Ordinary Differential Equations
MATH 308 - 523
A. Bonito
April 21
Spring 2015
Last Name:
First Name:
Exam 2
• 75 minute individual exam;
• Answer the questions in the space provided. If you run out of space, continue
onto the back of the page. Additional space is provided at the end;
• Show and explain all work;
• Underline the answer of each steps;
• The use of books, personal notes, calculator, cellphone, laptop, and communication with others is forbidden;
• By taking this exam, you agree to follow the university’s code of academic
integrity.
Ex 1
Ex 2
Ex 3
Ex 4
Total
Some Laplace Transforms
L(f )
f
L(f )
f
1
1
s
s>0
tn
n!
1
sn+1
s>0
e−αt
1
s+α
s > −α
e−αt tn
n!
(s+α)n+1
s > −α
sin(ωt)
ω
s2 +ω 2
s>0
cos(ωt)
s
s2 +ω 2
s>0
eαt sin(ωt)
ω
(s−α)2 +ω 2
s>α
eαt cos(ωt)
s−α
(s−α)2 +ω 2
s>α
sinh(ωt)
ω
s2 −ω 2
s > |ω|
cosh(ωt)
s
s2 −ω 2
s > |ω|
Hα (t)
e−αs
s
s>0
δα (t)
e−αs
s > −∞
Some Properties of the Laplace Transforms
Let f, g : [0, +∞) → R be piecewise continuous functions with piecewise continuous derivatives.
Assume there exists K > 0 and a1 , a2 ∈ R such that
|f (t)| 6 Kea1 t ,
|g(t)| 6 M ea2 t ,
∀t ∈ [0, +∞).
Then there holds
dn
dn−2
dn−1
n
n−1
(ı.) L
f
(t)
(s)
=
s
L
(f
(t))
−
s
f
(0)
−
...
−
s
f
(0)
−
f (0),
dtn
dtn−2
dtn−1
n
d
∀s > a1 , (f ∈ C n−1 ([0, ∞)),
f piecewise continuous)
dtn
Z t
1
(ıı.) L
f (τ )dτ (s) = L (f (t)) (s),
∀s > a1 ,
s
0
(ııı.) L ((−1)n tn f (t)) (s) =
dn
L (f (t)) (s),
dsn
(ıv.) L e−αt f (t) (s) = L (f (t)) (s + α),
∀s > a1 ,
∀s > a1 + α, α > 0,
(v.) L (Hα (t)f (t − α)) (s) = e−αs L (f (t)) (s),
(vı.) L ((f ∗ g)(t)) (s) = L(f (t))(s) · L(g(t))(s),
∀s > a1 , α > 0,
∀s > max(a1 , a2 ).
Exercise 1
25%
Solve the following initial value problem using the Laplace Transforms
y 00 + y = sin(t),
y(0) = 1,
y 0 (0) = 1.
(Hint: use the convolution product and the identity 2 sin(a) sin(b) = cos(a − b) − cos(a + b).)
Exercise 2
25%
• Write the definition of the laplace transform of a function f (t).
• Without using the table, compute the laplace transform of g(t) = 5δ3π (t). Do not forget to
mention for which values of s the laplace transform L(g)(s) is well defined.
• Find the solution of
y 00 + 2y 0 + 3y = 5δ3π ,
(you can use the table for this part.)
y(0) = y 0 (0) = 0.
Exercise 3
25%
Solve the following system
y0 =
and draw the phase portrait.
−2
−5
1
4
y
Exercise 4
25%
Solve the following system
y0 =
and draw the phase portrait.
3
4
−2
−1
y
Ordinary Differential Equations
MATH 308 - 523
A. Bonito
April 21
Spring 2015
Exam 2: solutions
Exercise 1
25%
The Laplace transform of both sides of the ODE yields
(s2 + 1)Y − s − 1 = L(sin(t)(s),
where Y (s) = L(y)(s). Thus we obtain an expression for Y
Y =
1
L(sin(t)(s)
s
+
+
.
s2 + 1 s2 + 1
s2 + 1
Noting that
s2
1
= L(sin(t))(s),
+1
we obtain
s
1
+ 2
+ L(sin(t)(s)L(sin(t)(s).
