Ecuaciones Diferenciales · 2009. 2. 4. · Ecuaciones Diferenciales, Laboratorio de Transformada de Laplace, Tipo: 1 5 F 9 4 e 6t 5 4 e 2t 21.Seleccionar la opci on que contiene

Ecuaciones DiferencialesLaboratorio de Transformada de Laplace

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Grupo: Matrıcula: Nombre: Tipo:1

1. Cual de las siguientes opciones representa la transformada de Laplace Inversa de la funcion:

F (s) =10

3 + s

A 10 e−3 t

B 10 e3 t

C − 103 e−3 t

D −30 e3 t

2. Indique cual de las opciones siguientes es la ecuacion subsidiaria del problema con condiciones iniciales y(0) = 0 y y′(0) = 3con ecuacion:

18 y − 9 y′ + y′′ = 3 e6 t

A Y (s) = −15+3 s(−6+s) (18−9 s+s2)

B Y (s) = 15+3 s18−9 s+s2

C Y (s) = 15+3 s(−6+s) (18−9 s+s2)

D Y (s) = −15+3 s(−6+s) (18+9 s+s2)

3. Cual de las siguientes opciones representa la transformada de Laplace de la funcion: (Sugerencia: use las definiciones delas funciones de las funciones hiperbolicas mediante exponenciales)

f(t) = t cosh(8 t)

A F (s) = 16 s(−64+s2)2

B F (s) = (64 + s2)−1

C F (s) = 64+s2

−64+s2

D F (s) = 64+s2

(−64+s2)2

4. Cual de las siguientes opciones representa la transformada de Laplace de la funcion:

f(t) = −3 cos(4 t)− 4 sen(4 t)

A F (s) = −12+4 s16+s2

B F (s) = −16+3 s16+s2

C F (s) = −16−3 s16+s2

D F (s) = −12−4 s16+s2


f(t) = e4 t − sen(3 t)

A F (s) = −3−3 s+s2

(4+s) (9+s2)

B F (s) = 3(4−s) (9+s2)

Ecuaciones Diferenciales, Laboratorio de Transformada de Laplace, Tipo: 1 2

C F (s) = −3(4+s) (9+s2)

D F (s) = −21+3 s−s2(4−s) (9+s2)


f(t) ={

2 t si 0 ≤ t < 10 si 1 ≤ t

A 2s2 −

1+ss2 e−s

B 2s2 −

2 (1+s)s2 e−s

C 2s2 −

2s2 e−s

D − 2s2 + 2 (1+s)

s2 e−s

7. Determine A, B, C, D y E para que

Y (s) =As+B

(s2 + C s+D) (s+ E)

sea la ecuacion subsidiaria del problema con condiciones iniciales y(0) = 0 y y′(0) = 4 con ecuacion:

8 y − 6 y′ + y′′ = 4 e2 t

Respuesta:


f(t) =

0 si 0 ≤ t < 45 t si 4 ≤ t < 60 si 6 ≤ t

A −(e−6 s + e−4 s + 6 s e−6 s + 4 s e−4 s

)5s

B −(e−6 s − e−4 s + 6 s e−6 s − 4 s e−4 s

)5s2

C(e−6 s − e−4 s + 6 s e−6 s − 4 s e−4 s

)5s2

D −(e−6 s + e−4 s + 6 s e−6 s + 4 s e−4 s

)5s2

9. Cual de las siguientes opciones representa la transformada de Laplace de la funcion: (Sugerencia:Use identidades trigo-nometricas para convertir un producto en una suma)

f(t) = sen(2 t) sen(7 t)

A F (s) = 12 s(

125+s2 −

181+s2

)B F (s) = s2

(4+s2) (49+s2)

C F (s) = s(

14+s2 + 1

49+s2

)D F (s) = s

(1

25+s2 −1

81+s2

)10. Cual de las siguientes opciones representa la transformada de Laplace de la funcion:

f(t) ={

3 si 0 ≤ t < 50 si 5 ≤ t

A −3 + 3 e−5 s


B 3 s− 3 s e−5 s

C 3s −

3s e−5 s

D − 3s + 3

s e−5 s

E 3− 3 e−5 s

11. Indique cual de las siguientes opciones representa la transformada inversa de la funcion:

F (s) =8 + 6 s

−20 + s+ s2

A f(t) = 229 e−5 t + 32

9 e4 t

B f(t) = 229 e−5 t + 32

9 e−4 t

C f(t) = 329 e4 t + 22

9 e5 t

D f(t) = − 229 e−5 t + 32

9 e−4 t


f(t) = e−5 t + 3 e−t

A F (s) = −16+4 s(−5+s) (−1+s)

B F (s) = 14+2 s(−5−s) (1+s)

C F (s) = −16−4 s(−5−s) (1+s)

D F (s) = 14−2 s(−5+s) (−1+s)

13. Indique cual opcion contiene las ecuaciones subsidiarias que se obtienen cuando se pretende resolver el sistema de EDs:

x′ = −5 + 5x+ 9 y

y′ = 2 t+ 9x+ 5 y

sujeto a las condiciones iniciales:x(0) = 2 y y(0) = −5

A {X(s) = 18+25 s−60 s2+2 s3

s2 (−56−10 s+s2) , Y (s) = 10+43 s−43 s2+5 s3

s2 (56+10 s−s2) }

B {X(s) = 18+25 s−40 s2−2 s3

s2 (−56−10 s+s2) , Y (s) = 10+43 s−7 s2+5 s3

s2 (56+10 s−s2) }

C {X(s) = 18+25 s+50 s2−2 s3

s2 (−56−10 s+s2) , Y (s) = 10+43 s+43 s2−5 s3

s2 (56+10 s−s2) }

D {X(s) = 18+25 s+30 s2+2 s3

s2 (−56−10 s+s2) , Y (s) = 10+43 s+7 s2−5 s3

s2 (56+10 s−s2) }

14. Use la transformada de Laplace para resolver la siguiente ED con condiciones iniciales y(0) = 5 y y′(0) = 0 :

49 y + y′′ = sen(4 t)U2π(t)

A y(t) = 5 sen(7 t) + 133 cos(4 t)U2π(t) + 4

231 cos(7 t)U2π(t)

B y(t) = 5 cos(7 t) + 133 sen(4 t)U2π(t) + 4

231 sen(7 t)U2π(t)

C y(t) = 5 cos(7 t)− 133 sen(4 t)U2π(t) + 4

231 sen(7 t)U2π(t)

D y(t) = 5 cos(7 t) + 133 sen(4 t)U2π(t)− 4

231 sen(7 t)U2π(t)


f(t) = cos(8 t) sen(2 t)


A F (s) = 12 ( −6

36+s2 + 10100+s2 )

B F (s) = 8 s(4+s2) (64+s2)

C F (s) = 636+s2 + 10

100+s2

D F (s) = −636+s2 + 10

100+s2


F (s) =−1 + 9 s−s+ s3

A f(t) = 1 + 5 e−t − 4 et

B f(t) = 1− 5 e−t + 4 et

C f(t) = 1 + 4 e−t − 5 et

D f(t) = 1− 4 e−t + 5 et

17. Indique cual opcion contiene la transformada inversa de Laplace de:

F (s) = (65 + 16 s+ s2)−1e−π s

A f(t) = − e(−8π−8 t) sen(t)Uπ(t)

B f(t) = − e(8π−8 t) cos(t)Uπ(t)

C f(t) = − e(8π+8 t) sen(t)Uπ(t)

D f(t) = − e(8π−8 t) sen(t)Uπ(t)

18. Determine los valores de A, B, C y D para que

F (s) =A

s2+ (

B

s+C

s2) eD s

sea la transformada de Laplace de la funcion:

f(t) ={t si 0 < t ≤ 20 si t > 2

Respuesta:

19. Indique cual de las siguientes opciones es la transformada de Laplace de

f(t) =16t2 sen(3 t)

A F (s) = −9+3 s2

(3+s2)3

B F (s) = 9+3 s2

(9+s2)3

C F (s) = −9+3 s2

(9+s2)3

D F (s) = −9+s2

(9+s2)3

20. Cual de las siguientes opciones representa la transformada Inversa de Laplace de la funcion:

F (s) =3 + s

(−6 + s) (−2 + s)

A 74 e−6 t − 3

4 e−2 t

B 34 e

2 t + 74 e

6 t

C − 54 e

2 t + 94 e

6 t

D 34 e

2 t − 74 e

6 t

E − 94 e−6 t + 5

4 e−2 t


F 94 e−6 t − 5

4 e−2 t

21. Seleccionar la opcion que contiene la transformada de Laplace de

f(t) =−1 e−7 t + e−6 t

t

A F (s) = ln( 7+s6+s )

B F (s) = 7+s6+s

C F (s) = 16+s −

17+s

D F (s) = − 1−7+s + 1

−6+s


−14 y − 5 y′ + y′′ = t5 e7 t

A Y (s) = 120(−2+s) (7+s)7

B Y (s) = 5(−7+s)7 (2+s)

C Y (s) = 120(−7+s)7 (2+s)

D Y (s) = 120(−7+s)6 (2+s)


f(t) = cosh(t) senh(t)

A F (s) = − 1(2−s) (2+s)

B F (s) = 1(2−s) (2+s)

C F (s) = − s(−1+s) (1+s) (1+s2)

D F (s) = 2−s2(2−2 s+s2) (2+2 s+s2)

E F (s) = s(−1+s)2 (1+s)2

24. Determine A, B, C , D y E para que

Y (s) =As3 +B s2 + C s+D

s2 (s2 + E)


64 y + y′′ = 64 t

Respuesta:





36 y + y′′ = 4 cos(6 t)

A Y (s) =s (144+4 s2)

(36+s2)2

B Y (s) =s (148+4 s2)

(36+s2)2

C Y (s) =s (148−4 s2)

(36+s2)2

D Y (s) = 148+4 s2

(36+s2)2



A F (s) = 12 (−(1 + s2)−1 + 13

169+s2 )

B F (s) = 7 s(36+s2) (49+s2)

C F (s) = −(1 + s2)−1 + 13169+s2

D F (s) = (1 + s2)−1 + 13169+s2


x′ = 3x+ 6 y

y′ = 6x+ 3 y

sujeto a las condiciones iniciales:x(0) = 2 y y(0) = 4

A {X(s) = 2 (−15+s)−27−6 s+s2 , Y (s) = 4 (−6+s)

27+6 s−s2 }

B {X(s) = 2 (9+s)27+6 s−s2 , Y (s) = 4 s

27+6 s−s2 }

C {X(s) = 2 (9+s)−27−6 s+s2 , Y (s) = 4 s

−27−6 s+s2 }

D {X(s) = 2 (15−s)−27−6 s+s2 , Y (s) = 4 (−6+s)

−27−6 s+s2 }


4 y + y′′ = 4 t

A Y (s) = 16+5 s2+8 s3

s2 (4+s2)

B Y (s) = 4+5 s2+8 s3

4+s2

C Y (s) = 4+5 s2+8 s3

s2 (4+s2)

D Y (s) = 4−5 s2+8 s3

s2 (4+s2)


5. Use la transformada de Laplace para resolver la ecuacion diferencial

4 y + y′′ = f(t)

con condiciones iniciales y(0) = 4 y y′(0) = 0 y donde

f(t) =

0 si 0 ≤ t < π

1 si π ≤ t < 2π0 si 2π ≤ t

A y(t) = 4 cos(2 t) + 14 (1 + cos(2 t)) Uπ(t)− 1

4 (1 + cos(2 t)) U2π(t)

