This document provides examples of calculating the Laplace transform of various functions. It begins with 14 examples of taking the Laplace transform of functions like 2sen t + 3 cos t, t^2e^4t, and e^-2t sen t. It then provides 10 examples of taking the inverse Laplace transform of functions like 1/(s(s+1)), 25/(s^2+1)^2, and 1/(s^2-4s+5). Finally, it concludes with 3 additional examples of calculating the inverse Laplace transform.

1. Instituto Universitario Politécnico
“Santiago Mariño”
Extensión Maturín
Esc. Ingeniería Eléctrica y Electrónica
Ejercicios Transformada de Laplace
Profesora: RealizadoPor:
MariangelaPollonais RubénGonzález
C.I:25.453.370
Teléfono:0426-6895258
Enero 2017

2. En lossiguientesejerciciosdetermine laTransformadade Laplace de lassiguientesfunciones:
1.) F(t) = 2sen t + 3 cos 2
t
∫ {f(t)} =2 ∫ {sent}+ 3 ∫ {cos 2t}
F (s) = 2 1 + 3 s
s2
+ 1 s2
+ 4.
2.) g (t) = t2
e4t
∫ {g(t)} = ∫ {t2
e4t
}
g(s) = 1
(s– 4)3
3.) h(t) = e-2t
sens t
∫ {h(t)} =∫{e-2t
sens t}
h(s) = 5
(s+2)2
+ 25.
4.) p(t) = (t + a)3
t3
+ 3 t2
a + 3 t a2
+ a3
∫ {p(t)} =∫{t3
} + 3a ∫{t2
} + 3a2
∫ {t} + a3
∫{1}
p(s) = 3! + 3a 2! + 3a2
1! + a3
1
s4
s3
s2
s
p(s) = 6 + 6a + 3a2
+ a3
s4
s5
s2
s
5.) q(t) = sen2
a t
q(t) = ½ ( 1 – sen2 a t)
∫ {q(t)} = ½ ∫{1} - ½ ∫ {sen2 a t 1}
q(s) = ½ x ⅕ - ½ 2a
s2
+ 4 a2
6.) f(t) = 3 e-t
+ sen6t
∫ {f(t)} =3 ∫{e-3t
} + ∫ {sen6 t}
f(s) = 3 1 + 6
s + 3 s2
+ 36
7.) g(t) = t3
– 3 t + cos 4 t
∫{ g(t) } = ∫ {t3
} – ∫ {t} + ∫ {cos4 t}
g(s) = 3! - 3 1 + S
s4
s2
s2
+ 16
8.) h(t) = - 3 cos 2 t + s Sen4 t
∫ { h(t) } = - 3 ∫ { cos 2 t} + s ∫ {sen4t}
h(s) = - 3 S + S 4
s2 + 4 s2 + 16
9.) q(t) = 4 cos2
3 t.
q(t) = 4 . ½ ( 1 + cos 6 t)

3. q(t) = 2 + 2 cos 6 t
∫ { q(t) } = 2 ∫{ 1} + 2 { cos 6 t}
q(s) = 2 1 + 2 S
S S2
+ 36
q(s) = 2 + 2 S
S2
+ 36
10.) r(t) = cos3
t = cos2
t cos t
r(t) = ( 1 – Sen2
t) cost
r(t) = cos t – sen2
cost
11.) f(t) = e-3t
(t – 2)
t e-3t
– 2 e-3t
∫{ f(t) } = ∫{t e-3t
} – 2 ∫ { e-3t
}
f(s) = 1 - 2 1
(s+ 3)2
s + 3
12.) g(t) = e 4t
[ 1 – cos t ]
e 4t
- e 4t
cos t
∫ { g(t) } = ∫ { e 4t
} – ∫ { e 4t
cos t }
g(s) = 1 - S – 4
S – 4 ( S – 4)2
+ 1
13.)h(t) = e-5t
[ t4
+ 2t + 1]
h(t) = t4
e-5t
+ 2 t e-5t
+ e-5t
∫ { h(t) } = ∫ { t4
e-5t
} + 2 ∫ { e-5t
}
h(s) = 1 + 2 1 + 1
(s+ 5)5
(s+ 5)5
s + 5
14.)q(t) = t3
sen 2 t
∫ { q(t) } = ∫{ t3
Sen2 t }
q(s) = 2 . 2 S = 4 S
(S2
+ 4 )2
(S2
+ 4)4
En lossiguientesejercicioscalculela transformadainversade Laplace de lafunciónsdada
1.) f(s) = 2s + 3
S2
+ 4s + 20
S2
+ 4S + 20
(S2
- 4S + 22
) + 20 – (2)2
(S -2 )2
+ 16
f(s) = 2S + 3 + 4 - 4
(S– 2)2
+ 16
f(s) = 25 – 4 + 7 (4/4)
(S – 2)2
+ 16 (S– 2)2
+ 16

4. f(s) = 2 (S – 2) + 7/4 4
(S – 2)2
+ 16 (S – 2)2
+ 16
∫-1
{S} = 2 ∫-1
{ (S– 2) } + 7/4 ∫ -1
{ 4 }
(S– 2)2
+ 16 (S – 2)2
+ 16
f(s) = 2 e2t
cos 4t + 7/4 e2t
Sen4t
2.) G(s) = 1
S(s+ 1)
∫-1
{G(s)} = ∫ -1
{1/5} . ∫ -1
{ 1/ S + 1}
G(s) = 1 . e-t
= e-t
3.) H(s) = 25
(S2
+ 1)2
∫-1
H(s) = 2 ∫-1
{ s }
(S2
+ 1)2
H(t) = 2 t cos t
4.) P(s) = 1
S(S2
+ 16)
∫{p(s)} = ∫-1
{ ⅕} ∫-1
{ 1 }
(S + 2)2
p(t) = 1 . t e-2t
p(t) = t e-2t
5.) Q(s) = 1 , n€z
(s– a)
∫ -1
{ Q ( S !) } = ∫ -1
{ 1 }
(S– a)n
Q(t) = tn-1
eat
(n – 1)!
6.) R(s) = 3s2
(s + 1)2
∫-1
{R(s)} = ∫-1
{ 3s2
}
(s+ 1)2
R(t) = 3 ∫-1
{ S2
}
(s+ 1)2
R(t) = 3 . t et
7.) R(s) = 2 [ 1 + 3 + 4 ]
s s2
s6
R(s) = 2 + 6 + 8
S5
56
510
∫-1
{ R(s) } = 2 ∫-1
{ 1/s5
} + 6 ∫-1
{1 /s6
} + 8 ∫-1
{1 /s10
}
R(t) = 2 t4
/4! + 6 t5
/s! + 8 t9
/9!

