Math 215
Spring 2010
Assignment 7
You are highly recommended to own a textbook, Boyce-DiPrima or another, to see a comprehensive treatment of the subjects and more examples.
§6.6: 6, 14, 18, §7.1: 6, §7.5: 4a, 5a, 13, §7.6: 3a, 4a, 9
Z
• (§6.6: 6) Find the Laplace transform of the function f (t) =
t
(t − τ )eτ dτ .
0
Answer.
Since f (t) = t ∗
et ,
Lf = (Lt) ·
(Let )
=
1
s2
·
1
s−1 .
• (§6.6: 14) Express the solution of the initial value problem
y 00 + 2y 0 + 2y = sin αt;
y(0) = 0,
y 0 (0) = 0,
in terms of a convolution integral.
Answer. Taking Laplace transform, we have (s2 + 2s + 2)Y (s) = F (s) with F (s) =
1
1
· F . Since s2 +2s+2
= (s+1)1 2 +1 = L[e−t sin t], we get
L[sin αt]. Thus Y (s) = s2 +2s+2
Rt
y(t) = [e−t sin t] ∗ sin αt = 0 e−(t−τ ) sin(t − τ ) ∗ sin ατ dτ .
• (§6.6: 18) Express the solution of the initial value problem
y 00 + 3y 0 + 2y = cos αt;
y(0) = 1,
y 0 (0) = 0,
in terms of a convolution integral.
Answer.
Taking Laplace transform, we have L[y] = Y (s), L[y 0 ] = sY − 1, L[y 00 ] =
s2 Y − s, and the equation becomes
(s2 + 3s + 2)Y (s) = s + 3 + F (s),
F (s) = L[cos αt].
Since s2 + 3s + 2 = (s + 1)(s + 2), we get
Y (s) =
s+3
1
+
L[cos αt].
(s + 1)(s + 2) (s + 1)(s + 2)
By partial fraction,
s+3
2
1
=
−
,
(s + 1)(s + 2)
s+1 s+2
Thus
Z
−t
y(t) = 2e
−e
−2t
+
0
t
1
1
1
=
−
.
(s + 1)(s + 2)
s+1 s+2
[e−(t−τ ) − e−2(t−τ ) ] ∗ cos ατ dτ.
• (§7.1: 6) Transform the given initial value problem into an initial value problem for
two first order equations.
u00 + p(t)u0 + q(t)u = g(t);
Answer.
u(0) = u0 ,
u0 (0) = u00
Let x1 (t) = u(t) and x2 (t) = u0 (t). We have
x01 = x2 ,
x02 = −q(t)x1 − p(t)x2 + g(t),
with initial conditions
x1 (0) = u0 ,
x2 (0) = u00 .
µ
¶
1 1
• (§7.5: 4a) Find the general solution of x =
x and describe the behavior of
4 −2
the solution as t → ∞.
µ
¶
1−r
1
Answer.
det(A − rI) = det
= r2 + r − 6 = (r + 3)(r − 2). Thus
4
−2 − r
µ
¶
−1 1
eigenvalues are r = 2, −3. For eigenvalue r = 2, A − rI =
whose nullspace
4 −4
µ ¶
µ
¶
1
4 1
is spanned by
. For eigenvalue r = −3, A − rI =
whose nullspace is
1
4 1
µ ¶
1
spanned by
. Thus the general solution is
−4
0
µ ¶
µ ¶
1 2t
1
x(t) = c1
e + c2
e−3t .
1
−4
µ ¶
1 2t
As t → ∞, the second part vanishes and x(t) ∼ c1
e is asymptotic to the
1
µ ¶
1
directions ±
.
1
µ
¶
−2 1
• (§7.5: 5a) Find the general solution of x =
x and describe the behavior
1 −2
of the solution as t → ∞.
µ
¶
−2 − r
1
Answer.
det(A − rI) = det
= r2 + 4r + 3 = (r + 3)(r + 1).
1
−2 − r
µ
¶
−1 1
Thus eigenvalues are r = −1, −3. For eigenvalue r = −1, A − rI =
1 −1
0
2
¶
µ
µ ¶
1 1
1
whose
. For eigenvalue r = −3, A − rI =
whose nullspace is spanned by
1
1 1
µ ¶
1
nullspace is spanned by
. Thus the general solution is
−1
µ ¶
µ ¶
1 −t
1
x(t) = c1
e + c2
e−3t .
1
−1
As t → ∞,
µ both
¶ parts vanish. However the second part vanishes faster if c1 6= 0 and
1 −t
x(t) ∼ c1
e .
1