B y(t) = 4 cos(2 t) + 14 (1− cos(4 t)) Uπ(t)− 1

4 (1− cos(4 t)) U2π(t)

C y(t) = 2 sen(2 t) + 14 (1− cos(2 t)) Uπ(t)− 1

4 (1− cos(2 t)) U2π(t)

D y(t) = 4 cos(2 t) + 14 (1− cos(2 t)) Uπ(t)− 1

4 (1− cos(2 t)) U2π(t)

E y(t) = 4 cos(2 t) + 12 (1− cos(2 t)) Uπ(t)− 1

2 (1− cos(2 t)) U2π(t)


f(t) =116t sen(8 t)

A F (s) = s(64+s2)2

B F (s) = 1(64+s2)2

C F (s) = s(−64+s2)2

D F (s) = 1(−64+s2)2


f(t) ={

4 si 0 ≤ t < 30 si 3 ≤ t

A −4 + 4 e−3 s

B 4 s− 4 s e−3 s

C − 4s + 4

s e−3 s

D 4s −

4s e−3 s

E 4− 4 e−3 s


F (s) =2

(9 + s)3

A f(t) = t2 e−9 t

B f(t) = 2 t2 e−9 t

C f(t) = t2 e9 t

D f(t) = 2 t2 e9 t


f(t) = e−3 t − sen(4 t)

A F (s) = −4+4 s−s2(−3−s) (16+s2)

B F (s) = 28−4 s+s2

(−3+s) (16+s2)


C F (s) = −4(−3+s) (16+s2)

D F (s) = 4(−3−s) (16+s2)


F (s) =5

34 + 6 s+ s2e−π s

A f(t) = − e(−3π−3 t) sen(5 t)Uπ(t)

B f(t) = − e(3π−3 t) cos(5 t)Uπ(t)

C f(t) = − e(3π+3 t) sen(5 t)Uπ(t)

D f(t) = − e(3π−3 t) sen(5 t)Uπ(t)


f(t) = e−4 t − 2 e3 t

A F (s) = 11+s(−4−s) (−3+s)

B F (s) = −5−3 s(−4−s) (−3+s)

C F (s) = −5+3 s(−4+s) (3+s)

D F (s) = 11−s(−4+s) (3+s)


f(t) =

0 si 0 ≤ t < 45 t si 4 ≤ t < 90 si 9 ≤ t

A(e−9 s − e−4 s + 9 s e−9 s − 4 s e−4 s

)5s2

B −(e−9 s − e−4 s + 9 s e−9 s − 4 s e−4 s

)5s2

C −(e−9 s + e−4 s + 9 s e−9 s + 4 s e−4 s

)5s

D −(e−9 s + e−4 s + 9 s e−9 s + 4 s e−4 s

)5s2


f(t) = 4 cosh(3 t) senh(3 t)

A F (s) = 34

1(6−s) (6+s)

B F (s) = −12 s(−3+s) (3+s) (9+s2)

C F (s) = 12 s(−3+s)2 (3+s)2

D F (s) = −12(6−s) (6+s)

E F (s) =4 (54−3 s2)

(18−6 s+s2) (18+6 s+s2)


f(t) = − cos(4 t) + 3 sen(4 t)

A F (s) = −4+3 s16+s2


B F (s) = 12−s16+s2

C F (s) = −4−3 s16+s2

D F (s) = 12+s16+s2


F (s) =A

s2+ (

B

s+C

s2) eD s


f(t) ={t si 0 < t ≤ 60 si t > 6

Respuesta:


f(t) =sen(7 t)

t

A F (s) = Atan( 17 s)

B F (s) = Atan(s)

C F (s) = Atan( 7s )

D F (s) = 17 Atan(s)


F (s) =1 + 6 s+ 9 s2

s7

A 95 t

5 + t6 + 17 t

7

B 9 t5 + 6 t6 + t7

C 340 t

5 + 1120 t

6 + 15040 t

7

D 38 t

4 + 120 t

5 + 1720 t

6

E 94 t

4 + 65 t

5 + 16 t

6

F 9 t4 + 6 t5 + t6


Y (s) =As3 +B s2 + C s+D

(s2 + E)2


25 y + y′′ = 6 cos(5 t)

Respuesta:


Y (s) =As3 +B s2 + C s+D

s2 (s2 + E)


16 y + y′′ = 16 t

Respuesta:


F (s) =7 + 6 s

−21− 4 s+ s2

A f(t) = 4910 e−7 t − 11

10 e−3 t

B f(t) = 4910 e−7 t + 11

10 e−3 t

C f(t) = 1110 e−3 t + 49

10 e7 t

D f(t) = 1110 e

3 t + 4910 e

7 t


21. Cual es la transformada de Laplace de la funcion:

f(t) = senh(5 t) e3 t

A F (s) = −3+s−16−6 s+s2

B F (s) = 3+s34+6 s+s2

C F (s) = 5−16−6 s+s2

D F (s) = 534+6 s+s2


F (s) =8

9 + s2

A 83 sen(9 t)

B 8 cos(3 t)

C 8 sen(3 t)

D 83 sen(3 t)

E 83 cos(3 t)

F 83 cos(9 t)



A F (s) = s2

(49+s2) (64+s2)

B F (s) = 12 s(

11+s2 −

1225+s2

)C F (s) = s

(1

1+s2 −1

225+s2

)D F (s) = s

(1

49+s2 + 164+s2


f(t) ={

4 t si 0 ≤ t < 50 si 5 ≤ t

A 4s2 −

4 (1+5 s)s2 e−5 s

B 4s2 −

1+5 ss2 e−5 s

C − 4s2 + 4 (1+5 s)

s2 e−5 s

D 4s2 −

4s2 e−5 s




1. Utilice el teorema de la derivada de la transformada y que

L(√t) =

12

√π

s32

para obtener la transformada de:f(t) = t

32

A F (s) = 34

√π

s52

B F (s) = 3√π

s52

C F (s) = 3√π

s32

D F (s) = 34

√π

s32


x′ = 3 t+ 5x+ 7 y

y′ = 4 + 7x+ 5 y


A {X(s) = −15+31 s+13 s2+3 s3

s2 (−24−10 s+s2) , Y (s) = 21−20 s+5 s2+4 s3

s2 (−24−10 s+s2) }

B {X(s) = −15+31 s−43 s2+3 s3

s2 (−24−10 s+s2) , Y (s) = 21−20 s+45 s2−4 s3

s2 (−24−10 s+s2) }

C {X(s) = −15+31 s−13 s2−3 s3

s2 (−24−10 s+s2) , Y (s) = 21−20 s+3 s2−4 s3

s2 (−24−10 s+s2) }

D {X(s) = −15+31 s+43 s2−3 s3

s2 (−24−10 s+s2) , Y (s) = 21−20 s−37 s2+4 s3

s2 (−24−10 s+s2) }



A F (s) = 12 ((1 + s2)−1 + 5

25+s2 )

B F (s) = 2 s(4+s2) (9+s2)

C F (s) = (1 + s2)−1 + 525+s2

D F (s) = −(1 + s2)−1 + 525+s2


F (s) =5 + s

(−10 + s) (−3 + s)

A 47 e

3 t − 117 e10 t

B − 157 e−10 t + 8

7 e−3 t

C 157 e−10 t − 8

7 e−3 t

D 117 e−10 t − 4

7 e−3 t


E 47 e

3 t + 117 e10 t

F − 87 e

3 t + 157 e10 t


f(t) = e−3 t − sen(t)

A F (s) = 1(−3−s) (1+s2)

B F (s) = 2+s−s2(−3−s) (1+s2)

C F (s) = 4−s+s2(−3+s) (1+s2)

D F (s) = − 1(−3+s) (1+s2)


f(t) ={

3 si 0 ≤ t < 50 si 5 ≤ t

A 3− 3 e−5 s

B − 3s + 3

s e−5 s

C 3s −

3s e−5 s

D −3 + 3 e−5 s

E 3 s− 3 s e−5 s


−32 y − 4 y′ + y′′ = t6 e8 t

A Y (s) = 720(−8+s)7 (4+s)

B Y (s) = 720(−4+s) (8+s)8

C Y (s) = 6(−8+s)8 (4+s)

D Y (s) = 720(−8+s)8 (4+s)



A F (s) = s(

116+s2 −

1144+s2

)B F (s) = s

(1

16+s2 + 164+s2

)C F (s) = 1

2 s(

116+s2 −

1144+s2

)D F (s) = s2

(16+s2) (64+s2)


24 y + 10 y′ + y′′ = U3(t)


A y(t) = − 32 e−6 t + 3

2 e−4 t − 1

24 U3(t) + 112 e

(18−6 t) U3(t) + 18 e

(12−4 t) U3(t)

B y(t) = − 32 e−6 t + 3

2 e−4 t + 1

24 U3(t)− 18 e

(−12+4 t) U3(t) + 112 e

(−18+6 t) U3(t)

C y(t) = − 32 e−6 t + 3

2 e−4 t + 1

24 U3(t) + 112 e

(18−6 t) U3(t)− 18 e

(12−4 t) U3(t)

D y(t) = 32 e−6 t − 3

2 e−4 t + 1

24 U3(t) + 112 e

(18−6 t) U3(t)− 18 e

(12−4 t) U3(t)


f(t) = − sen(t) senh(6 t)

A F (s) = −12 s(37−12 s+s2) (37+12 s+s2)

B F (s) = 6(6−s) (−1+s) (1+s) (6+s)

C F (s) = −6(1+s2) (36+s2)

D F (s) = 6(6−s) (6+s) (1+s2)


f(t) =

0 si 0 ≤ t < 33 t si 3 ≤ t < 70 si 7 ≤ t

A(e−7 s − e−3 s + 7 s e−7 s − 3 s e−3 s

)3s2

B −(e−7 s + e−3 s + 7 s e−7 s + 3 s e−3 s

)3s2

C −(e−7 s − e−3 s + 7 s e−7 s − 3 s e−3 s

)3s2

D −(e−7 s + e−3 s + 7 s e−7 s + 3 s e−3 s

)3s


28 y − 11 y′ + y′′ = 7 e4 t

A Y (s) = −21+7 s(−4+s) (28−11 s+s2)

B Y (s) = 21+7 s28−11 s+s2

C Y (s) = −21+7 s(−4+s) (28+11 s+s2)

D Y (s) = 21+7 s(−4+s) (28−11 s+s2)


f(t) = cosh(4 t) senh(4 t)

A F (s) = 128−4 s2

(32−8 s+s2) (32+8 s+s2)

B F (s) = 4(8−s) (8+s)

C F (s) = −4(8−s) (8+s)

D F (s) = −4 s(−4+s) (4+s) (16+s2)


E F (s) = 4 s(−4+s)2 (4+s)2


Y (s) =As+B

(s2 + C s+D) (s+ E)


32 y − 12 y′ + y′′ = 8 e4 t

Respuesta:


F (s) =3 + 4 s+ 6 s2

s5

A 3 t2 + 23 t

3 + 18 t

4

B 2 t3 + t4 + 35 t

5

C 6 t3 + 4 t4 + 3 t5

D 3 t2 + 43 t

3 + 34 t

4

E t3 + 16 t

4 + 140 t

5

F 6 t2 + 4 t3 + 3 t4

16. Determine A, B, C y D para quef(t) = eA t (B cosh(C t) +D senh(C t))

sea la transformada inversa de la funcion:F (s) =

s

−15 + 2 s+ s2

Respuesta:

