extra 1125
TRANSCRIPT
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8/11/2019 Extra 1125
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ay +by +cy = g(t), y(0) =y0, y(0) =z0
a b c
s y = y (t) Y(s) y y
Y(s)
y(0)
y(0)
Y(s)
y(t)
f(t)
f(t)
s
limA
A0
estf(t)
f(t) L {f(t)} F(s)
s
f(t) =e2t
limA A0 e
st
e2t
limA
A0
e(2s)t
s= 2
limA
A0
e(22)t = limA
A0
1 = limA
t|A0= limA
A=
s
2
s= 2
limA
A0
e(2s)t = limA
e(2s)t
2 s |A0= lim
Ae(2s)A
2 s
e(2s)0
2 s
2 s >0 2 s 2
limA
e(2s)A
2 s
e(2s)0
2 s
= 0
1
2 s=
1
s 2
e2t
L {f(t)}= F(s) = 1
s 2
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8/11/2019 Extra 1125
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eat
cos(bt)
sin(bt)
t
tn
g(t)
f(t) L {f(t)}
eat 1sa
cos(bt) ss2+b2
sin(bt) bs2+b2
tn n!sn+1
n! = n(n 1)(n 2) . . . (2)(1) n 0! 1
bi s
b
tn
s
f(t) =t
limA
A0
estt
u= t
v = est
u = 1
v = 1s
est
limA
1
stest
A0
1
sest(1)
= limA
1
stest +
A0
1
sest
= limA
1
stest
1
s2est
A0
= limA
1
sAesA
1
s2esA +
1
s0e0 +
1
s2e0
s
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8/11/2019 Extra 1125
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f1(t) f2(t) f1(t) +f2(t)
L {f1(t) +f2(t)}= L {f1(t)} + L {f2(t)}
c
c f1(t)
L {c f1(t)}= c L {f1(t)}
f(t) = 3e2t + cos(4t) + 5t2
L3e2t + cos(4t) + 5t2
=L
3e2t
+ L {cos(4t)} + L
5t2
= 3L
e2t
+ L {cos(4t)} + 5L
t2
= 3 1
s 2+
s
s2 + 42+ 5
2!
s(2+1) =
3
s 2+
s
s2 + 42+
10
s(2+1)
L
3e2t + cos(4t) + 5t2
= 3
s 2+
s
s2 + 42+
10
s(2+1)
L {f1(t)f2(t)} =L {f1(t)} L {f2(t)}
L {eat cos(bt)}= limA
A0
esteat cos(bt) = limA
A0
e(as)t cos(bt)
s=a
u= cos(bt)
v = e(as)t
u =b sin(bt) v= e(as)t
as
limA
e(as)t
a s cos(bt)
A0
e(as)t
a s (b sin(bt))
= limA
e(as)t
a s
cos(bt) + b
a s A
0
e(as)t(sin(bt)) u = sin(bt) v = e(as)t u =
b cos(bt) v= e(as)t
as
= limA
e(as)t
a s cos(bt) +
b
a s
e(as)t
a s sin(bt)
A0
e(as)t
a s (b cos(bt))
= limA
e(as)t
a s cos(bt) +
b
a s
e(as)t
a s sin(bt)
b2
(a s)2
A0
e(as)t(cos(bt))
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8/11/2019 Extra 1125
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limA A0
e(as)t
cos(bt)
= limAe(as)t
a s cos(bt) + b
a s
e(as)t
a s sin(bt) b2
(a s)2 A0
e(as)t
(cos(bt))
b
2
(as)2
A0
e(as)t(cos(bt))
limA
1 +
b2
(a s)2
A0
e(as)t cos(bt)
= limA
e(as)t
a s cos(bt) +
b
a s
e(as)t
a s sin(bt)
limA
(a s)2 +b2
(a s)2
A0
e(as)t cos(bt)
= limA
e(as)t
a s cos(bt) +
b
a s
e(as)t
a s sin(bt)
(a s)2 = (s a)2 (as)2+b2
(as)2
limA
A0
e(as)t cos(bt) = (s a)2
(s a)2 +b2 limA
e
(as)t
a s cos(bt) + b
a se
(as)t
a s sin(bt)