5. 8.) Q(s) ={ s – 4 } + { s }
(S2
+ 5)2
(S2
+ 2
∫-1
{Q(s)} = ∫ -1
{ { s – 4 } + ∫ -1
{ s }
(S2
+ 5)2
(S2
+ 2)
Q(t) = ∫-1
{ s } - 4 ∫-
{ 1 } + ∫-1
{ s }
(S2
+ 5)2
(S2
+ 5)2
(S2
+ 2)
Q(t) = t cos √5 t - 4 t e-5t
+ cos √2 t
2√5
9.) P(s) = 1
S2
– 4s + 5
(s2
– 4S + 22
) + 5 - 22
(s – 2)2
+ 1
∫-1
{ P(s) } = ∫-1
{ 1 }
(5 – 2)2
+ 1
P(t) = e 2t
sent
10.)H(s) = S – 3
S2
+ 10s + 9
S – 3
(s + 9) (s + 1)
∫-1
{ H(s) } = ∫ -1
{ s – 3 }
(s+ 9) (s+ 1)
s – 3 = A + B
(s + 9) (s+ 1) s + 9 s + 1
S – 3 = A(s+ 1) + B(s + 9)
(s – 9)( s + 1) (s – 9)( s + 1)
S – 3 = A(s+ 1) + B(s+ 9)
Si s – 1
- 1 – 3 = A (-1 +1) + B(- 1 + 9)
- 4 = 8B
B = - ¼
Si s = - 9
- 9 – 3 = A( -9 + 1) + B(-9+ 9) – 12 = - 8A
A = 12/8 = 3/2
∫-1
{ H(s) } = 3 / 2 ∫ -1
{ 1 } - ¼ ∫-1
{ 1 }
S + 9 s + 1
H(s) = 3/2 e-9t
- ¼ e-6
11.) G(s) = s
S2
- 14s + 1
(S2
– 14s + (7)2
) + 1 - (7)2
(s – 7)2
– 48
∫-1
{ G(s) } = ∫ -1
{ s } ( 7 – 7)
(s – 7)2
– 48

6. ∫-1
{ s – 7 } - 7 ∫-1
{ 1 } √48
(s – 7)2
– 48 ( s – 7)2
– 48 √48
∫-1
{ s – 7 } - 7/48 ∫ -1
{ √48 }
(s– 7)2
– 48 ( s – 7)2
- 48
G(s) = e7t
cos √48 t – 7/√48 e7t
sen√48 t
12.)F(s) = e-4s
S ( s2
+16)
∫-1
{f(s)} = ∫-1
{ e+4s
}
S( s2
+ 16)
F(t) = 1
( s + 4)
13.) F(s) = { e-s
}
(s – s)3
∫-1
F(s) = ∫-1
{ e-s
}
(s– s)3
F(t) = u (t2
+ s)
14.) G(s) = s e-cos
(s2
+ 4)2
∫-1
{G(s)} = ∫ -1
s e-10s
(s2
+ 4)2
G(t) = t2
cos2
+ 2 t
( s + 10)
15.)H(s) = s2
-2s + 3
S(s2
– 3s + 2)
S2
– 2s + 3 = A + B + C
S( s – 2) (s – 1) S s-2 S-1
S2
– 2s + 3 = A( s-2)(s-1) + Bs(S-1) + cs(s-2)
S = 2
4 – 4 + 3 = 2B B = 3/2
S = 1
1 – 2 + 3 = - C C=2
S = 0
3 = 2A A = 3/2
∫-1
{ F(s)}=3/2 ∫ -1
{1/5} + 3/2 ∫ -1
{ 1 } + 2 ∫-1
{ 1 }
S – 2 s – 1
F(t) = 3/2 + 3/2 e2t
+ 2et

7. 16.) P(s) = 4s – 5
S3
– S2
– 5s – 3
∫-1
{ P(s) } = ∫-1
{ 4s – 5 }
S3
– S2
– 5s – 3
S3
– S2
– 5s – 3
1 -1 -5 -3
-1 -1 2 3
1 -2 -3 0
-1 -1 3
1 -3 0
3 3
1 0
4s – 5 4s – 5 = A + B + C
(s + 1) (s + 1)(s – 2) = (s+ 1)2
(s– 2) s + 1 (s + 1) s – 2
4s – 5 = A(s+1) (s – 2 + (s - 2) + C (s + 1)2
4s – 5 = A (S2
– s + s – 2) + B5 – 2B + c (S2
+ 2s + 1)
4s – 5 = A s2
– 2s +2A + B5 – 2B – 2B + CS2
+ 2cs + c
4s – 5 = (A+ C) s2
+ (-A + B +2C)S + (-2A – 2B +C)
A + C = 0
- A + B + 2c + 4 - 2A + 2B + 4C = 8
- 2A – 2B + C = - 5 - 2A – 2B + C = -5
- 4A + 5 C = 5
A + C = 0 4A + 4C = 0
-4A +5C + 2s - 4A + 5C = 5
9C = 5 C = 5/9
A = - C = – 5/9
B = A – 2C + 4
B = - 5/9 – 10/9 + 4 = 7/3
∫-1
{P(s) } = 5/9 ∫-1
{ 1 } + 7/3 ∫ -1
{ 1 } + 5/9 ∫-1
{ 1 }
S + 1 (s+ 1)2
S – 2
P(t) = - 5/9 e-t
7/3 t e-t
5/9 e2t