17. Indique cual opcion contiene la transformada inversa de Laplace de :

F (s) =s

58 + 14 s+ s2e−π s

A f(t) = − e(7π−7 t)(cos(3 t)− 7

3 sen(3 t))Uπ(t)

B f(t) = − e(7π−7 t) (cos(3 t)− 7 sen(3 t)) Uπ(t)

C f(t) = − e(7π−7 t) cos(3 t)Uπ(t)

D f(t) = − cos(3 t)Uπ(t) e−7 t


f(t) ={

6 t si 0 ≤ t < 40 si 4 ≤ t

A 6s2 −

6s2 e−4 s

B 6s2 −

6 (1+4 s)s2 e−4 s

C 6s2 −

1+4 ss2 e−4 s

D − 6s2 + 6 (1+4 s)

s2 e−4 s


Y (s) =As3 +B s2 + C s+D

s2 (s2 + E)


36 y + y′′ = 36 t

Respuesta:



f(t) = e−5 t − 3 e4 t

A F (s) = 19−2 s(−5+s) (4+s)

B F (s) = 19+2 s(−5−s) (−4+s)

C F (s) = −11+4 s(−5+s) (4+s)

D F (s) = −11−4 s(−5−s) (−4+s)


f(t) = 2 cos(3 t) + 3 sen(3 t)

A F (s) = 9+2 s9+s2

B F (s) = 9−2 s9+s2

C F (s) = 6+3 s9+s2

D F (s) = 6−3 s9+s2

22. Determine A, B, C y D para quef(t) = eA t (B cos(C t) +D sen(C t))


s

82 + 2 s+ s2

Respuesta:


F (s) =A

s2+ (

B

s+C

s2) eD s


f(t) ={t si 0 < t ≤ 60 si t > 6

Respuesta:


f(t) = cos(6 t)− 3 t sen(6 t)

A F (s) = s3

(36+s2)2

B F (s) = s3

(−36+s2)2

C F (s) = s2

(−36+s2)2

D F (s) = s2

(36+s2)2





f(t) =12t cos(8 t) +

116

sen(8 t)

A F (s) = s2

(64+s2)2

B F (s) = s2

(−64+s2)2

C F (s) = s−64+s2

D F (s) = s2

−64+s2


F (s) =A

s2+ (

B

s+C

s2) eD s


f(t) ={t si 0 < t ≤ 20 si t > 2

Respuesta:

3. Determine A, B, C y D (con B > D) para que

f(t) = AeB t + C eD t


7 + 9 s−24− 5 s+ s2

Respuesta:


Y (s) =As3 +B s2 + C s+D

(s2 + E)2


9 y + y′′ = 4 sen(3 t)

Respuesta:


f(t) ={

3 si 0 ≤ t < 20 si 2 ≤ t

A 3− 3 e−2 s

B − 3s + 3

s e−2 s

C 3 s− 3 s e−2 s

D −3 + 3 e−2 s

E 3s −

3s e−2 s



36 y + y′′ = f(t)


f(t) =

0 si 0 ≤ t < π

1 si π ≤ t < 2π0 si 2π ≤ t

A y(t) = 6 cos(6 t) + 136 (1− cos(6 t)) Uπ(t)− 1

36 (1− cos(6 t)) U2π(t)

B y(t) = sen(6 t) + 136 (1− cos(6 t)) Uπ(t)− 1

36 (1− cos(6 t)) U2π(t)

C y(t) = 6 cos(6 t) + 118 (1− cos(6 t)) Uπ(t)− 1

18 (1− cos(6 t)) U2π(t)

D y(t) = 6 cos(6 t) + 136 (1− cos(12 t)) Uπ(t)− 1

36 (1− cos(12 t)) U2π(t)

E y(t) = 6 cos(6 t) + 136 (1 + cos(6 t)) Uπ(t)− 1

36 (1 + cos(6 t)) U2π(t)

7. Indique cual de las opciones siguientes es la ecuacion subsidiaria del problema con condicion inicial y(0) = 3 y ED:

5 y + y′ = f(t)

donde

f(t) =

0 t ≤ 05 + t t ≤ 10 1 < t

A Y (s) = 35+s

B Y (s) = 3+s−2−s−2 e−s+ 5s−

6s e−s

5+s

C Y (s) = 3+s−2+ 5s

5+s

D Y (s) = 3+ 1s−

1s e−s

5+s


64 y + y′′ = 64 t

A Y (s) = 64−4 s2+5 s3

s2 (64+s2)

B Y (s) = 64+4 s2+5 s3

64+s2

C Y (s) = 64+4 s2+5 s3

s2 (64+s2)

D Y (s) = 40+4 s2+5 s3

s2 (64+s2)


f(t) = e−5 t + sen(t)

A F (s) = − 1(−5−s) (1+s2)

B F (s) = −4+s+s2

(−5+s) (1+s2)

C F (s) = −6−s−s2(−5−s) (1+s2)

D F (s) = 1(−5+s) (1+s2)



f(t) = −3 cosh(5 t) senh(5 t)

A F (s) = −15 s(−5+s)2 (5+s)2

B F (s) = 15 s(−5+s) (5+s) (25+s2)

C F (s) = − 53

1(−10−s) (−10+s)

D F (s) = 15(−10−s) (−10+s)

E F (s) =3 (−250+5 s2)

(50−10 s+s2) (50+10 s+s2)


F (s) =8

4 + s2

A 4 cos(2 t)

B 4 sen(4 t)

C 8 sen(2 t)

D 4 sen(2 t)

E 4 cos(4 t)

F 8 cos(2 t)



A F (s) = s(

19+s2 + 1

64+s2

)B F (s) = s2

(9+s2) (64+s2)

C F (s) = s(

125+s2 −

1121+s2

)D F (s) = 1

2 s(

125+s2 −

1121+s2

)13. Indique cual de las siguientes opciones representa la transformada inversa de la funcion:

F (s) =−1 + 11 s−s+ s3

A f(t) = 1 + 5 e−t − 6 et

B f(t) = 1− 6 e−t + 5 et

C f(t) = 1 + 6 e−t − 5 et

D f(t) = 1− 5 e−t + 6 et


f(t) = e−5 t + 2 e5 t

A F (s) = 15−s(−5+s) (5+s)

B F (s) = −5+3 s(−5+s) (5+s)

C F (s) = −5−3 s(−5−s) (−5+s)

D F (s) = 15+s(−5−s) (−5+s)



L(t52 ) =

158

√π

s72


72

A F (s) = 10516

√π

s72

B F (s) = 10516

√π

s92

C F (s) = 1054

√π

s92

D F (s) = 1054

√π

s72


f(t) ={

6 t si 0 ≤ t < 50 si 5 ≤ t

A 6s2 −

6s2 e−5 s

B 6s2 −

1+5 ss2 e−5 s

C 6s2 −

6 (1+5 s)s2 e−5 s

D − 6s2 + 6 (1+5 s)

s2 e−5 s


x′ = x+ 7 y

y′ = 7x+ y


A {X(s) = −10+3 s−48−2 s+s2 , Y (s) = −22+s

48+2 s−s2 }

B {X(s) = 10−3 s−48−2 s+s2 , Y (s) = 22−s

48+2 s−s2 }

C {X(s) = 4+3 s−48−2 s+s2 , Y (s) = 20+s

−48−2 s+s2 }

D {X(s) = 4+3 s48+2 s−s2 , Y (s) = 20+s

48+2 s−s2 }



A F (s) = 3+s45+6 s+s2

B F (s) = −3+s−27−6 s+s2

C F (s) = 6−27−6 s+s2

D F (s) = 645+6 s+s2


f(t) = − cos(2 t)− 4 sen(2 t)

A F (s) = −2−4 s4+s2

B F (s) = −2+4 s4+s2

C F (s) = −8+s4+s2

D F (s) = −8−s4+s2



f(t) =

0 si 0 ≤ t < 42 t si 4 ≤ t < 80 si 8 ≤ t

A −(e−8 s + e−4 s + 8 s e−8 s + 4 s e−4 s

)2s

B −(e−8 s + e−4 s + 8 s e−8 s + 4 s e−4 s

)2s2

C(e−8 s − e−4 s + 8 s e−8 s − 4 s e−4 s

)2s2

D −(e−8 s − e−4 s + 8 s e−8 s − 4 s e−4 s

)2s2



A F (s) = 8 s(36+s2) (64+s2)

B F (s) = −24+s2 + 14

196+s2

C F (s) = 12 ( −2

4+s2 + 14196+s2 )

D F (s) = 24+s2 + 14

196+s2


F (s) =6

72 + 12 s+ s2e−π s

A f(t) = e(−6π−6 t) sen(6 t)Uπ(t)

B f(t) = e(6π−6 t) cos(6 t)Uπ(t)

C f(t) = e(6π−6 t) sen(6 t)Uπ(t)

D f(t) = e(6π+6 t) sen(6 t)Uπ(t)


Y (s) =As+B

(s2 + C s+D) (s+ E)


12 y − 8 y′ + y′′ = 6 e2 t

Respuesta:


F (s) =5 + 6 s+ 9 s2

s7

A 340 t

5 + 1120 t

6 + 11008 t

7

B 9 t4 + 6 t5 + 5 t6

C 94 t

4 + 65 t

5 + 56 t

6

D 9 t5 + 6 t6 + 5 t7

E 38 t

4 + 120 t

5 + 1144 t

6

F 95 t

5 + t6 + 57 t

7





Y (s) =As3 +B s2 + C s+D

s2 (s2 + E)


64 y + y′′ = 64 t

Respuesta:


F (s) =7− 2 s16 + s2

A 74 cos(4 t)− 2 sen(4 t)

B −2 cos(4 t) + 7 sen(4 t)

C −2 cos(4 t) + 74 sen(4 t)

D 7 cos(4 t)− 2 sen(4 t)

E −2 cos(4 t)− 74 sen(4 t)

F cos(4 t) + sen(4 t)


2 y + y′ = f(t)

donde

f(t) =

0 t ≤ 02 + t t ≤ 10 1 < t

A Y (s) = 8+ 1s−

1s e−s

2+s

B Y (s) = 82+s

C Y (s) = 8+s−2+ 2s

2+s

D Y (s) = 8+s−2−s−2 e−s+ 2s−

3s e−s

2+s

4. Determine A, B y C para que

Y (s) =A

(s+B) (s+ C)5


−15 y + 2 y′ + y′′ = t3 e3 t

Respuesta:

5. Determine A, B, C, D y E (B > D) para que

f(t) = AeB t + C eD t + E


−8 + 8 s−4 s+ s3

Respuesta:




A F (s) = 45

1(−8−s) (−8+s)

B F (s) = 20 s(−4+s)2 (4+s)2

C F (s) = −20(−8−s) (−8+s)

D F (s) = −20 s(−4+s) (4+s) (16+s2)

E F (s) =−5 (−128+4 s2)

(32−8 s+s2) (32+8 s+s2)


f(t) = cos(2 t)− t sen(2 t)