1
1
1
1 −1 x.
• (§7.5: 13) Find the general solution of x0 =  2
−8 −5 −3


1−r
1
1
1−r
−1  = r3 + r2 − 4r − 4 = (r +
Answer.
det(A − rI) = det  2
−8
−5 −3 − r
2)(r − 2)(r + 1). Thus eigenvalues are r = 2, −2, −1.
For eigenvalue r = 2, (A − rI)ξ = 0 is reduced to ξ1 = 0 and ξ2 + ξ3 = 0. We can take
ξ (1) = (0, 1, −1)T .
For eigenvalue r = −2, (A − rI)ξ = 0 is reduced to 7ξ1 + 4ξ3 = 0 and 7ξ2 − 5ξ3 = 0.
We can take ξ (2) = (4, −5, −7)T .
For eigenvalue r = −1, (A − rI)ξ = 0 is reduced to 2ξ1 + 3ξ3 = 0 and ξ2 − 2ξ3 = 0.
We can take ξ (3) = (3, −4, −2)T .
The general solution is


 
 
0
4
3
x(t) = c1  1  e2t + c2 −5 e−2t + c3 −4 e−t .
−1
−7
−2
µ
¶
2 −5
• (§7.6: 3a) Express the general solution of x =
x in terms of real-valued
1 −2
functions.
µ
¶
2−r
−5
Answer.
det(A − rI) = det
= r2 + 1. Thus eigenvalues are
1
−2 − r
µ
¶
2−i
−5
r = ±i. For eigenvalue r = i, A − rI =
whose nullspace is spanned
1
−2 − i
µ
¶
2+i
by
. The corresponding solution is (using eit = cos t + i sin t)
1
µ
¶
µ
¶
µ
¶
µ
¶
2 + i it
2+i
2 cos t − sin t
cos t + 2 sin t
x(1) (t) =
e =
() =
+i
.
1
1
cos t
sin t
0
3
The general solution is
µ
µ
¶
¶
2 cos t − sin t
cos t + 2 sin t
x(t) = k1
+ k2
.
cos t
sin t
Remark. If we let 5c1 = 2k1 + k2 and 5c2 = −k1 + 2k2 , then we get the solution on
page 758 of the textbook.
µ
¶
2 −5/2
0
• (§7.6: 4a) Express the general solution of x =
x in terms of real-valued
9/5 −1
functions.
µ
¶
2 − r −5/2
Answer.
det(A − rI) = det
= r2 − r + 5/2 = (r − 1/2)2 + 9/4.
9/5 −1 − r
Thus eigenvalues are r = 12 ± 23 i. For eigenvalue r = 12 + 32 i, (A − rI)ξ = 0 reduce to
the single equation (3 − 3i)ξ1 − 5ξ2 = 0. The corresponding eigenvector is (5, 3 − 3i)T ,
and the solution is
µ
¶
µ
¶
µ
¶
5
5 cos 32 t
5 sin 32 t
(1)
( 21 + 32 i)t
t/2
t/2
x (t) =
e
=e
+ ie
.
3 − 3i
3 cos 32 t + 3 sin 32 t
−3 cos 32 t + 3 sin 23 t
The general solution is
x(t) = c1 e
t/2
¶
µ
¶
5 sin 23 t
5 cos 23 t
t/2
+ c2 e
.
3 cos 32 t + 3 sin 32 t
−3 cos 23 t + 3 sin 32 t
µ
µ
¶
µ ¶
1 −5
1
• (§7.6: 9) Find the solution of x =
x, x(0) =
. Describe the behavior
1 −3
1
of the solution as t → ∞.
µ
¶
1−r
−5
Answer.
det(A − rI) = det
= r2 + 2r + 2 = (r + 1)2 + 1. Thus
1
−3 − r
eigenvalues are r = −1 ± i. For eigenvalue r = −1 + i, (A − rI)ξ = 0 reduce to the
single equation ξ1 = (2 + i)ξ2 . The corresponding eigenvector is (2 + i, 1)T , and the
solution is
µ
¶
µ
¶
µ
¶
2 + i (−1+i)t
(1)
−t 2 cos t − sin t
−t cos t + 2 sin t
x (t) =
e
=e
+ ie
.
1
cos t
sin t
0
The general solution is
µ
¶
µ
¶
2 cos t − sin t
−t cos t + 2 sin t
x(t) = c1 e
+ c2 e
.
cos t
sin t
µ ¶
µ ¶
µ ¶
1
2
1
The initial data give
= c1
+ c2
. Thus c1 = 1, c2 = −1, and
1
1
0
µ
¶
µ
¶
µ
¶
−t 2 cos t − sin t
−t cos t + 2 sin t
−t cos t − 3 sin t
x(t) = e
−e
=e
.
cos t
sin t
cos t − sin t
−t
This solution spirals to zero as t → ∞ due to the e−t factor.
4