8. 17.) Q(s) = -S
(s – 4)2
(s– 5)
- S = A + B + C
(s– 4)2
(s – 5) s – 4 (s-4)2
s – 5
-S = A(S– 4(S – 5) + B(s– 5) + C(S– 4)2
-S = A(S2
- 5S -4S +20) + B5 – 5B – C(S2
– 8S + 16)
-S = A S2
– 9 AS + 20A + B5 – 5B + Cs2
– 8cs + 16 C
-S = (A +C) S2
+ ( -9A +B – 8C) + S (20A – 5B +16C)
A +C = 0
-9A + B – 8C = -1 - 45A + 5B – 40C = -5
20A – 5B+ 16C = 0 20A – 5B + 16 C = 0
-25 A -24C = -5
25 A + C = 0
C = - 5
A = - C = - ( - 5) = 5
B = - 1 + 8C + 9A = - 1 + 8 (-5) + 9 (5) + 9(5) = 4
∫-1
{Q(s) } = S ∫-1
{ 1 } +4 ∫ -1
{ 1 } -5 ∫-1
{ 1 }
S - 4 (S – 4)2
s – 5
Q(t) = 5 e4t
+ 4 t e4t
-5 e5t
18.) R(S) = S2
+ 4s + 1
(S – 2)2
( 5+ 3)
S2
+ 4s +1 = A + B + C
(s – 2)2
(s + 3) s – 2 (s – 2)2
s + 3
S2
+ 4s + 1 = A( s – 2) (s + 3) + B( s + 3) + (s + 2)2
S2
+ 4s + 1 = A S2
+ As – 6 A + B5 + 3B +CS2
– 4Cs +4C
S2
+ 4s + 1 = (A +C) S2
+ (A + B – 4C)S + (- 6A + 3B + 4C)
A+C = 1
-3 A + B – 4C = 4 - 3A – 3B +12C = -12
-6 A + 3B +4C = 1 - 6A + 3B + 4C = 1
- 9A + 16 C = - 11
9 A + C = 1
-9A + 16 C = -11
9A + 9C = 9
25 C = 2
C = 2/25

9. A = 1 – C
A = 1 – 2/25 = 23/25
B = 4 + 4C – A
B = 4 + 4 ( 2/25) -25/25 =17/5
∫ -1
{R(s)} = 25/25 ∫ -1
{ 1 } + 17/5 ∫ -1
{ 1 } 2/25 ∫-1
{ 1 }
S – 2 (S – 2)2
S + 3
R(t) = 23/25 e2t
+ 17/5 e2t
+ 2/25 e-3t
19.) R(s) = S
(S2
+ a2
(S2
– b2)
S = A5+ B + C5 + D
(S2
+ A2
)(S2
–b2
) (S2
+ A2
) (S2
– b2
)
S = (As+ B) (S2
– B2
) + (CS+ D)(S2
+ A2
)
S = AS3
– B2
A S + B S2
– b2
B+ CS3
+ C A2
S + Ds2
+ A2
D
S = (A + C) S2
+ (B +D) S2
+ (- b2
A + C A2
) S ( - b2
B + A2
D)
A + C = 0
B + D = 0
-b2
A + C a2
= 1
-b2
B + A2
D = 0
B2
A + C = b2
A + b2
C = 0
-b2
A + CA2
= 1 -b2
A + C A2
= 1
C = ( B2
+ a) = 1
A = - C = - 1 C = 1
b2
+ a2
b2
+ a2
B2
B + D = 0 b2
B + b2
D = 0
-B2
+ A2
D = 0 -B2
B + a2
D = 0
D(B2
+ a2
) = 0
D = 0
B = -D = 0
∫-1
{R(s)} = - 1 ∫ -1
{ 1 } ∫-1
{ a/b + 1 } ∫-1
{ 1 }
B2
a2
s2
+ a2
b2
+ a2
S2
– B2
R(t) = - { 1 } ∫-1
{ a } + { 1 } ∫ -1
{ S }
a (b2
+ a2
) S2
+ a2
b2
+ a2
S2
– b2
R(t) = - 1 sena t + 1 cos h (bt)
A(b2
+ a2
) b2
+ a2

10. 20.) Q(s) = 1
(s + 2) (S2
– 9)
1 = A + B + C
(s+ 2) ( s – 3) ( s + 3) s + 2 s – 3 s + 3
1 = A(S– 3) (s + 3) + B (S + 2) (S+ 3) + C( S + 2) (S – 3)
S = 3
1 = 30B B = 1/30
S = - 3
1 = 6C C = 1/6
S = - 2
1 = - 5A A = -1/5
∫-1
{R(s)} = -1/5 ∫-1
{ 1 } + 1/30 ∫ -1
{ 1 } 1/6 ∫-1
{ 1 }
S + 2 S – 3 S – 3
Q(t) = -1/5 e-2t
+ 1/30 e3t
+ 1/6 e-3t
21.) P(s) = 2
S3
(S2
+ 5)
2 = A + B + C + Ds + E
S3
(S2
+ 5) S S2
S3
S2
+ 5
2 = AS2
(S2
+ 5) + B5 (S2
+ 5) + C(S2
+ 5) + (Ds + E) S3
2 = AS4
+ 5 As2
+ bs3
+ 5Bs + CS2
+ 5C + DS4
+ ES3
2
2 = (A +D) S4
+ (B + E) S3
+ ( SA + C) S2
(5B) S +(5C)
A + D = D = - A D = 2/25
B + E = 0 E = - B = 0
5 A + C = A = - C/5 = -2/5/5 = 2/25
5B = 0 B = 0
5C = 2 E = 2/5
∫-1
{R(s)} = - 2/25 ∫ -1
{ 1/5 } + 2/5 ∫-1
{ 2/2 } + ∫-1
{ 4 25 S }
S2
+ 5
P(t) = - 2/25 ∫ -1
{ 1/5 } + 1/5 ∫-1
{ 2/s3
} + 2/25 ∫-1
{ S }
S2
+ 5
P(t) = -2/25 + 1/5 t2
+ 2/25 cos √5 t