A F (s) = s2

(−4+s2)2

B F (s) = s3

(−4+s2)2

C F (s) = s3

(4+s2)2

D F (s) = s2

(4+s2)2


f(t) =

0 si 0 ≤ t < 36 t si 3 ≤ t < 50 si 5 ≤ t

A −(e−5 s + e−3 s + 5 s e−5 s + 3 s e−3 s

)6s2

B −(e−5 s − e−3 s + 5 s e−5 s − 3 s e−3 s

)6s2

C −(e−5 s + e−3 s + 5 s e−5 s + 3 s e−3 s

)6s

D(e−5 s − e−3 s + 5 s e−5 s − 3 s e−3 s

)6s2


F (s) =A

s2+ (

B

s+C

s2) eD s


f(t) ={t si 0 < t ≤ 30 si t > 3

Respuesta:


f(t) = (1 + 7 t) e−t

A F (s) = 6+s(−1+s)2

B F (s) = 8+s(−1−s)2

C F (s) = 7+ss2 (1+s)

D F (s) = 8−s(−1+s)2



36 y + y′′ = 7 cos(6 t)

A Y (s) =s (252+7 s2)

(36+s2)2

B Y (s) =s (259−7 s2)

(36+s2)2

C Y (s) =s (259+7 s2)

(36+s2)2

D Y (s) = 259+7 s2

(36+s2)2



s

4 s+ s2

Respuesta:


f(t) = e−5 t − 3 e−t

A F (s) = 14+2 s(−5−s) (1+s)

B F (s) = −16−4 s(−5−s) (1+s)

C F (s) = −16+4 s(−5+s) (−1+s)

D F (s) = 14−2 s(−5+s) (−1+s)


21 y + 10 y′ + y′′ = U2(t) + U4(t)

A Y (s) = e−2 s + 1s (3+s) (7+s) e

−4 s

B Y (s) = e−4 s+e−2 s

s (3+s) (7+s)

C Y (s) = e2 s+e4 s

s (3+s) (7+s)

D Y (s) = e2 s + 1s (3+s) (7+s) e

4 s


F (s) =7 + s

(−7 + s) (−5 + s)

A −7 e−7 t + 6 e−5 t

B 3 e5 t − 4 e7 t

C 4 e−7 t − 3 e−5 t

D 3 e5 t + 4 e7 t

E 7 e−7 t − 6 e−5 t


F −6 e5 t + 7 e7 t



A F (s) = 12 s(

11+s2 −

1225+s2

)B F (s) = s2

(49+s2) (64+s2)

C F (s) = s(

11+s2 −

1225+s2

)D F (s) = s

(1

49+s2 + 164+s2


f(t) = 3 cos(4 t)− sen(4 t)

A F (s) = 12+s16+s2

B F (s) = −4−3 s16+s2

C F (s) = 12−s16+s2

D F (s) = −4+3 s16+s2

18. Un sistema de ecuaciones diferenciales resuelto mediante la transformada de Laplace ha conducido al siguiente sistema

(−10 + s) X(s)− 5Y (s) =10s

X(s) + (−4 + s) Y (s) = 0

Obtenga las ecuaciones subsidiarias y posteriormente aplique la transformada inversa para determinar x(1)

Respuesta:


f(t) ={

5 si 0 ≤ t < 30 si 3 ≤ t

A 5s −

5s e−3 s

B − 5s + 5

s e−3 s

C −5 + 5 e−3 s

D 5 s− 5 s e−3 s

E 5− 5 e−3 s


f(t) = e4 t + sen(t)

A F (s) = 1(4+s) (1+s2)

B F (s) = 3−s−s2(4−s) (1+s2)

C F (s) = − 1(4−s) (1+s2)


D F (s) = 5+s+s2

(4+s) (1+s2)


F (s) = (26 + 10 s+ s2)−1e−π s

A f(t) = − e(5π−5 t) sen(t)Uπ(t)

B f(t) = − e(−5π−5 t) sen(t)Uπ(t)

C f(t) = − e(5π−5 t) cos(t)Uπ(t)

D f(t) = − e(5π+5 t) sen(t)Uπ(t)


f(t) =senh(5 t)

t

A F (s) = 110 ln( 5+s

−5+s )

B F (s) = ln(−5+s5+s )

C F (s) = 12 ln( 5+s

−5+s )

D F (s) = ln( 5+s−5+s )


f(t) ={

3 t si 0 ≤ t < 10 si 1 ≤ t

A 3s2 −

3s2 e−s

B − 3s2 + 3 (1+s)

s2 e−s

C 3s2 −

1+ss2 e−s

D 3s2 −

3 (1+s)s2 e−s



A F (s) = 12 ( −2

4+s2 + 12144+s2 )

B F (s) = 24+s2 + 12

144+s2

C F (s) = −24+s2 + 12

144+s2

D F (s) = 7 s(25+s2) (49+s2)





F (s) =7

58 + 6 s+ s2e−π s

A f(t) = − e(3π−3 t) sen(7 t)Uπ(t)

B f(t) = − e(3π+3 t) sen(7 t)Uπ(t)

C f(t) = − e(3π−3 t) cos(7 t)Uπ(t)

D f(t) = − e(−3π−3 t) sen(7 t)Uπ(t)


f(t) = − cosh(5 t) senh(5 t)

A F (s) = −5 s(−5+s)2 (5+s)2

B F (s) = 5(−10−s) (−10+s)

C F (s) = −5(−10−s) (−10+s)

D F (s) = −250+5 s2

(50−10 s+s2) (50+10 s+s2)

E F (s) = 5 s(−5+s) (5+s) (25+s2)


f(t) = cos(7 t)− 72t sen(7 t)

A F (s) = s2

(49+s2)2

B F (s) = s2

(−49+s2)2

C F (s) = s3

(−49+s2)2

D F (s) = s3

(49+s2)2




−64−16 s+ s3

Respuesta:


Y (s) =As3 +B s2 + C s+D

(s2 + E)2


64 y + y′′ = 3 cos(8 t)

Respuesta:



f(t) = −3 e−2 t + e2 t

A F (s) = 4−4 s(2−s) (2+s)

B F (s) = −8+2 s(2−s) (2+s)

C F (s) = 4+4 s(−2+s) (2+s)

D F (s) = −8−2 s(−2+s) (2+s)


64 y + y′′ = 8 cos(8 t)

A Y (s) = 520+8 s2

(64+s2)2

B Y (s) =s (520−8 s2)

(64+s2)2

C Y (s) =s (520+8 s2)

(64+s2)2

D Y (s) =s (512+8 s2)

(64+s2)2


Y (s) =As3 +B s2 + C s+D

s2 (s2 + E)


9 y + y′′ = 9 t

Respuesta:



A F (s) = 12 (−(1 + s2)−1 + 5

25+s2 )

B F (s) = (1 + s2)−1 + 525+s2

C F (s) = −(1 + s2)−1 + 525+s2

D F (s) = 3 s(4+s2) (9+s2)


x′ − y′ = x+ 6 y

x′ + y′ = 6x+ y


A {X(s) = −1+8 s35+2 s−2 s2 , Y (s) = −41+6 s

−35−2 s+2 s2 }

B {X(s) = −1+8 s−35−2 s+2 s2 , Y (s) = 41−6 s

−35−2 s+2 s2 }

C {X(s) = 41+8 s35+2 s−2 s2 , Y (s) = 1−6 s

−35−2 s+2 s2 }


D {X(s) = 41+8 s−35−2 s+2 s2 , Y (s) = −1+6 s

−35−2 s+2 s2 }


f(t) ={

6 si 0 ≤ t < 40 si 4 ≤ t

A − 6s + 6

s e−4 s

B 6 s− 6 s e−4 s

C −6 + 6 e−4 s

D 6− 6 e−4 s

E 6s −

6s e−4 s



s

65 + 2 s+ s2

Respuesta:


F (s) =A

s2+ (

B

s+C

s2) eD s


f(t) ={t si 0 < t ≤ 30 si t > 3

Respuesta:


f(t) =

0 si 0 ≤ t < 52 t si 5 ≤ t < 80 si 8 ≤ t

A −(e−8 s + e−5 s + 8 s e−8 s + 5 s e−5 s

)2s2

B −(e−8 s + e−5 s + 8 s e−8 s + 5 s e−5 s

)2s

C(e−8 s − e−5 s + 8 s e−8 s − 5 s e−5 s

)2s2

D −(e−8 s − e−5 s + 8 s e−8 s − 5 s e−5 s

)2s2


F (s) =6

9 + s2

A 2 cos(3 t)

B 2 sen(9 t)

C 2 sen(3 t)

D 2 cos(9 t)

E 6 cos(3 t)

F 6 sen(3 t)



25 y + y′′ = 6 sen(5 t)

A Y (s) = 30+150 s+6 s3

(−25+s2)2

B Y (s) = 30+30 s+6 s2

(−25+s2)2

C Y (s) = 30+150 s+6 s3

(25+s2)2

D Y (s) = 30+30 s+6 s3

(25+s2)2


f(t) ={

5 t si 0 ≤ t < 20 si 2 ≤ t

A 5s2 −

5 (1+2 s)s2 e−2 s

B 5s2 −

5s2 e−2 s

C − 5s2 + 5 (1+2 s)

s2 e−2 s

D 5s2 −

1+2 ss2 e−2 s


F (s) =4 + s

(−7 + s) (−1 + s)

A 13 e

t − 43 e

7 t

B 116 e−7 t − 5

6 e−t

C 43 e−7 t − 1

3 e−t

D 13 e

t + 43 e

7 t

E − 56 e

t + 116 e7 t

F − 116 e−7 t + 5

6 e−t


36 y + y′′ = f(t)


f(t) =

0 si 0 ≤ t < π

1 si π ≤ t < 2π0 si 2π ≤ t

A y(t) = 5 cos(6 t) + 136 (1− cos(6 t)) Uπ(t)− 1

36 (1− cos(6 t)) U2π(t)

B y(t) = 5 cos(6 t) + 136 (1− cos(12 t)) Uπ(t)− 1

36 (1− cos(12 t)) U2π(t)

C y(t) = 5 cos(6 t) + 118 (1− cos(6 t)) Uπ(t)− 1

18 (1− cos(6 t)) U2π(t)

D y(t) = 56 sen(6 t) + 1

36 (1− cos(6 t)) Uπ(t)− 136 (1− cos(6 t)) U2π(t)


E y(t) = 5 cos(6 t) + 136 (1 + cos(6 t)) Uπ(t)− 1

36 (1 + cos(6 t)) U2π(t)


f(t) = e−4 t − sen(2 t)

A F (s) = 4+2 s−s2(−4−s) (4+s2)

B F (s) = 2(−4−s) (4+s2)

C F (s) = −2(−4+s) (4+s2)

D F (s) = 12−2 s+s2

(−4+s) (4+s2)



A F (s) = s2

(4+s2) (36+s2)

B F (s) = s(

116+s2 −

164+s2

)C F (s) = 1

2 s(

116+s2 −

164+s2

)D F (s) = s

(1

4+s2 + 136+s2


f(t) = 2 cos(3 t)− sen(3 t)

A F (s) = 6+s9+s2

B F (s) = −3+2 s9+s2

C F (s) = −3−2 s9+s2

D F (s) = 6−s9+s2


f(t) =− cos(3 t) + cos(8 t)

t

A F (s) = − s9+s2 + s

64+s2

B F (s) = 12 ln( 9+s2

64+s2 )

C F (s) = s9+s2 + s

64+s2

D F (s) = ln( 8+s3+s )


f(t) = (1 + 5 t) e5 t

A F (s) = s(5−s)2

B F (s) = 10+s(5+s)2

C F (s) = 5+s(−5+s) s2

D F (s) = − s(5+s)2






A F (s) = s(

14+s2 + 1

49+s2

)B F (s) = s2

(4+s2) (49+s2)

