(5t))
(b) Q(s) =
e−s (2s−1)
s2 (s+1)2
Solution: q(t) = U(t − 1) 4 − (t − 1) − 4e−(t−1) − 3(t − 1)e−(t−1)
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One More Thing
There are a few topics that are not included on this review, but that doesn’t
mean they won’t be on the exam. You are still responsible for all material
covered in lecture!
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