11. En losSiguientesproblemasresuelvalassiguientesecuacionesdiferenciales.
25.) y” + y = e-2t
sen t y(0) = 0
y´ (0) = 0
∫ { y”} * ∫-1
{ y } = ∫ { e-2t
sent}
s y s – s (0) – y´(0) + y s = 1
(s+ 2)2
+ 1
ys (s+ 1) = 1
(s + 2)2
+ 1
ys = 1 = a + bs + c
(s+ 1) [(s+ 1)2
+1] S + 1 ( S + 2)2
+ 1
1 = a [ ( s + 2)2
+ 1] + (bs+ c) (S+1)
1 = a(s + 2)2
+ a + bs2
+ BS + cs + + c
1 = a( S2
+ 4s + 4) + a + BS2
+ bs + cs + c
1 = aS2
+ 4as + 4a + a bs2
+ bs + cs + c
1 = (a + b) s2
+ (4a + b + c) S + (5a +c)
A + B = 0
-1 4A + B + C = 0 -4A – B – C = 0
5A + C = 1 5A + C = 1
A - B = 1
A + B = 0
2A = 1 A = ½
B = - A = -1/2
C = 1-5A = 1 – S (1/2) = 1 – s/2 = -3/2
∫-1
{F(s)} = ½ ∫ -1
{1 / s + 1} + ∫ -1
-1/5 – 3/2
(s + 2) + 2}
F(s) = ½ ∫ -1
{ 1 } -1/2 ∫-1
{ S – 3 }
s + 1 s + 2 )2
+ 1
F(t) = ½ ∫-1
{ 1 } -1/2 ∫-1
{ s + 2 – 2 } + 3/2 ∫-1
{ 1 }
S + 1 ( s + 2)2
+ 1 (s+ 2)2
+ 1
F(t) = ½ ∫-1
{ 1 } -1/2 ∫-1
{ S + 2 } - ½ ∫-1
{ 1 }
( S + 2)2
+ 1 (s + 2)2
+ 1
F(t) = ½ e-t
-1/2 e-t
cos t – ½ e-t
Sent
26.) y” + 4 y´ + 5 y + 2y = 10 cos t s (0) = y´ (0) =
s” (0) = 3
∫ { y”} + 4 ∫ { y”} + 5 ∫ { y´ } + ∫ { y } = 10 ∫cos E

12. s3
ys– S2
y(0) – S y´(0) – y“(0) + 4 { S2
ys - S y (0) – y´(0) } + 5 [ 5 y s – y(0)] + 2 y s = 10 * s / s2
+ 1
s3
y s – 3 + 4 S2
y s + 5 s y s + 2 y s + 2 ys = 10 S / s2
+ 1
s (S3
+ 4 S2
+ 5s + 2) = 10S / s2
+1 + 3
ss = 10S / S2
+1 +3 = 10s + 3 (s2
+ 1) / s2
+ 1
S3
+ 4S2
+ 5 S + 2 s3
+ 4s2
5s + 2
ss = 10S + 3S2
+ 3
(s2
+ 1) (s3
+ 4s2
+ 5s + 2)
1 4 5 3
-2 -2 -4 -2
1 2 1 0
-1 -1 -1
1 1 0
-1 -1
0
10S + 3S2
+3 = AS + B + C + D + t
(S2
+ 1) ( s+ 2) ( S + 1)2
S2
+ 1 S + 2 (S+ 1) (S+ 1)2
10S + 3S2
+3 = (AS + B) ( s+ 2) ( S + 1)2 + C (S2 + 1) ( S + 1)2 + D (S2 + 1) ( s+ 2) ( S + 1) E (S2 + 1) ( s+ 2)
(S2
+ 1) ( s+ 2) ( S + 1)2
(S2
+ 1) ( s+ 2) ( S + 1)2
10S + 3S2 +3 =(AS + B) ( s+ 2) (S2 + 2s + 1) + C (s2 + m1 (S2 + 2s + 1) + D (S3 + 2s2 + S + 2) + E (S2 + 2s2 + S + 2)
10S + 3S2 +3 = (AS2
+ 2AS + BS + 2B) (S2
+ 2s + 1) + C (S4
+ 2S3
+ s2
+ S6
+ 2S + 1) + D (S4
+ S3
+ 2S3
+ 2s2
+ S2
+ S + 25 + 2) + E (S2
+ 2S2
+ S + 2).
10S + 3S2 +3 = AS4
+ 2AS3
+AS2
+2AS3
+ 2AS + BS3
+ 2BS2
+ 2B+ 2BS2
+ 2BS + 2B + CS4
+ 2C S3
+CS2
+ CS2
+ 2CS + C + DS4
+ DS3
+2DS3
+2DS2
+DS2
+ DS + 2DS+ 2D + ES2
+ 2 ES2
+ ES+ 2E.
10S + 3S2 +3 = ( A + C +D) S4
+ ( 4A + B + 2C + 3D) S3
+ (3A + 4B + 2C +3D + 3E) S2
+ (2A + 2B+ 2C+
3D+ E) S + (4B + C+ 2D +2E)
A + C + D = 0
4A + B+ 2C+ 3A= 0
3A+ 4B+ 2C + 3D + 3E =3
2A + 2B + 2C + 3C + 3ª E = 10
4B + C +2D + 2E = 3
-4 4A + B + 2C+ 3D + = 0