+1 s +1
It remains to compute the inverse Laplace transforms of the three terms in the right hand side of
the above equation. Using the Laplace table for the two first terms we have
1
s
−1
=
sin(t),
and
L
= cos(t).
L−1
s2 + 1
s2 + 1
Y =
s2
For the last term, property (vi) yields
L−1 (L(sin(t)(s)L(sin(t)(s)) = sin(t) ∗ sin(t).
We compute
Z
sin(t) ∗ sin(t) =
t
sin(t − u) sin(u)du =
0
1
2
Z
t
(cos(t − 2u) − cos(t))
=
0
Adding all the inverse Laplace transforms computed leads to
y(t) =
3
t
sin(t) + cos(t) − cos(t).
2
2
Exercise 2
25%
R∞
• L(f )(s) := 0 f (t)e−st dt.
• We start from the definition of the laplace transform
Z ∞
L(g)(s) := 5
δ(t − 3π)e−st ds = 5e−3π
0
which holds for any s ∈ R.
s
1
t
sin(t) − cos(t).
2
2
• We take the laplace transform and use the previous item to arrive at
(s2 + 2s + 3)Y = 5e−3πs
or
5
e−3πs .
+ 2s + 3
We first find the Laplace transform inverse of
Y =
s2
√
5
5
2
F (s) := 2
=√
,
2+2
s + 2s + 3
(s
+
1)
2
which is given by
√
5
L−1 (F )(t) = √ e−t sin( 2t).
2
Hence,using the formula
L−1 (F (s)e−cs ) = L−1 (F )(t − c)uc (t)
we arrive at
y(t) = L−1 (
Exercise 3
√
5
5 −(t−3π)
−3πs
√
2(t − 3π))u3π (t).
e
)(t)
=
e
sin(
s2 + 2s + 3
2
25%
We find the eigenvalues-eigenvectors of the matrix. The eigenvalues are found solving
−2 − λ
1
det
= 0,
−5
4−λ
i.e.
λ2 − 2λ − 3 = 0
or λ1 = 3 and λ2 = −1.
The eigenvector associated to λ1 is given by
−2 1
ξ1
ξ1
=3
−5 4
ξ2
ξ2
or
−5ξ1 + ξ2 = 0.
Hence all eigenvectors are given by
ξ1
ξ2
=α
1
5
,
∀α 6= 0.
The eigenvector associated to λ2 is given by
−2 1
ξ1
ξ1
=−
−5 4
ξ2
ξ2
or
−ξ1 + ξ2 = 0.
Hence all eigenvectors are given by
ξ1
ξ2
=α
1
1
,
∀α 6= 0.
Gathering the above results, we find that the general solution reads
1
1
3t
y = C1
e + C2
e−t .
5
1
The phase portrait is provided in Fig. 1.
Figure 1: Phase portrait: The red lines correspond to the directions given by the eigenvectors
Exercise 4
25%
We find the eigenvalues-eigenvectors of the matrix. The eigenvalues are found solving
3−λ
−2
det
= 0,
4
−1 − λ
i.e.
λ2 − 2λ + 5 = 0
or λ = 1 ± 2i.
We only consider λ = 1 + 2i and find the associated eigenvector:
3 −2
ξ1
ξ1
= (1 + 2i)
4 −1
ξ2
ξ2
or
(2 − 2i)ξ1 − 2ξ2 = 0.
Hence all eigenvectors are given by
ξ1
1
=α
,
ξ2
1−i
∀α 6= 0.
As a consequence two linearly independent solution are given by
1
cos(2t)
(1+2i)t
t
y=< e
=e
1−i
cos(2t) + sin(2t)
and
y = = e(1+2i)t
1
1−i
= et
sin(2t)
sin(2t) − cos(2t)
.
Gathering the above results, we find that the general solution reads
cos(2t)
sin(2t)
t
y = e C1
+ C2
.
cos(2t) + sin(2t)
sin(2t) − cos(2t)
The phase portrait is provided in Fig. 2.
Figure 2: Phase portrait