C F (s) = 12 s(

125+s2 −

181+s2

)D F (s) = s

(1

25+s2 −1

81+s2

)2. Cual de las siguientes opciones representa la transformada de Laplace Inversa de la funcion:

F (s) =2 + 2 s36 + s2

A cos(6 t) + sen(6 t)

B 2 cos(6 t) + 2 sen(6 t)

C 2 cos(6 t)− 13 sen(6 t)

D 2 cos(6 t) + 13 sen(6 t)

E 13 cos(6 t) + 2 sen(6 t)

F 2 cos(6 t) + 2 sen(6 t)


16 y + y′′ = sen(7 t)U2π(t)

A y(t) = 4 cos(4 t)− 7132 sen(4 t)U2π(t) + 1

33 sen(7 t)U2π(t)

B y(t) = 4 cos(4 t)− 7132 sen(4 t)U2π(t)− 1

33 sen(7 t)U2π(t)

C y(t) = 4 cos(4 t) + 7132 sen(4 t)U2π(t)− 1

33 sen(7 t)U2π(t)

D y(t) = 4 sen(4 t)− 7132 cos(4 t)U2π(t)− 1

33 cos(7 t)U2π(t)


F (s) =A

s2+ (

B

s+C

s2) eD s


f(t) ={t si 0 < t ≤ 50 si t > 5

Respuesta:


36 y + y′′ = 6 cos(6 t)

A Y (s) =s (216+6 s2)

(36+s2)2


B Y (s) =s (222+6 s2)

(36+s2)2

C Y (s) = 222+6 s2

(36+s2)2

D Y (s) =s (222−6 s2)

(36+s2)2


f(t) = cos(5 t) e−7 t

A F (s) = 574+14 s+s2

B F (s) = 7+s74+14 s+s2

C F (s) = 574−14 s+s2

D F (s) = −7+s74−14 s+s2


f(t) =sen(5 t)

t

A F (s) = 15 Atan(s)

B F (s) = Atan( 15 s)

C F (s) = Atan(s)

D F (s) = Atan( 5s )


Y (s) =As3 +B s2 + C s+D

s2 (s2 + E)


4 y + y′′ = 4 t

Respuesta:


F (s) =s

40 + 4 s+ s2

A f(t) = (cos(6 t)− 13 sen(6 t)) e−2 t

B f(t) = cos(6 t)− 13 sen(6 t)

C f(t) = (cos(6 t)− 2 sen(6 t)) e−2 t

D f(t) = cos(6 t)− 2 sen(6 t)


F (s) =s

53 + 14 s+ s2e−π s

A f(t) = e(7π−7 t) (cos(2 t)− 7 sen(2 t)) Uπ(t)

B f(t) = cos(2 t)Uπ(t) e−7 t

C f(t) = e(7π−7 t) cos(2 t)Uπ(t)

D f(t) = e(7π−7 t)(cos(2 t)− 7

2 sen(2 t))Uπ(t)


f(t) = −3 et + e2 t

A F (s) = 7+4 s(1+s) (2+s)


B F (s) = −5−2 s(1+s) (2+s)

C F (s) = −5+2 s(2−s) (−1+s)

D F (s) = 7−4 s(2−s) (−1+s)


f(t) =

0 si 0 ≤ t < 55 t si 5 ≤ t < 80 si 8 ≤ t

A −(e−8 s + e−5 s + 8 s e−8 s + 5 s e−5 s

)5s

B(e−8 s − e−5 s + 8 s e−8 s − 5 s e−5 s

)5s2

C −(e−8 s − e−5 s + 8 s e−8 s − 5 s e−5 s

)5s2

D −(e−8 s + e−5 s + 8 s e−8 s + 5 s e−5 s

)5s2


f(t) = e−3 t + sen(3 t)

A F (s) = −18−3 s−s2(−3−s) (9+s2)

B F (s) = 3(−3+s) (9+s2)

C F (s) = 3 s+s2

(−3+s) (9+s2)

D F (s) = −3(−3−s) (9+s2)


f(t) = −5 cos(2 t)− 2 sen(2 t)

A F (s) = −10−2 s4+s2

B F (s) = −4−5 s4+s2

C F (s) = −4+5 s4+s2

D F (s) = −10+2 s4+s2


f(t) ={

3 t si 0 ≤ t < 50 si 5 ≤ t

A 3s2 −

1+5 ss2 e−5 s

B 3s2 −

3s2 e−5 s

C 3s2 −

3 (1+5 s)s2 e−5 s

D − 3s2 + 3 (1+5 s)

s2 e−5 s



F (s) =6 + 6 s

−10− 3 s+ s2

A f(t) = 367 e−5 t + 6

7 e−2 t

B f(t) = 367 e−5 t − 6

7 e−2 t

C f(t) = 67 e−2 t + 36

7 e5 t

D f(t) = 67 e

2 t + 367 e5 t


Y (s) =As3 +B s2 + C s+D

(s2 + E)2


49 y + y′′ = 6 sen(7 t)

Respuesta:


f(t) = 4 cosh(t) senh(t)

A F (s) = 4 s(−1+s)2 (1+s)2

B F (s) =4 (2−s2)

(2−2 s+s2) (2+2 s+s2)

C F (s) = 14

1(2−s) (2+s)

D F (s) = −4 s(−1+s) (1+s) (1+s2)

E F (s) = −4(2−s) (2+s)


f(t) ={

5 si 0 ≤ t < 30 si 3 ≤ t

A 5 s− 5 s e−3 s

B 5− 5 e−3 s

C − 5s + 5

s e−3 s

D −5 + 5 e−3 s

E 5s −

5s e−3 s


64 y + y′′ = 5 sen(8 t)

A Y (s) = 40+320 s+5 s3

(64+s2)2

B Y (s) = 40+320 s+5 s3

(−64+s2)2

C Y (s) = 40+40 s+5 s2

(−64+s2)2

D Y (s) = 40+40 s+5 s3

(64+s2)2



f(t) =114t2 sen(7 t)

A F (s) = −49+s2

(49+s2)3

B F (s) = 49+3 s2

(49+s2)3

C F (s) = −49+3 s2

(7+s2)3

D F (s) = −49+3 s2

(49+s2)3


x′ − y′ = 2x+ 7 y

x′ + y′ = 7x+ 2 y


A {X(s) = −17+4 s45+4 s−2 s2 , Y (s) = −37+6 s

−45−4 s+2 s2 }

B {X(s) = −17+4 s−45−4 s+2 s2 , Y (s) = 37−6 s

−45−4 s+2 s2 }

C {X(s) = 37+4 s−45−4 s+2 s2 , Y (s) = −17+6 s

−45−4 s+2 s2 }

D {X(s) = 37+4 s45+4 s−2 s2 , Y (s) = 17−6 s

−45−4 s+2 s2 }



A F (s) = (1 + s2)−1 + 13169+s2

B F (s) = 12 (−(1 + s2)−1 + 13

169+s2 )

C F (s) = 7 s(36+s2) (49+s2)

D F (s) = −(1 + s2)−1 + 13169+s2


F (s) =3

4 + s

A 3 e4 t

B 3 e−4 t

C −12 e4 t

D − 34 e−4 t





Y (s) =As3 +B s2 + C s+D

(s2 + E)2


4 y + y′′ = 7 cos(2 t)

Respuesta:


f(t) = t cosh(4 t)

A F (s) = 16+s2

(−16+s2)2

B F (s) = 16+s2

−16+s2

C F (s) = 8 s(−16+s2)2

D F (s) = (16 + s2)−1


F (s) =2

4 + s2

A cos(4 t)

B sen(4 t)

C 2 sen(2 t)

D 2 cos(2 t)

E cos(2 t)

F sen(2 t)


f(t) ={

6 si 0 ≤ t < 10 si 1 ≤ t

A 6s −

6s e−s

B −6 + 6 e−s

C 6 s− 6 s e−s

D 6− 6 e−s

E − 6s + 6

s e−s


f(t) ={

4 t si 0 ≤ t < 20 si 2 ≤ t


A 4s2 −

1+2 ss2 e−2 s

B 4s2 −

4 (1+2 s)s2 e−2 s

C − 4s2 + 4 (1+2 s)

s2 e−2 s

D 4s2 −

4s2 e−2 s


36 y + y′′ = sen(3 t)U2π(t)

A y(t) = 3 cos(6 t)− 127 sen(3 t)U2π(t) + 1

54 sen(6 t)U2π(t)

B y(t) = 3 cos(6 t) + 127 sen(3 t)U2π(t) + 1

54 sen(6 t)U2π(t)

C y(t) = 3 sen(6 t) + 127 cos(3 t)U2π(t) + 1

54 cos(6 t)U2π(t)

D y(t) = 3 cos(6 t) + 127 sen(3 t)U2π(t)− 1

54 sen(6 t)U2π(t)


F (s) =2

5 + 2 s+ s2e−π s

A f(t) = e(π−t) cos(2 t)Uπ(t)

B f(t) = e(π−t) sen(2 t)Uπ(t)

C f(t) = e(π+t) sen(2 t)Uπ(t)

D f(t) = e(−π−t) sen(2 t)Uπ(t)



A F (s) = 6+s+s2

(5+s) (1+s2)

B F (s) = 1(5+s) (1+s2)

C F (s) = 4−s−s2(5−s) (1+s2)

D F (s) = − 1(5−s) (1+s2)


f(t) = 4 cos(4 t)− 2 sen(4 t)

A F (s) = 16+2 s16+s2

B F (s) = −8−4 s16+s2

C F (s) = 16−2 s16+s2

D F (s) = −8+4 s16+s2


f(t) =sen(8 t)

t

A F (s) = Atan( 8s )

B F (s) = 18 Atan(s)

C F (s) = Atan(s)

D F (s) = Atan( 18 s)




s

−32 + 4 s+ s2

Respuesta:


Y (s) =A

(s+B) (s+ C)6


−14 y + 5 y′ + y′′ = t4 e2 t

Respuesta:



A F (s) = 43

1(8−s) (8+s)

B F (s) = 12 s(−4+s)2 (4+s)2

C F (s) =3 (128−4 s2)

(32−8 s+s2) (32+8 s+s2)

D F (s) = −12 s(−4+s) (4+s) (16+s2)

E F (s) = −12(8−s) (8+s)


f(t) =

0 si 0 ≤ t < 46 t si 4 ≤ t < 70 si 7 ≤ t

A −(e−7 s + e−4 s + 7 s e−7 s + 4 s e−4 s

)6s

B −(e−7 s − e−4 s + 7 s e−7 s − 4 s e−4 s

)6s2

C(e−7 s − e−4 s + 7 s e−7 s − 4 s e−4 s

)6s2

D −(e−7 s + e−4 s + 7 s e−7 s + 4 s e−4 s

)6s2


F (s) =s

−24 + 2 s+ s2

A f(t) = cosh(5 t)− 15 senh(5 t)

B f(t) = (cosh(5 t)− 15 senh(5 t)) e−t

C f(t) = (cosh(5 t)− senh(5 t)) e−t

D f(t) = cosh(5 t)− senh(5 t)