A
0
s > a limA
e(as)A
as cos(bA) = lim
A
b
as
e(as)A
as sin(bA) = 0
limA
A0
e(as)t cos(bt) = (s a)2
(s a)2 +b2
e(as)0
a s cos(0)
b
a s
e(as)0
a s sin(0)
= (s a)2
(s a)2 +b2
1
a s
=
(s a)2
(s a)2 +b2
1
s a
=
(s a)
(s a)2 +b2
s > a
Leat cos(bt) =
(s a)
(s a)2 +b2
L
eat sin(bt)
= b
(s a)2 +b2
s a bi
eat cos(bt)
s a
L {f(t)} f(t)
L {f(t)}= 1s2
L
e2t
= 1s2
t 1s2
1s2
e2t L1
L1
1s2
= e2t
s F(s)
F(s)
f(t) L {f(t)}= F(s)
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8/11/2019 Extra 1125
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L1 {F(s)} = f(t) t L {f(t)} = 1
s2
f(t) =e2t L1
1s2
=e2t
F1(s) F2(s) F1(s) +F2(s)
L1 {F1(s) +F2(s)}= L1 {F1(s)} + L
1 {F2(s)}
c c F1(s)
L1 {c F1(s)}= c L1 {F1(s)}
F(s) = 1s2
+ 8s2+4
+ 2s2s24s+13 f(t) L {f(t)}= F(s)
t F(s)
s 2
2
eat a a= 2
a= 2 F(s) e2t
s2 + 4 2i s = 2i s= 2i cos(2t) sin(2t)
s
s
s
8
sin(2t)
s2 4s+ 13
2 3i s2 4s + 13 = (s 2)2 + 32
s
a bi (s a)2 +b2
2s 2
(s 2)2 + 32
s
s 2
e2t cos(3t) e2t sin(3t)
L
e2t
= 1s2
1s2
1
s 2 =1
1
s 2=1L
e2t
=L
e2t
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8/11/2019 Extra 1125
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e2t
L1 1
s 2
=e2t
L {sin(2t)} = 2s2+22
8s2+22
= 4 2s2+22
4 = 82
4 sin(2t)
L1
8
s2 + 4
= 4 sin(2t)
L
e2t cos(3t)
= s2(s2)2+32
L
e2t sin(3t)
= 3(s2)2+32
e2t cos(3t)
e2t sin(3t)
s
2s 2
(s 2)2 + 32 = 2
s 1
(s 2)2 + 32
eat cos(bt) eat sin(bt)
s
s a+a
2 s 1
(s 2)2 + 32 = 2
s 2 + 2 1
(s 2)2 + 32 = 2
(s 2) + 1
(s 2)2 + 32
= 2
s 2
(s 2)2 + 32+
1
(s 2)2 + 32
e2t cos(3t)
e2t sin(3t)
1
(s 2)2 + 32 =
1
3
3
(s 2)2 + 32
13e
2t sin(3t)
2
s 2
(s 2)2 + 32 +
1
3
3
(s 2)2 + 32
2s2(s2)2+32
L1 2s 2
(s 2)2 + 32 = 2e2t cos(3t) +
1
3e2t sin(3t)
t
F(s) = 1s2
+ 8s2+4
+ 2s2s24s+13
L1
1
s 2+
8
s2 + 4+
2s 2
s2 4s+ 13
=e2t + 4 sin(2t) + 2
e2t cos(3t) +
1
3e2t sin(3t)
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8/11/2019 Extra 1125
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F(s)
1
s 2+ 4
2
s2 + 22 + 2 s 2
(s 2)2 + 32 +
1
3
3
(s 2)2 + 32
= 1
s 2+
8
s2 + 4+ 2
s 2
s2 4s+ 13+
1
s2 4s+ 13
= 1
s 2+
8
s2 + 4+
2(s 2 + 1)
s2 4s+ 13
= 1
s 2+
8
s2 + 4+
2s 2
s2 4s+ 13 =F(s)
f(t)
f(t) =
f1(t) a0 t < a1
f2(t) a1 t < a2
. . .
fn(t) an1 t < an
f1(t) f2(t) . . . fn(t) t a0 an
f(t) =f1(t) t < a1 f(t) =fn(t) an1 t
f(t) =
et t
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8/11/2019 Extra 1125
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t f(t)
t = 1
u1(t) u1(t)
0
1
t= 1
u1(t)
t= 1
et
sin(t)
sin(t) et u1(t)
f(t) =et +u1(t)
sin(t) et
+?
