Practice Test/Quiz 7 (Solutions)
You may work in groups of up to 3. Turn in only one set of answers for each group, but be sure
to do two things: everyone must write their name on the answer sheet and each person must write
the answer to at least one problem.
~ 0 (t) = AY
~ (t), with A =
1) Given the system of differential equations Y
5 α
0 β
, find the general
solution if
a) α = 5, β = 2
The eigenvalues are λ1 = 5 and λ2 = 2 with corresponding eigenvectors ~v1 =
−5
3
−5
1
5t
~ (t) = c1
e2t .
e + c2
Therefore, the solution is Y
3
0
1
0
and ~v2 =
b) α = β = 5
1
0
. To find the second
We get a repeated eigenvalue of λ = 5 with only one eigenvector ~v1 =
0
.
vector, we solve the system (A − λI)~v2 = ~v1 , which yields ~v2 =
1/5
0
1
1
5t
5t
~
e5t .
te +
e + c2
Therefore, the solution is Y (t) = c1
1/5
0
0
c) α = β = 0
The eigenvalues are λ1 = 5 and λ2 = 0 with corresponding eigenvectors ~v1 =
~ (t) = c1
Therefore, the solution is Y
1
0
e5t
+ c2
0
1
1
0
and ~v2 =
0
1
.
a
b
~
~ (t) has the given solution. Determine values of a, b, c,
2) Say that the system Y
=
Y
c d
and d that will yield the correct solution type. Hint: If either b = 0 or c = 0, then the eigenvalues
will both be real. Also, there is more than one correct answer for each part.
0 (t)
~ (t) = c1~v1 e6t + c2~v2 e−3t
a) Y
Any
choice
of matrix A will work, as long as the eigenvalues are λ = 6, 3. So we can choose
6 0
.
0 −3
~ (t) = c1
b) Y
3
0
e2t
+ c2
3
0
te2t
+
0
1
e2t
In this part, we have to be more careful since both the eigenvalues
AND eigenvectors are known.
3
λ = 2 is a repeated eigenvalue with only one eigenvector
, which is clear from the solution
0
form. The second vector results from the same process as in 1(b), so we can pick a matrix and
then see what needs to be true.
2 b
Say that we use A =
, which is the most general form we can use and still know that we
0 2
will have repeated eigenvalues
get one
but only eigenvector. All that remains is to solve for the
3
0
constant b. Using ~v1 =
and ~v2 =
, we see that we must have b = 3.
1
0
2 3
Therefore, we have A =
.
0 2
~ (t) has a phase portrait that is periodic with period π (remember that, for
c) The solution Y
example, the period of cos(nt) is equal to 2π/n)
In order for the phase portrait to be periodic, we must have that the eigenvalues are purely imaginary, namely λ = ±βi. In order for the period to be π, then we must have λ = ±2i.
0 2
.
Therefore, we need any matrix that has eigenvalues ±2i, so choose A =
−2 0
3) Find the general solution to the second-order equation ay 00 + by 0 + cy = 0, assuming that
a) b2 − 4ac > 0
The roots of the quadratic equation as2 + bs + c = 0 in this case are s =
both real.
√
−b+
So the solution is y(t) = c1 e
b2 −4ac
2a
t
√
−b−
+ c2 e
b2 −4ac
2a
t
√
−b± b2 −4ac
,
2a
which are
.
b) b2 − 4ac = 0
If b2 − 4ac = 0, then the root s =
−b
2a
−b
is a repeated root.
−b
So the solution is y(t) = c1 e 2a t + c2 te 2a t .
c) b2 − 4ac < 0
In this case, we have complex roots, which can be written as s =
−b
So the solution is y(t) = c1 e 2a t cos(
√
4ac−b2
t)
2a
−b
+ c2 e 2a t sin(
√
−b
2a
√
±
b2 −4ac
2a
=
−b
2a
√
±
4ac−b2
i.
2a
4ac−b2
t).
2a
4) For the equation my 00 + by 0 + ky = 0, assume that m > 0, b ≥ 0, k > 0, and b2 < 4km.
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a) What type(s) of solutions are possible? That is, what different solution forms could we get? You
may want to separately consider the cases b = 0 and b > 0.
q q q
√
−4km
k
k
k
If b = 0, then s = ± 2m = ± m i. So y(t) = c1 cos
m t + c2 sin
mt .
√
√
−b
−b
If b > 0, then y(t) = c1 e 2m t cos( 4km − b2 t) + c2 e 2m t sin( 4km − b2 t).
b) What value does y(t) approach as t → ∞, assuming that b > 0? Why does this make sense
(considering what this equation models)?
−b
Since 2m
< 0, then y(t) → 0 as t → ∞. This corresponds to the situation of a mass on a spring
being pulled, let go, and slowly returning to equilibrium over time.
3 α ~
Y (t), what different types of equilibrium points are possible
2 0
as the value of α varies? Find all possible value(s) of α for each type. Justify your answer.
~ 0 (t) =
5) For the system Y
The characteristic polynomial for this matrix is p(λ) = λ2 − 3λ − 2α, which yields eigenvalues of
√
3 ± 9 + 8α
λ=
.
2
The eigenvalues change from being complex to real when α = −9/8. If α < −9/8, then we will
always have complex eigenvalues with a positive real part. Therefore, the origin will always be a
spiral source.
√
√
If α > −9/8, the roots will both be real. However, if α > 0, then 9 + 8α > 3 and 3− 9 + 8α < 0.
This means that we will have two eigenvalues of opposite signs, hence the origin is a saddle.
Finally, if −9/8 < α < 0, we will have two positive eigenvalues, and the origin is a source. We
don’t really need to worry about the cases α = −9/8 or α = 0, but it is still a source in either case.
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