7 y + y′ = f(t)

donde

f(t) =

0 t ≤ 07 + 2 t t ≤ 10 1 < t


A Y (s) = 2+ 1s−

1s e−s

7+s

B Y (s) =2+ 2

s2+ 7

s

7+s

C Y (s) =2+ 2

s2− 2

s2e−s+ 7

s−9s e−s

7+s

D Y (s) = 27+s


F (s) =A

s2+ (

B

s+C

s2) eD s


f(t) ={t si 0 < t ≤ 30 si t > 3

Respuesta:


x′ = 2 t+ 2x+ 10 y

y′ = −5 + 10x+ 2 y


A {X(s) =2 (−2−24 s+27 s2−s3)s2 (−96−4 s+s2) , Y (s) =

5 (4+2 s−7 s2+s3)s2 (−96−4 s+s2) }

B {X(s) =2 (2+24 s+23 s2+s3)s2 (96+4 s−s2) , Y (s) =

5 (4+2 s−3 s2−s3)s2 (−96−4 s+s2) }

C {X(s) =2 (−2−24 s−27 s2+s3)s2 (−96−4 s+s2) , Y (s) =

5 (4+2 s+5 s2−s3)s2 (−96−4 s+s2) }

D {X(s) =2 (−2−24 s+23 s2+s3)s2 (−96−4 s+s2) , Y (s) =

5 (4+2 s+s2+s3)s2 (−96−4 s+s2) }


16 y − 10 y′ + y′′ = 2 e8 t

A Y (s) = 14+2 s16−10 s+s2

B Y (s) = −14+2 s(−8+s) (16−10 s+s2)

C Y (s) = 14+2 s(−8+s) (16−10 s+s2)

D Y (s) = −14+2 s(−8+s) (16+10 s+s2)


F (s) =3 + 5 s+ 3 s2

s6

A 3 t4 + 5 t5 + 3 t6

B 3 t3 + 5 t4 + 3 t5

C 18 t

4 + 124 t

5 + 1240 t

6

D t3 + 54 t

4 + 35 t

5

E 34 t

4 + t5 + 12 t

6

F 12 t

3 + 524 t

4 + 140 t

5



f(t) = e−4 t + 3 e4 t

A F (s) = 16+2 s(−4−s) (−4+s)

B F (s) = −8+4 s(−4+s) (4+s)

C F (s) = −8−4 s(−4−s) (−4+s)

D F (s) = 16−2 s(−4+s) (4+s)



A F (s) = 12 ( 5

25+s2 + 981+s2 )

B F (s) = −525+s2 + 9

81+s2

C F (s) = 525+s2 + 9

81+s2

D F (s) = 2 s(4+s2) (49+s2)


f(t) =12t2 cos(6 t)

A F (s) = −18 s+s3

(36+s2)3

B F (s) = −108 s+s3

(36+s2)3

C F (s) = 108 s+s3

(36+s2)3

D F (s) = 18 s+s3

(36+s2)3



A F (s) = s(

19+s2 + 1

49+s2

)B F (s) = s

(1

16+s2 −1

100+s2

)C F (s) = s2

(9+s2) (49+s2)

D F (s) = 12 s(

116+s2 −

1100+s2

)





f(t) =

0 si 0 ≤ t < 45 t si 4 ≤ t < 80 si 8 ≤ t

A −(e−8 s + e−4 s + 8 s e−8 s + 4 s e−4 s

)5s

B −(e−8 s + e−4 s + 8 s e−8 s + 4 s e−4 s

)5s2

C −(e−8 s − e−4 s + 8 s e−8 s − 4 s e−4 s

)5s2

D(e−8 s − e−4 s + 8 s e−8 s − 4 s e−4 s

)5s2


f(t) ={

3 si 0 ≤ t < 40 si 4 ≤ t

A − 3s + 3

s e−4 s

B 3− 3 e−4 s

C −3 + 3 e−4 s

D 3 s− 3 s e−4 s

E 3s −

3s e−4 s


F (s) =s

65 + 2 s+ s2

A f(t) = cos(8 t)− 18 sen(8 t)

B f(t) = (cos(8 t)− 18 sen(8 t)) e−t

C f(t) = cos(8 t)− sen(8 t)

D f(t) = (cos(8 t)− sen(8 t)) e−t



A F (s) = −250+5 s2

(50−10 s+s2) (50+10 s+s2)

B F (s) = 5 s(−5+s) (5+s) (25+s2)

C F (s) = −5 s(−5+s)2 (5+s)2

D F (s) = −5(−10−s) (−10+s)

E F (s) = 5(−10−s) (−10+s)



Y (s) =As3 +B s2 + C s+D

(s2 + E)2


64 y + y′′ = 3 cos(8 t)

Respuesta:


F (s) =10

16 + s2

A 52 cos(4 t)

B 10 sen(4 t)

C 52 sen(4 t)

D 52 sen(16 t)

E 10 cos(4 t)

F 52 cos(16 t)


f(t) = cos(5 t)− 52t sen(5 t)

A F (s) = s2

(25+s2)2

B F (s) = s3

(−25+s2)2

C F (s) = s3

(25+s2)2

D F (s) = s2

(−25+s2)2


f(t) = cos(t)− 3 sen(t)

A F (s) = −3+s1+s2

B F (s) = 1+3 s1+s2

C F (s) = −3−s1+s2

D F (s) = 1−3 s1+s2



A F (s) = 12 s(

125+s2 −

1121+s2

)B F (s) = s

(1

9+s2 + 164+s2

)C F (s) = s2

(9+s2) (64+s2)

D F (s) = s(

125+s2 −

1121+s2

)10. Seleccionar la opcion que contiene la transformada de Laplace de

f(t) =−1 e−7 t + e−4 t

t

A F (s) = 7+s4+s

B F (s) = − 1−7+s + 1

−4+s

C F (s) = ln( 7+s4+s )

D F (s) = 14+s −

17+s



F (s) =8

73 + 6 s+ s2e−π s

A f(t) = e(3π+3 t) sen(8 t)Uπ(t)

B f(t) = e(3π−3 t) sen(8 t)Uπ(t)

C f(t) = e(−3π−3 t) sen(8 t)Uπ(t)

D f(t) = e(3π−3 t) cos(8 t)Uπ(t)


F (s) =6

(6 + s)3

A f(t) = 3 t2 e6 t

B f(t) = 3 t2 e−6 t

C f(t) = 6 t2 e−6 t

D f(t) = 6 t2 e6 t


F (s) =A

s2+ (

B

s+C

s2) eD s


f(t) ={t si 0 < t ≤ 20 si t > 2

Respuesta:



A F (s) = −39+s2 + 11

121+s2

B F (s) = 4 s(16+s2) (49+s2)

C F (s) = 39+s2 + 11

121+s2

D F (s) = 12 ( 3

9+s2 + 11121+s2 )


f(t) = t cosh(4 t)

A F (s) = 8 s(−16+s2)2

B F (s) = 16+s2

−16+s2

C F (s) = (16 + s2)−1

D F (s) = 16+s2

(−16+s2)2


16 y + y′′ = f(t)


f(t) =

0 si 0 ≤ t < π

1 si π ≤ t < 2π0 si 2π ≤ t


A y(t) = 6 cos(4 t) + 116 (1− cos(8 t)) Uπ(t)− 1

16 (1− cos(8 t)) U2π(t)

B y(t) = 6 cos(4 t) + 18 (1− cos(4 t)) Uπ(t)− 1

8 (1− cos(4 t)) U2π(t)

C y(t) = 6 cos(4 t) + 116 (1− cos(4 t)) Uπ(t)− 1

16 (1− cos(4 t)) U2π(t)

D y(t) = 32 sen(4 t) + 1

16 (1− cos(4 t)) Uπ(t)− 116 (1− cos(4 t)) U2π(t)

E y(t) = 6 cos(4 t) + 116 (1 + cos(4 t)) Uπ(t)− 1

16 (1 + cos(4 t)) U2π(t)


f(t) ={

3 t si 0 ≤ t < 40 si 4 ≤ t

A − 3s2 + 3 (1+4 s)

s2 e−4 s

B 3s2 −

1+4 ss2 e−4 s

C 3s2 −

3 (1+4 s)s2 e−4 s

D 3s2 −

3s2 e−4 s


F (s) =4

16 + s2

A cos(4 t)

B 4 cos(4 t)

C sen(4 t)

D 4 sen(4 t)

E sen(16 t)

F cos(16 t)


f(t) = et + sen(t)

A F (s) = 2+s+s2

(1+s) (1+s2)

B F (s) = 1(1+s) (1+s2)

C F (s) = −s−s2(1−s) (1+s2)

D F (s) = − 1(1−s) (1+s2)


Y (s) =As3 +B s2 + C s+D

(s2 + E)2


4 y + y′′ = 6 sen(2 t)

Respuesta:


16 y + y′′ = 6 cos(4 t)

A Y (s) = 102+6 s2

(16+s2)2


B Y (s) =s (102−6 s2)

(16+s2)2

C Y (s) =s (102+6 s2)

(16+s2)2

D Y (s) =s (96+6 s2)(16+s2)2


24 y − 11 y′ + y′′ = 3 e8 t

A Y (s) = −21+3 s(−8+s) (24+11 s+s2)

B Y (s) = −21+3 s(−8+s) (24−11 s+s2)

C Y (s) = 21+3 s24−11 s+s2

D Y (s) = 21+3 s(−8+s) (24−11 s+s2)


f(t) = −5 e−4 t + e−t

A F (s) = −9+6 s(−4+s) (−1+s)

B F (s) = 1−4 s(−4+s) (−1+s)

C F (s) = −9−6 s(−1−s) (4+s)

D F (s) = 1+4 s(−1−s) (4+s)


x′ = 3x+ 8 y

y′ = 8x+ 3 y


A {X(s) = −39+5 s−55−6 s+s2 , Y (s) = 49−3 s

−55−6 s+s2 }

B {X(s) = 9+5 s−55−6 s+s2 , Y (s) = 31+3 s

−55−6 s+s2 }

C {X(s) = 39−5 s−55−6 s+s2 , Y (s) = 49−3 s

55+6 s−s2 }

D {X(s) = 9+5 s55+6 s−s2 , Y (s) = 31+3 s

55+6 s−s2 }





Y (s) =As+B

(s2 + C s+D) (s+ E)


40 y − 13 y′ + y′′ = 8 e5 t

Respuesta:


64 y + y′′ = 5 cos(8 t)

A Y (s) = 325+5 s2

(64+s2)2

B Y (s) =s (325−5 s2)

(64+s2)2

C Y (s) =s (325+5 s2)

(64+s2)2

D Y (s) =s (320+5 s2)

(64+s2)2


F (s) =3

34 + 10 s+ s2e−π s

A f(t) = − e(5π−5 t) sen(3 t)Uπ(t)

B f(t) = − e(−5π−5 t) sen(3 t)Uπ(t)

C f(t) = − e(5π+5 t) sen(3 t)Uπ(t)

D f(t) = − e(5π−5 t) cos(3 t)Uπ(t)



A F (s) = 5+s+s2

(4+s) (1+s2)

B F (s) = 3−s−s2(4−s) (1+s2)

C F (s) = − 1(4−s) (1+s2)

D F (s) = 1(4+s) (1+s2)


F (s) =s

53 + 4 s+ s2

A f(t) = cos(7 t)− 2 sen(7 t)

B f(t) = (cos(7 t)− 27 sen(7 t)) e−2 t

C f(t) = (cos(7 t)− 2 sen(7 t)) e−2 t

D f(t) = cos(7 t)− 27 sen(7 t)



f(t) = −4 cosh(t) senh(t)

A F (s) = −4 s(−1+s)2 (1+s)2

B F (s) = − 14

1(−2−s) (−2+s)