13. 3A + 4B + 2C + 3D + 3E = 3
-12 A – 4B – 8C – 12D = 0
3A + 4B + 2C + 3D + 3E = 3
-13A – 6C – 9D + 3E = 3
-2 2A + 2B +2C +3D + E =10
4B + C + 2D +2E = 3
-4A – 4B – 4C – 6 D – 2E = - 20
- 4B + C + 2D + 2E = 3
-4A – 3C – 4D = -17
-1 3A + 4B +2C + 3D + 3E = 3
4B + C + 2D + 2E = 3
-3A + 4B + 2C +3D – 3E = -3
4B + C + 2D + 2E = 3
3 -3A – C – D – E = 0
-13A – 6C – 9D + 3E =3
- 9A – 3 C 3D -3E = 0
- 13A – 6C – 9d + 3E = 3
- 22A – 9C – 12D = 3
-3 -4A – 3C – 4D = -17
- 22A – 9C -12D = 3
12A + 9C + 12D = 51
10A = 54 A = - 27/5
-12 A + C + D = 0 - 12A -12C -12D = 0
12A + 9C + 12D = 51 12ª + 9C + 12D = 51
-3C = 51
C = - 17
D = - A –C = 27/5 + 17 = 112/5
B = 4A -2C – 3D
B = -4 (-27/5) – 2(-17) – 3 (112/5) = -58/5
E = 3 – 4B – C – 2D = 3 – 4 ( -58/5) – (-17) – 2(112/5)
2 2
E = 54/5
∫-1
-27/s s – 58/5 -17 ∫ -1
{1 / S+2} + 112/5 ∫-1
{1 / S+1} + 54/5 ∫-1
{1 / S+1}2
s2
+ 1
-27/5 ∫ -1
{S/ S2
+2} - 58/5 ∫-1
{1 / S2
+2} -17 ∫-1
{1 / S+2}
y (t) = -27/5 Cos t – 58/5 Sent – 17 e-t
+ 112/5 et
+ 54/5 t et
27) y ‘’+ 4 y’+8y= sen t y (0) =1

14. y (0) =0
∫ {y’’}+4∫{y ‘}+ 8∫ {y}= ∫ {sen t}
s2y - sy (0) – y’(0)+ 4 [sy - y (0)] + 8ys = 1
s2 + 1
s2∫s - s∫(0) - ∫1(0) + 4s∫ - 4∫(0) + 8ys = 1
s2 + 1
∫s (s2 – s + 4 + 8) = 1
s2 + 1
∫s= 1
(s2 + 1) (s2 – s + 12)
1 = as + b + cs + d
(s2 + 1) (s2 + 12) s2 + 1 s2 –s + 12
1 = (as + b) (s2 – s + 12) + (cs + d) (s2 + 1)
1 = as3 – as2 + 12 as + bs2 – bs + 12 b + cs3 + cs + ds2 + d
1 = (a + c) s3 + (- a + b + d) s2 + (12a – b + c) s + (12b + d)
{
a + c = 0
−a + b + d = 0
} (-1) {– 𝑎 + 𝑏 + 𝑑 = 0
12𝑏 + 𝑑 = 1
}
12𝑎 − 𝑏 + 𝑐 = 0 𝐴 − 𝑏 − 𝑑 = 0
12𝑏 + 𝑑 = 1 12𝑏 + 𝑑 = 1
11 {
𝑎 + 11𝑏 = 1
12𝑎 − 𝑏 + 𝑐 = 0
}
𝑎 + 11𝑏 = 1
132a – 11b + 11c =0
133a + 11c =1
-11{
a + c = 0
133a + 11c = 1
} -11a – 11c =0
133a + 11c =0
122a=1
a= 1
122
c= -a= -1
122
-12{
−a + b + d = 0
12𝑏 + 𝑑 = 1
}
12𝑎 − 12𝑏 − 12𝑑 = 0
12𝑏 + 𝑑 = 1
12𝑎 − 11𝑑 = 1
d=12a-1 = 12(1/122) – 1= - 5
11 11 61
𝑏 = 12𝑎 + 𝑐 = 12 (
1
122
) −
1
122
=
11
122
∫-1 {
1
122
s +
11
122
s2 + 1
} + ∫-1{
−
1
122
s −
s
61
s2 − s + 12
}

15. 1/122 ∫-1 {
𝑆
s2 + 1
} + 11/122 ∫-1 {
1
s2 + 1
} - 1/22 ∫-1 {
𝑆 −
1
2
+
1
2
(𝑆 −
1
2
)
2
+
47
4
}
s/61 ∫-1 {
1
(S−
1
2
)
2
+
47
4
}
1/122 ∫-1 {
𝑆
s2 + 1
} + 1/122 ∫-1 {
𝑆
s2 + 1
} – 1/22 ∫-1 {
s − 1
(s−
1
2
)
2
+
47
4
}
71/122 ∫-1 {
1
(s−
1
2
)
2
+
47
4
}
1/122 cos t + 1/122 sen t -1/22 t cos √
47
2
𝑡 + 71/122 t sen √
47
2
𝑡
28) 𝒚′ − 𝟐𝒚 = 𝟏 − 𝒕 𝒚( 𝟎) = 𝟏
∫ { 𝑦′}- 2 ∫{ 𝑦} = ∫ {1} - ∫{ 𝑡}
s y s – y(0) – 2ys =
1
5
−
1
52
y s (s – 2) =
𝑆−1
𝑆2
+ 1 ys (s – 2) =
𝑆−1+𝑆2
𝑆2
ys =
𝑆2+𝑆−1
𝑆2 ( 𝑆−2)
𝑠2 + 𝑠 − 1
𝑠2 ( 𝑠 − 2)
=
𝑎
𝑠
+
𝑏
𝑠2 +
𝑐
𝑠 − 2
𝑠2 + 𝑠 − 1 = 𝑎𝑠( 𝑠 − 2) + 𝑏( 𝑠 − 2) + 𝑐𝑠2
s=0
-1 = -2b b=1/2
s=2
4+2-1= 4c c=5/4
𝑠2 + 𝑠 − 1 = 𝑎𝑠2 − 2𝑎𝑠 + 𝑏𝑠 − 2𝑏 + 𝑐𝑠2
𝑠2 + 𝑠 − 1 = ( 𝑎 + 𝑐) 𝑠2 + (−2𝑎 + 𝑏) 𝑠 − 2𝑐
a+c=1 = a=1-c
-2a+b=1 a=1-
𝑆
4
= −
1
4
-2b= -1
−
1
4
∫
−1
{
1
𝑠
} +
1
2
∫
−1
{
1
s2
}+
5
4
∫
−1
{
1
s − 2
}