t f(t)
t= 2
t2
sin(t)
f(t) =et +u1(t)
sin(t) et
+u2(t)
t2 sin(t)
t < 1 u1(t) = 0 u2(t) = 0 f(t) =et
f(t) =et + (0)
sin(t) et
+ (0)
t2 sin(t)
=et
1 t
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8/11/2019 Extra 1125
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B u
uc(t)
f
ecs limB
B0
esuf(u) =ecsL {f(u)}= ecsL {f(t)}= ecsF(s)
L {f(u)}= L {f(t)} f
c 0 c c A
limA
c0
estuc(t)f(t c) +
Ac
estuc(t)f(t c)
uc(t) = 0 0 t < c c0
est(0)f(t c) = 0
uc(t) = 1 t c
limA
A
c
est(1)f(t c) = limA
A
c
estf(t c)
uc(t) 1
u= t c t= u+c t =
limA
Accc
es(u+c)f(u)
= limA
Ac0
escesuf(u)
=ecs limA
Ac0
esuf(u)
B= A c A B A
B
=ecs limB
B0
esuf(u)
f(u) F(s)
F(s) = limA
A0
estf(t)
u B u
t
B
A
uc(t)f(t c) ecsF(s) ecs
uc(t)f(t c) f
t
f(t c) u+c f(u+c c) =f(u)
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8/11/2019 Extra 1125
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L {f(u)} f(u) L {uc(t)f(t c)}= ecsL {f(u)}
u4(t)e2t
c= 4
f(t c) =f(t 4) =e2t
t u+c= u+ 4 f(u)
f(u) =e2(u+4) =e2u+8 =e8e2u
f(u) =e8e2u
L
e8e2u
=e8L
e2u
=e8 1
s 2
u4(t)e2t
L
u4(t)e2t
=e4s e8 1s 2
L {uc(t)}= ecs
1
s
uc(t) =uc(t)e0(tc)
f(t c) =e0(tc)
f(u) =e0u
e0u
1s0
= 1s
uc(t) =
uc(t)e0(tc)
L {uc(t)}= L
uc(t)e0(tc)
=ecsL
e0u
=ecs1
s
u2(t)(t 1) u4(t)et
u2(t)(t 1) c= 2 f(t 2) =t 1
f(u) = (u+ 2) 1 =u+ 1
f(u) =u+ 1
L {u+ 1}= 1
s2+
1
s
u2(t)(t 1)
L {u2(t)(t 1)}= L {u2(t)(u+ 1)}= e2s
1
s2 +
1
s
u4(t)et =u4(t) (e
t) c= 4 f(t 4) =et
f(u) =eu+4 =e4eu
f(u) =e4eu
L
e4eu
=e4L {eu}= e4 1
s 1
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8/11/2019 Extra 1125
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u4(t)(et)
L
u4(t)(et
)
=L
u4(t)(e4
eu
)
=e4s
e4 1
s 1
u2(t)(t 1) u4(t)et
e2s
1
s2+
1
s
+e4s
e4
1
s 1
ecsF(s) L1 {ecsF(s)}
u= t c L1 {F(s)}= f(u) L1 {ecsF(s)}= uc(t)f(u) =uc(t)f(t c)
t
ecsF(s) c
u
e2s ss2+32
c= 2
u= t 2
s
s2+32 u
t
s2 + 32 3i
cos(3u) sin(3u) s cos(3u)
L1
e2s s
s2 + 32
=u2(t) cos(3u) =u2(t) cos(3(t 2)) =u2(t) cos(3t 6)
e2s s(s2)2+32
c= 2 u= t 2
s
(s2)2+32 u
t
(sa)2+b2 e2u cos(3u)
e2u sin(3u)
s 2 3 s s= s 2 + 2
s 2 + 2
(s 2)2 + 32 =
s 2
(s 2)2 + 32 +
2
(s 2)2 + 32
e2u cos(3u)
3
33
3
s 2
(s 2)2 + 32 +
3
3
2
(s 2)2 + 32
= s 2
(s 2)2 + 32+
2
3
3
(s 2)2 + 32
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8/11/2019 Extra 1125
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23e
2u sin(3u)
e2u cos(3u) + 23
e2u sin(3u)
e2s s(s2)2+32
u2(t)
e2u cos(3u) + 23e2u sin(3u)
u= t 2
L1
e2s s
(s 2)2 + 32
=u2(t)
e2(t2) cos(3(t 2)) +
2
3e2(t2) sin(3(t 2))