C F (s) = 4 s(−1+s) (1+s) (1+s2)

D F (s) = 4(−2−s) (−2+s)

E F (s) =4 (−2+s2)

(2−2 s+s2) (2+2 s+s2)


f(t) =sen(2 t)

t

A F (s) = Atan( 12 s)

B F (s) = 12 Atan(s)

C F (s) = Atan(s)



F (s) =4

−6 + s

A 23 e

6 t

B 4 e−6 t

C 24 e−6 t

D 4 e6 t


f(t) =112t sen(6 t)

A F (s) = 1(−36+s2)2

B F (s) = s(36+s2)2

C F (s) = s(−36+s2)2

D F (s) = 1(36+s2)2


F (s) =2

3 + s

A 2 e−3 t

B −6 e3 t

C − 23 e−3 t

D 2 e3 t


64 y + y′′ = 64 t

A Y (s) = 48+5 s2+6 s3

s2 (64+s2)

B Y (s) = 64+5 s2+6 s3

s2 (64+s2)

C Y (s) = 64+5 s2+6 s3

64+s2


D Y (s) = 64−5 s2+6 s3

s2 (64+s2)


F (s) =−5

36− 13 s+ s2

A f(t) = e4 t + e9 t

B f(t) = −1 e−9 t + e−4 t

C f(t) = e4 t − e9 t

D f(t) = e−9 t + e−4 t


f(t) = e−2 t + 4 e2 t

A F (s) = 10−3 s(−2+s) (2+s)

B F (s) = −6+5 s(−2+s) (2+s)

C F (s) = −6−5 s(−2−s) (−2+s)

D F (s) = 10+3 s(−2−s) (−2+s)


f(t) = −3 cos(2 t)− 5 sen(2 t)

A F (s) = −6+5 s4+s2

B F (s) = −10+3 s4+s2

C F (s) = −6−5 s4+s2

D F (s) = −10−3 s4+s2


f(t) ={

5 t si 0 ≤ t < 30 si 3 ≤ t

A 5s2 −

1+3 ss2 e−3 s

B − 5s2 + 5 (1+3 s)

s2 e−3 s

C 5s2 −

5s2 e−3 s

D 5s2 −

5 (1+3 s)s2 e−3 s


F (s) =A

s2+ (

B

s+C

s2) eD s


f(t) ={t si 0 < t ≤ 50 si t > 5

Respuesta:



21 y + 10 y′ + y′′ = U4(t)

A y(t) = −1 e−7 t + e−3 t − 121 U4(t)− 1

28 e(28−7 t) U4(t)− 1

12 e(12−3 t) U4(t)

B y(t) = −1 e−7 t + e−3 t + 121 U4(t) + 1

28 e(28−7 t) U4(t)− 1

12 e(12−3 t) U4(t)

C y(t) = −1 e−7 t + e−3 t + 121 U4(t)− 1

12 e(−12+3 t) U4(t) + 1

28 e(−28+7 t) U4(t)

D y(t) = e−7 t − e−3 t + 121 U4(t) + 1

28 e(28−7 t) U4(t)− 1

12 e(12−3 t) U4(t)


f(t) =

0 si 0 ≤ t < 23 t si 2 ≤ t < 40 si 4 ≤ t

A −(e−4 s + e−2 s + 4 s e−4 s + 2 s e−2 s

)3s

B −(e−4 s + e−2 s + 4 s e−4 s + 2 s e−2 s

)3s2

C −(e−4 s − e−2 s + 4 s e−4 s − 2 s e−2 s

)3s2

D(e−4 s − e−2 s + 4 s e−4 s − 2 s e−2 s

)3s2


x′ = 2x+ 7 y

y′ = 7x+ 2 y


A {X(s) = 18+5 s−45−4 s+s2 , Y (s) = 27+4 s

−45−4 s+s2 }

B {X(s) = 38−5 s−45−4 s+s2 , Y (s) = 43−4 s

45+4 s−s2 }

C {X(s) = −38+5 s−45−4 s+s2 , Y (s) = 43−4 s

−45−4 s+s2 }

D {X(s) = 18+5 s45+4 s−s2 , Y (s) = 27+4 s

45+4 s−s2 }


f(t) ={

5 si 0 ≤ t < 20 si 2 ≤ t

A 5− 5 e−2 s

B − 5s + 5

s e−2 s

C 5 s− 5 s e−2 s

D 5s −

5s e−2 s

E −5 + 5 e−2 s




A F (s) = 12 s(

11+s2 −

1121+s2

)B F (s) = s

(1

1+s2 −1

121+s2

)C F (s) = s

(1

25+s2 + 136+s2

)D F (s) = s2

(25+s2) (36+s2)


f(t) = − sen(2 t) senh(2 t)

A F (s) = −8 s(8−4 s+s2) (8+4 s+s2)

B F (s) = 4(2−s) (2+s) (4+s2)

C F (s) = 4(2−s) (−2+s) (2+s)2

D F (s) = −4(4+s2)2


Y (s) =As3 +B s2 + C s+D

(s2 + E)2


4 y + y′′ = 7 cos(2 t)

Respuesta:



A F (s) = 12 ( −5

25+s2 + 11121+s2 )

B F (s) = −525+s2 + 11

121+s2

C F (s) = 8 s(9+s2) (64+s2)

D F (s) = 525+s2 + 11

121+s2





F (s) =9

4 + s

A 9 e4 t

B − 94 e−4 t

C 9 e−4 t

D −36 e4 t


16 y + y′′ = f(t)


f(t) =

0 si 0 ≤ t < π

1 si π ≤ t < 2π0 si 2π ≤ t

A y(t) = 5 cos(4 t) + 116 (1 + cos(4 t)) Uπ(t)− 1

16 (1 + cos(4 t)) U2π(t)

B y(t) = 5 cos(4 t) + 18 (1− cos(4 t)) Uπ(t)− 1

8 (1− cos(4 t)) U2π(t)

C y(t) = 54 sen(4 t) + 1

16 (1− cos(4 t)) Uπ(t)− 116 (1− cos(4 t)) U2π(t)

D y(t) = 5 cos(4 t) + 116 (1− cos(4 t)) Uπ(t)− 1

16 (1− cos(4 t)) U2π(t)

E y(t) = 5 cos(4 t) + 116 (1− cos(8 t)) Uπ(t)− 1

16 (1− cos(8 t)) U2π(t)


Y (s) =As3 +B s2 + C s+D

s2 (s2 + E)


9 y + y′′ = 9 t

Respuesta:


9 y + y′′ = 5 sen(3 t)

A Y (s) = 15+45 s+5 s3

(9+s2)2

B Y (s) = 15+15 s+5 s2

(−9+s2)2

C Y (s) = 15+45 s+5 s3

(−9+s2)2

D Y (s) = 15+15 s+5 s3

(9+s2)2




A F (s) = s(

19+s2 −

1121+s2

)B F (s) = 1

2 s(

19+s2 −

1121+s2

)C F (s) = s2

(16+s2) (49+s2)

D F (s) = s(

116+s2 + 1

49+s2

)6. Determine A, B, C , D y E para que

Y (s) =As3 +B s2 + C s+D

(s2 + E)2


25 y + y′′ = 4 sen(5 t)

Respuesta:


F (s) =2

(9 + s)3

A f(t) = t2 e−9 t

B f(t) = 2 t2 e−9 t

C f(t) = 2 t2 e9 t

D f(t) = t2 e9 t


F (s) =5

9 + s2

A 53 cos(9 t)

B 53 sen(9 t)

C 5 sen(3 t)

D 53 cos(3 t)

E 5 cos(3 t)

F 53 sen(3 t)



A F (s) = 416+s2 + 8

64+s2

B F (s) = 2 s(4+s2) (36+s2)

C F (s) = 12 ( 4

16+s2 + 864+s2 )

D F (s) = −416+s2 + 8

64+s2


f(t) = e2 t + 2 e4 t

A F (s) = 8+3 s(2+s) (4+s)

B F (s) = 8−3 s(2−s) (−4+s)

C F (s) = − s(2+s) (4+s)


D F (s) = s(2−s) (−4+s)


40 y − 13 y′ + y′′ = 8 e5 t

A Y (s) = −32+8 s(−5+s) (40−13 s+s2)

B Y (s) = −32+8 s(−5+s) (40+13 s+s2)

C Y (s) = 32+8 s40−13 s+s2

D Y (s) = 32+8 s(−5+s) (40−13 s+s2)


f(t) =e−6 t − e−3 t

t

A F (s) = 1−6+s −

1−3+s

B F (s) = ln( 3+s6+s )

C F (s) = 3+s6+s

D F (s) = − 13+s + 1

6+s



A F (s) = 2+s85+4 s+s2

B F (s) = 9−77−4 s+s2

C F (s) = 985+4 s+s2

D F (s) = −2+s−77−4 s+s2


F (s) =s

5 + 2 s+ s2e−π s

A f(t) = e(π−t) cos(2 t)Uπ(t)

B f(t) = cos(2 t)Uπ(t) e−t

C f(t) = e(π−t) (cos(2 t)− sen(2 t)) Uπ(t)

D f(t) = e(π−t)(cos(2 t)− 1

2 sen(2 t))Uπ(t)


f(t) ={

5 t si 0 ≤ t < 50 si 5 ≤ t

A 5s2 −

5 (1+5 s)s2 e−5 s

B 5s2 −

1+5 ss2 e−5 s

C 5s2 −

5s2 e−5 s

D − 5s2 + 5 (1+5 s)

s2 e−5 s


f(t) ={

6 si 0 ≤ t < 10 si 1 ≤ t

A 6− 6 e−s

B 6 s− 6 s e−s


C 6s −

6s e−s

D −6 + 6 e−s

E − 6s + 6

s e−s


x′ − y′ = 5x+ 8 y

x′ + y′ = 8x+ 5 y


A {X(s) = 4 (−5+s)−39−10 s+2 s2 , Y (s) = 4 (8−s)

−39−10 s+2 s2 }

B {X(s) = 4 (8+s)−39−10 s+2 s2 , Y (s) = 4 (−5+s)

−39−10 s+2 s2 }

C {X(s) = 4 (−5+s)39+10 s−2 s2 , Y (s) = 4 (−8+s)

−39−10 s+2 s2 }

D {X(s) = 4 (8+s)39+10 s−2 s2 , Y (s) = 4 (5−s)

−39−10 s+2 s2 }


F (s) =A

s2+ (

B

s+C

s2) eD s


f(t) ={t si 0 < t ≤ 20 si t > 2

Respuesta:


f(t) = − cos(4 t) + 5 sen(4 t)

A F (s) = −4−5 s16+s2

B F (s) = −4+5 s16+s2

C F (s) = 20+s16+s2

D F (s) = 20−s16+s2


f(t) =

0 si 0 ≤ t < 36 t si 3 ≤ t < 60 si 6 ≤ t

A −(e−6 s + e−3 s + 6 s e−6 s + 3 s e−3 s

)6s

B −(e−6 s − e−3 s + 6 s e−6 s − 3 s e−3 s

)6s2

C −(e−6 s + e−3 s + 6 s e−6 s + 3 s e−3 s

)6s2

D(e−6 s − e−3 s + 6 s e−6 s − 3 s e−3 s

)6s2




s

65 + 2 s+ s2

Respuesta:


f(t) =12t cos(3 t) +

16

sen(3 t)