16. −
1
4
+
1
2
𝑡 +
𝑠
4
𝑒2𝑡
29) 𝒚′′ − 𝟒𝒚′+ 𝟒𝒚 = 𝟏 𝒚( 𝟎) = 𝟏 𝒚′( 𝟎) = 𝟒
∫ { 𝑦′′}− 4∫ { 𝑦′}+ 4∫ { 𝑦} = ∫ {1}
𝑠2 𝑦𝑠 − 𝑠𝑦(0) − 𝑦′(0) − 4( 𝑠𝑦𝑠 − 𝑦(0)) + 4𝑦𝑠 =
1
5
𝑠2 𝑦𝑠 − 𝑠 − 4 − 4𝑠𝑦𝑠 + 4 + 4𝑦𝑠 =
1
5
𝑦𝑠( 𝑠2 − 4𝑠 + 4) =
1
5
+ 𝑠
𝑦𝑠 =
1 + 𝑠2
𝑠( 𝑠2 − 4𝑠 + 4)
1 + 𝑠2
𝑠( 𝑠 − 2)(𝑠 − 2)
=
1 + 𝑠2
𝑠( 𝑠 − 2)2 =
𝑎
5
+
𝑏
𝑠 − 2
+
𝑐
( 𝑠 − 2)2
1 + 𝑠2 = 𝑎( 𝑠 − 2)2 + 𝑏𝑠( 𝑆 − 2) + 𝑐𝑠
1 + 𝑠2 = 𝑎( 𝑠2 − 4𝑠 + 4) + 𝑏𝑠2 − 2𝑏𝑠 + 𝑐𝑠
1 + 𝑠2 = 𝑎𝑠2 − 4𝑎𝑠 + 4𝑎 + 𝑏𝑠2 − 2𝑏𝑠 + 𝑐𝑠
1 + 𝑠2 = ( 𝑎 + 𝑏) 𝑠2 + (−4𝑎 − 2𝑏 + 𝑐) 𝑠 + 4𝑎
𝑎 + 𝑏 = 1 𝑏 = 1 − 𝑎 = 1 −
1
4
=
3
4
−4𝑎 − 2𝑏 + 𝑐 = 0
4𝑎 = 1 𝑐 = 4𝑎 + 2𝑏
𝑎 =
1
4
𝑐 = 4(
1
4
) + 2(
3
4
)
𝑐 = 1 +
6
4
=
10
4
1
4
∫
−1
{
1
5
} +
3
4
∫
−1
{
1
s − 2
} +
10
4
∫
−1
{
1
(s − 2)2
}
1
4
+
3
4
𝑒2𝑡 +
10
4
𝑡 𝑒2𝑡
30) 𝒚′′ + 𝟗𝒚 = 𝒕 𝒚( 𝟎) = 𝒚′( 𝟎) = 𝟎
∫ {y′′}+ 9 ∫ {y} = ∫ {t}
𝑠2 𝑦𝑠 − 𝑠𝑦(0) − 𝑦′(0) + 9𝑦𝑠 =
1
𝑆2
𝑦𝑠( 𝑠2 + 9) =
1
𝑠2

17. 𝑦𝑠 =
1
𝑠2( 𝑠2 + 9)
=
𝑎
𝑠
+
𝑏
𝑠2 +
𝑐𝑠 + 𝑑
𝑠2 + 9
1 = 𝑎𝑠( 𝑠2 + 9) + 𝑏( 𝑠2 + 9) + (𝑐𝑠 + 𝑑)𝑠2
1 = 𝑎𝑠3 + 9𝑎𝑆 + 𝑏𝑠2 + 9𝑏 + 𝑐𝑠3 + 𝑑𝑠2
1 = ( 𝑎 + 𝑐) 𝑆3 + ( 𝑏 + 𝑑) 𝑠2 + 9𝑎𝑠 + 9𝑏
𝑎 + 𝑐 = 0 𝑐 = 0
𝑏 + 𝑑 = 0 𝑑 = −
1
9
9𝑎 = 0 𝑎 = 0
9𝑏 = 1
𝑏 =
1
9
0∫
−1
{
1
s
}+
1
9
∫
−1
{
1
s2
}−
1
9
∫
−1
{
1
s2 + 9
}3/3
1
9
∫
−1
{
1
s2
}−
1
2t
∫
−1
{
3
s2 + 9
}
1
9
𝑡 −
1
2𝑡
𝑆𝑒𝑛 3𝑡
31) 𝒚′′ − 𝟏𝟎𝒚′+ 𝟐𝟔𝒚 = 𝟒 𝒚( 𝟎) = 𝟑 𝒚′( 𝟎) = 𝟏𝟓
∫ {y′′}− 10∫ {y′}+ 26∫ {y} = 4∫ {1}
𝑠2 𝑦𝑠 − 𝑠𝑦(0) − 𝑦2(0) − 10{ 𝑠𝑦𝑠 − 𝑦(0)} + 26𝑦𝑠 = 4
1
5
𝑠2 𝑦𝑠 − 35 − 15 − 10𝑆𝑦𝑠 + 30 + 26𝑦𝑠 =
4
5
𝑠2 𝑦𝑠 − 10𝑠𝑦𝑠 + 26𝑦𝑠 =
4
5
+ 35 + 15
𝑦𝑠( 𝑠2 − 10𝑆 + 26) =
4 + 352 + 15𝑠
𝑠
𝑦𝑠 =
4 + 352 + 15𝑠
𝑠( 𝑠2 − 10𝑠 + 26)
=
𝑎
𝑠
+
𝑏𝑠 + 𝑐
𝑠2 − 10𝑠 + 26
4 + 352 + 15𝑠 = 𝑎( 𝑠2 − 10𝑠 + 26) + ( 𝑏𝑠 + 𝑐) 𝑠
4 + 352 + 15𝑠 = 𝑎𝑠2 − 10𝑎𝑠 + 26𝑎 + 𝑏𝑠2 + 𝑐𝑠
4 + 352 + 15𝑠 = ( 𝑎 + 𝑏) 𝑠2 + (−10𝑎 + 𝑐) 𝑠 + 26𝑎
𝑎 + 𝑏 = 3 𝑏 = 3 − 𝑎
−10𝑎 + 𝑐 = 15 𝑏 = 3 −
2
13
=
37
13
26𝑎 = 4
𝑎 =
2
13

