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Differential Equations and Matrix Algebra II
Answers to Worksheet
Professor Broughton
Name: Section #: Box #: Due: Tuesday, January 15
- (^) ∫ ∞ 1
x−^3 dx = lim b→∞
∫ (^) b
0
x−^3 dx = lim b→∞
[
x−^2 − 2
]b
1
= lim b→∞( x−^2 − 1 − 2
0
dx 1 + x^2 = lim b→∞
∫ (^) b
0
dx 1 + x^2 = lim b→∞
[
tan−^1 (x)
]b 0 = lim b→∞(tan
− (^1) (b)) = π 2
5
e−^3 tdt = lim b→∞
∫ (^) b
5
e−^3 tdt = lim b→∞
[
e−^3 t − 3
]b
51
blim→∞(e−^3 b^ −^ e−^15 ) =^
e−^15 3
0
te−^3 tdt = lim b→∞
∫ (^) b
0
te−^3 tdt = lim b→∞
∫ (^) b
0
td e−^3 t − 3
= lim b→∞
([
t e−^3 t − 3
]b
0
∫ (^) b
0
e−^3 t − 3
dt
= lim b→∞
b e−^3 b − 3
[
e−^3 t − 3
]b
0
= lim b→∞
b e−^3 b − 3
e−^3 b − 9
L{t^2 } =
0
te−stdt = lim b→∞
∫ (^) b
0
t^2 e−stdt = lim b→∞
∫ (^) b
0
t^2 d e−st −s
= lim b→∞
[
t^2 e−st −s
]b
0
− lim b→∞
∫ (^) b
0
e−st −s
dt^2 = 0 +
s
lim b→∞
∫ (^) b
0
te−stdt
=
s L{t} =
s
s^2
s^3
L{e−^5 t} =
0
e−^5 te−stdt = lim b→∞
∫ (^) b
0
e−^5 te−stdt = lim b→∞
∫ (^) b
0
e−^5 te−stdt
= lim b→∞
∫ (^) b
0
e−(s+5)tdt = lim b→∞
[
e−(s+5)t −(s + 5)
]b
0 = lim b→∞
e−(s+5)b −(s + 5)
s + 5
s + 5
L{sinh(kt)} = L{ ekt^ − e−kt 2
L{ekt} −
L{e−kt}
=
s − k
s + k
2 k s^2 − k^2
k s^2 − k^2
- Use integration by parts:
sin(kt)e−stdt =
sin(kt)de −−sts = sin(kt)e −−sts − ∫ (^) e−st −s d^ sin(kt) =^ · · ·^ and^ L{cos(kt)}^ in the book.
L{sin(kt)} =
0
sin(kt)e−stdt = lim b→∞
∫ (^) b
0
sin(kt)e−stdt = lim b→∞
∫ (^) b
0
sin(kt)d e−st −s
= lim b→∞
[
sin(kt)e−st −s
]b
0
− lim b→∞
∫ (^) b
0
e−st −s
d sin(kt)
= lim b→∞ sin(kb)e−sb −s
k s blim→∞
∫ (^) b
0
cos(kt)e−stdt
= 0 + k s L{cos(kt)} = k s
s s^2 + k^2
k s^2 + k^2