18. 𝑐 = 15 + 10𝑎 = 15 + 10(
2
13
) =
215
13
2
13
∫
−1
{
1
5
} + ∫
−1
{
37
13
s +
215
13
(s − 5)2 + 1
}
2
13
∫
−1
{
1
5
} +
37
13
∫
−1
{
s − 5 + 5
(s − 5)2 + 1
} +
215
13
∫
−1
{
1
(s − 5)2 + 1
}
2
13
∫
−1
{
1
5
} + 37∫
−1
{
s − 5
(s − 5)2 + 1
} +
280
13
∫
−1
{
1
(s − 5)2 + 1
}
2
13
+
37
13
𝑡 𝐶𝑜𝑠 𝑡 +
280
13
𝑡 𝑆𝑒𝑛 𝑡
32) 𝒚′′ − 𝟔𝒚′+ 𝟖𝒚 = 𝒆𝒕 𝒚( 𝟎) = 𝟑 𝒚′( 𝟎) = 𝟗
∫ {y′′}− 6∫ {y′}+ 8∫ {y} = ∫ {et}
𝑆2 𝑦𝑠 − 𝑆𝑦(0) − 𝑦′(0) − 6{ 𝑆𝑦𝑠 − 𝑦(0)} + 8𝑦𝑠 =
1
( 𝑠 − 1)
𝑆2 𝑦𝑠 − 3𝑠 − 9 − 6𝑠𝑦𝑠 + 18 + 8𝑦𝑠 =
1
( 𝑠 − 1)
𝑆2 𝑦𝑠 − 6𝑆𝑦𝑠 + 8𝑦𝑠 − 3𝑠 + 9 =
1
( 𝑠 − 1)
𝑦𝑠( 𝑠2 − 6𝑠 − 8) =
1
𝑆 − 1
+ 35 − 9
𝑦𝑠( 𝑠2 − 6𝑠 + 8) =
1 + 3𝑠( 𝑠 − 1) − 9( 𝑠 − 1)
𝑠 − 1
𝑦𝑠 =
1 + 3𝑠2 − 3𝑠 − 9𝑠 + 9
( 𝑠 − 1)( 𝑠2 − 6𝑠 + 8)
𝑦𝑠 =
3𝑠2 − 12𝑠 + 10
( 𝑠 − 1)( 𝑠 − 4)( 𝑠 − 2)
=
𝑎
𝑠 − 1
+
𝑏
𝑠 − 4
+
𝑐
𝑠 − 2
3𝑠2 − 12𝑠 + 10 − 𝑎( 𝑠 − 4)( 𝑠 − 2) + 𝑏( 𝑠 − 1)( 𝑠 − 2) + 𝑐( 𝑠 − 1)( 𝑠 − 4)
𝑠 = 4
10 = 6𝑏 = 𝑏 =
10
6
𝑠 = 2
−2 = −2𝑐 = 𝑐 = 1
𝑠 = 1
1 = 6𝑎 = 𝑎 =
1
6

19. 1
6
∫
−1
{
1
s − 1
} +
10
6
∫
−1
{
1
s − 4
} + ∫
−1
{
1
s − 2
}
1
6
𝑒 𝑡 +
10
6
𝑒4𝑡 + 𝑒2𝑡
33) 𝒚′′ + 𝟒𝒚 = 𝒆−𝒕 𝑺𝒆𝒏 𝒕 𝒚( 𝟎) = 𝟏 𝒚′( 𝟎) = 𝟒
∫ {y′′}+ 4∫ {y} = ∫ {e−t Sen t}
𝑠2 𝑦𝑠 − 𝑠𝑦(0) − 𝑦′(0) + 4𝑦𝑠 =
1
( 𝑠 + 1)2 + 1
𝑦𝑠( 𝑠2 + 4) =
1 + 𝑠{( 𝑠 + 1)2 + 1} + 4{( 𝑠 + 1)2 + 1}
( 𝑠 + 1)2 + 1
𝑦𝑠 =
1 + 𝑠( 𝑠2 + 25 + 1) + 𝑠 + 4( 𝑠2 + 2𝑠 + 1) + 4
{( 𝑠 + 1)2 + 1}( 𝑠2 + 4)
𝑦𝑠 =
1 + 𝑠3 + 2𝑠2 + 𝑠 + 𝑠 + 4𝑠2 + 8𝑠 + 4 + 4
{( 𝑠 + 1)2 + 1}( 𝑠2 + 4)
𝑌𝑠 =
𝑠3 + 6𝑠2 + 10𝑠 + 9
{( 𝑠 + 1)2 + 1}( 𝑠2 + 4)
𝑠3 + 6𝑠2 + 10𝑠 + 9
{( 𝑠 + 1)2 + 1}( 𝑠2 + 4)
=
𝑎𝑠 + 𝑏
( 𝑠 + 1)2 + 1
+
𝑐𝑠 + 𝑑
𝑠2 + 4
𝑠3 + 6𝑠2 + 10𝑠 + 9 = ( 𝑎𝑠 + 𝑏)( 𝑠2 + 4) + ( 𝑐𝑠 + 𝑑){( 𝑠 + 1)2 + 1}
𝑠3 + 6𝑠2 + 10𝑠 + 9 = 𝑎𝑠3 + 4𝑎𝑠 + 𝑏𝑠2 + 4𝑏 + ( 𝑐𝑠 + 𝑑)( 𝑠2 + 2𝑠 + 2)
𝑠36𝑠2 + 10𝑠 + 9 = 𝑎𝑠3 + 4𝑎𝑠 + 𝑏𝑠2 + 4𝑐 + 𝑐𝑠3 + 2𝑐𝑠2 + 2𝑐𝑠 + 𝑑𝑠2 + 2𝑑𝑠 + 2𝑑
𝑠36𝑠2 + 10𝑠 + 9 = ( 𝑎 + 𝑐) 𝑠3 + ( 𝑏 + 2𝑐 + 𝑑) 𝑠2 + (4𝑎 + 2𝑐 + 2𝑑) 𝑠 + (4𝑏 + 2𝑑)
𝑎 + 𝑐 = 1
𝑏 + 2𝑐 + 𝑑 = 6
4𝑎 + 2𝑐 + 2𝑑 = 10
4𝑏 + 2𝑑 = 9
−4 {
𝑏 + 2𝑐 + 𝑑 = 6
4𝑏 + 2𝑑 = 9
}
−4𝑏 − 8𝑒 − 4𝑑 = −24
4𝑏 + 2𝑑 = 9
−8𝑐 − 2𝑑 = −15
{
𝑎 + 𝑐 = 1
4𝑎 + 2𝑐 + 2𝑑 = 10
} − 4
−4𝑎 − 4𝑐 = −4

20. 4𝑎 + 2𝑐 + 2𝑑 = 10
2𝑐 + 2𝑑 = 6
−8𝑐 − 2𝑑 = −15
−6𝑐 = −9
𝑐 =
9
6
𝑎 = 1 − 𝑐 = 1 −
9
6
= −
3
6
𝑑 =
−8𝑐 + 15
2
=
−8 (
9
6
) + 15
2
=
3
2
𝑏 =
9 − 2𝑑
4
=
9 − 2(
3
2
)
4
=
6
4
∫
−1
{
3
6𝑠
+
6
4
( 𝑠 + 1)2 + 1
}+ ∫
−1
{
9
6s
+
3
2
s2 + 4
}
−
3
6
∫
−1
{
s + 1 − 1
(s + 1)2 + 1
} +
6
4
∫
−1
{
1
(s + 1)2 + 1
} +
9
6
∫
−1
{
s
s2 + 4
} +
3
2
∫
−1
{
1
s2 + 4
}
2
2
−
3
6
∫
−1
{
s + 1
(s + 1)2 + 1
} +
1
2
∫
−1
{
1
(s + 1)2 + 1
} +
9
6
∫
−1
{
s
s2 + 4
} +
3
4
∫
−1
{
2
s2 + 4
}
−
3
6
𝑒 𝐶𝑜𝑠 𝑡 +
1
2
𝑒−𝑡 𝑆𝑒𝑛 𝑡 +
9
6
𝐶𝑜𝑠 2𝑡 +
3
4
𝑆𝑒𝑛 2𝑡
Dado el siguiente circuito determine la corriente I aplicandoTransformada de Laplace
Malla I:
𝑉𝑅 + 𝑉𝐿 = 𝐸 𝐼𝑅1 + 𝐿
𝐷𝐼𝐿
𝐷𝑇
= 𝐸 𝐿
𝐷𝐼𝐿
𝐷𝑇
+ 𝐼𝑅 = 𝐸 (/𝐿)
∫ { 𝐼1}+ 𝑅 𝐿 ∫
−1
{ 𝐼} = ∫ {
𝐸
𝐿
}
𝑆𝐼 +
𝑅
𝐿
𝐼 =
𝐸
𝐿
.
1
𝑆

21. 𝐼 [ 𝑆 +
𝑅
𝐿
] =
𝐸
𝐿
.
1𝐼
𝑆
𝐼1 =
𝐸
𝐿
𝑆(𝑆 +
𝑅
𝐿
)
=
𝐴
𝑆
+
𝐵
𝑆 +
𝑅
𝐿
𝐸
𝐿
= 𝐴 ( 𝑆 +
𝑅
𝐿
) + 𝐵𝑆
𝑆 = 0
𝐸
𝐿
= 𝐴
𝑅
𝐿
𝐴 =
𝐸
𝑅
𝑆 = −
𝑅
𝐿
𝐸
𝐿
= 𝐵(−
𝑅
𝐿
) 𝐵 = −
𝐸
𝑅
𝐼1 =
𝐸
𝑅
∫
−1
{
1
S
} −
E
R
∫
−1
{
1
S +
R
L
}
𝐼1 =
𝐸
𝑅
−
𝐸
𝑅
𝑒−
𝑅
𝐿
𝑡
Malla II:
𝑉𝐶 + 𝑉𝑅 = 0
𝑉𝐶 + 𝐼2 𝑅 = 0 𝐼𝐶 = 𝐼𝑅2
𝐼
𝐶
𝐼2 + 𝐼2
1
𝑅 = 0
𝐼2
1
𝑅 +
𝐼
𝐶
𝐼2 = 0
𝐼2
1
+
1
𝐶𝑅
𝐼2 = 0
∫
−1
{ 𝐼2
1} +
1
𝐶𝑅
∫{I2} = ∫{0}
𝑆 𝐼2 +
1
𝐶𝑅
𝐼2 = 0

22. 𝐼2 ( 𝑆 +
1
𝐶𝑅
) = 1
∫
−1
{ 𝐼2}= ∫
−1
{
1
𝑆 +
1
𝐶𝑅
}
𝐼2 = 𝑒−
1
𝐶𝑅
𝑡
𝐼 = 𝑒−
1
𝐶𝑅
𝑡
−
𝐸
𝑅
+
𝐸
𝑅
𝑒−
𝑅
𝐿
𝑡