A F (s) = s2

−9+s2

B F (s) = s2

(−9+s2)2

C F (s) = s−9+s2

D F (s) = s2

(9+s2)2



A F (s) = 16(8−s) (8+s)

B F (s) = 16 s(−4+s) (4+s) (16+s2)

C F (s) = −16 s(−4+s)2 (4+s)2

D F (s) = − 1(8−s) (8+s)

E F (s) =−4 (128−4 s2)

(32−8 s+s2) (32+8 s+s2)


f(t) = e−2 t + sen(5 t)

A F (s) = 15+5 s+s2

(−2+s) (25+s2)

B F (s) = −5(−2−s) (25+s2)

C F (s) = −35−5 s−s2(−2−s) (25+s2)

D F (s) = 5(−2+s) (25+s2)





F (s) =5 + 5 s+ 4 s2

s6

A t4 + t5 + 56 t

6

B 4 t3 + 5 t4 + 5 t5

C 43 t

3 + 54 t

4 + t5

D 16 t

4 + 124 t

5 + 1144 t

6

E 23 t

3 + 524 t

4 + 124 t

5

F 4 t4 + 5 t5 + 5 t6


F (s) =A

s2+ (

B

s+C

s2) eD s


f(t) ={t si 0 < t ≤ 60 si t > 6

Respuesta:


Y (s) =As3 +B s2 + C s+D

(s2 + E)2


64 y + y′′ = 2 cos(8 t)

Respuesta:


f(t) = 4 cos(2 t)− sen(2 t)

A F (s) = 8+s4+s2

B F (s) = −2+4 s4+s2

C F (s) = −2−4 s4+s2

D F (s) = 8−s4+s2


f(t) ={

4 si 0 ≤ t < 40 si 4 ≤ t

A −4 + 4 e−4 s

B 4 s− 4 s e−4 s


C 4− 4 e−4 s

D − 4s + 4

s e−4 s

E 4s −

4s e−4 s


64 y + y′′ = 7 sen(8 t)

A Y (s) = 56+56 s+7 s3

(64+s2)2

B Y (s) = 56+448 s+7 s3

(64+s2)2

C Y (s) = 56+56 s+7 s2

(−64+s2)2

D Y (s) = 56+448 s+7 s3

(−64+s2)2


f(t) = (−5 + 2 t) e−2 t

A F (s) = −8+5 s(−2+s)2

B F (s) = −8−5 s(−2−s)2

C F (s) = 2−5 ss2 (2+s)

D F (s) = 12−5 s(−2+s)2


36 y + y′′ = 3 cos(6 t)

A Y (s) =s (111−3 s2)

(36+s2)2

B Y (s) =s (111+3 s2)

(36+s2)2

C Y (s) = 111+3 s2

(36+s2)2

D Y (s) =s (108+3 s2)

(36+s2)2


Y (s) =As+B

(s2 + C s+D) (s+ E)


18 y − 9 y′ + y′′ = 6 e3 t

Respuesta:


21 y + 10 y′ + y′′ = U2(t) + U4(t)

A Y (s) = e2 s+e4 s

s (3+s) (7+s)

B Y (s) = e2 s + 1s (3+s) (7+s) e

4 s


C Y (s) = e−4 s+e−2 s

s (3+s) (7+s)

D Y (s) = e−2 s + 1s (3+s) (7+s) e

−4 s


f(t) =− cos(3 t) + cos(4 t)

t

A F (s) = s9+s2 + s

16+s2

B F (s) = ln( 4+s3+s )

C F (s) = 12 ln( 9+s2

16+s2 )

D F (s) = − s9+s2 + s

16+s2


f(t) = −5 cosh(t) senh(t)

A F (s) = 5 s(−1+s) (1+s) (1+s2)

B F (s) = −5 s(−1+s)2 (1+s)2

C F (s) = 5(2−s) (2+s)

D F (s) =−5 (2−s2)

(2−2 s+s2) (2+2 s+s2)

E F (s) = − 15

1(2−s) (2+s)


f(t) = cos(8 t)− 4 t sen(8 t)

A F (s) = s2

(−64+s2)2

B F (s) = s2

(64+s2)2

C F (s) = s3

(64+s2)2

D F (s) = s3

(−64+s2)2



s

82 + 2 s+ s2

Respuesta:


f(t) ={

5 t si 0 ≤ t < 20 si 2 ≤ t

A 5s2 −

5s2 e−2 s

B 5s2 −

5 (1+2 s)s2 e−2 s

C − 5s2 + 5 (1+2 s)

s2 e−2 s

D 5s2 −

1+2 ss2 e−2 s



x′ = 5x+ 8 y

y′ = 8x+ 5 y


A {X(s) = 2 (21−s)−39−10 s+s2 , Y (s) = 36−4 s

39+10 s−s2 }

B {X(s) = 2 (−21+s)−39−10 s+s2 , Y (s) = −36+4 s

39+10 s−s2 }

C {X(s) = 2 (11+s)39+10 s−s2 , Y (s) = −4+4 s

39+10 s−s2 }

D {X(s) = 2 (11+s)−39−10 s+s2 , Y (s) = 4−4 s

39+10 s−s2 }


f(t) =

0 si 0 ≤ t < 22 t si 2 ≤ t < 70 si 7 ≤ t

A −(e−7 s + e−2 s + 7 s e−7 s + 2 s e−2 s

)2s

B −(e−7 s + e−2 s + 7 s e−7 s + 2 s e−2 s

)2s2

C −(e−7 s − e−2 s + 7 s e−7 s − 2 s e−2 s

)2s2

D(e−7 s − e−2 s + 7 s e−7 s − 2 s e−2 s

)2s2


F (s) =9

4 + s2

A 92 sen(2 t)

B 92 cos(4 t)

C 92 sen(4 t)

D 92 cos(2 t)

E 9 cos(2 t)

F 9 sen(2 t)


F (s) =8 + 6 s

−8 + 2 s+ s2

A f(t) = − 83 e−4 t + 10

3 e−2 t

B f(t) = 103 e2 t + 8

3 e4 t

C f(t) = 83 e−4 t + 10

3 e−2 t

D f(t) = 83 e−4 t + 10

3 e2 t



A F (s) = 12 ( −6

36+s2 + 10100+s2 )

B F (s) = 636+s2 + 10

100+s2

C F (s) = 8 s(4+s2) (64+s2)


D F (s) = −636+s2 + 10

100+s2


F (s) =s

89 + 10 s+ s2e−π s

A f(t) = e(5π−5 t)(cos(8 t)− 5

8 sen(8 t))Uπ(t)

B f(t) = e(5π−5 t) cos(8 t)Uπ(t)

C f(t) = cos(8 t)Uπ(t) e−5 t

D f(t) = e(5π−5 t) (cos(8 t)− 5 sen(8 t)) Uπ(t)


f(t) = 2 e2 t + e4 t

A F (s) = 10−3 s(4−s) (−2+s)

B F (s) = −6+s(4−s) (−2+s)

C F (s) = −6−s(2+s) (4+s)

D F (s) = 10+3 s(2+s) (4+s)


f(t) = e−2 t + sen(2 t)

A F (s) = −8−2 s−s2(−2−s) (4+s2)

B F (s) = 2 s+s2

(−2+s) (4+s2)

C F (s) = 2(−2+s) (4+s2)

D F (s) = −2(−2−s) (4+s2)



A F (s) = s(

19+s2 −

149+s2

)B F (s) = s

(1

4+s2 + 125+s2

)C F (s) = 1

2 s(

19+s2 −

149+s2

)D F (s) = s2

(4+s2) (25+s2)





Y (s) =As+B

(s2 + C s+D) (s+ E)


20 y − 9 y′ + y′′ = 4 e5 t

Respuesta:


f(t) =senh(5 t)

t

A F (s) = 110 ln( 5+s

−5+s )

B F (s) = ln(−5+s5+s )

C F (s) = ln( 5+s−5+s )

D F (s) = 12 ln( 5+s

−5+s )


F (s) =2 + s

(−9 + s) (−1 + s)

A 54 e−9 t − 1

4 e−t

B 118 e−9 t − 3

8 e−t

C 14 e

t + 54 e

9 t

D − 118 e−9 t + 3

8 e−t

E 14 e

t − 54 e

9 t

F − 38 e

t + 118 e9 t


−21 y + 4 y′ + y′′ = t3 e3 t

A Y (s) = 6(−3+s)5 (7+s)

B Y (s) = 3(−3+s)5 (7+s)

C Y (s) = 6(−7+s) (3+s)5

D Y (s) = 6(−3+s)4 (7+s)


2 y + y′ = f(t)

donde

f(t) =

0 t ≤ 02 + 3 t t ≤ 10 1 < t


A Y (s) = 32+s

B Y (s) = 3+ 1s−

1s e−s

2+s

C Y (s) =3+ 3

s2+ 2

s

2+s

D Y (s) =3+ 3

s2− 3

s2e−s+ 2

s−5s e−s

2+s


x′ = 2 t+ 3x+ 7 y

y′ = −5 + 7x+ 3 y


A {X(s) = −6−33 s+41 s2−2 s3

s2 (−40−6 s+s2) , Y (s) = 14+15 s−34 s2+5 s3

s2 (−40−6 s+s2) }

B {X(s) = −6−33 s−41 s2+2 s3

s2 (−40−6 s+s2) , Y (s) = 14+15 s+24 s2−5 s3

s2 (−40−6 s+s2) }

C {X(s) = −6−33 s+29 s2+2 s3

s2 (−40−6 s+s2) , Y (s) = 14+15 s−6 s2+5 s3

s2 (−40−6 s+s2) }

D {X(s) = 6+33 s+29 s2+2 s3

s2 (40+6 s−s2) , Y (s) = 14+15 s−4 s2−5 s3

s2 (−40−6 s+s2) }



A F (s) =−2 (128−4 s2)

(32−8 s+s2) (32+8 s+s2)

B F (s) = −2(8−s) (8+s)

C F (s) = 8 s(−4+s) (4+s) (16+s2)

D F (s) = −8 s(−4+s)2 (4+s)2

E F (s) = 8(8−s) (8+s)


f(t) ={

3 t si 0 ≤ t < 30 si 3 ≤ t

A − 3s2 + 3 (1+3 s)

s2 e−3 s

B 3s2 −

1+3 ss2 e−3 s

C 3s2 −

3 (1+3 s)s2 e−3 s

D 3s2 −

3s2 e−3 s


f(t) = et − sen(3 t)

A F (s) = 6−3 s+s2

(1+s) (9+s2)

B F (s) = −12+3 s−s2(1−s) (9+s2)


C F (s) = 3(1−s) (9+s2)

D F (s) = −3(1+s) (9+s2)


F (s) =8

113 + 14 s+ s2e−π s



C f(t) = e(−7π−7 t) sen(8 t)Uπ(t)



f(t) = 2 cos(2 t) + sen(2 t)

A F (s) = 4+s4+s2

B F (s) = 2−2 s4+s2

C F (s) = 4−s4+s2

D F (s) = 2+2 s4+s2


F (s) =−5

24− 11 s+ s2

A f(t) = e3 t − e8 t

B f(t) = −1 e−8 t + e−3 t

C f(t) = e−8 t + e−3 t

D f(t) = e3 t + e8 t


f(t) ={

3 si 0 ≤ t < 20 si 2 ≤ t

A 3s −

3s e−2 s

B −3 + 3 e−2 s

C 3 s− 3 s e−2 s

D − 3s + 3

s e−2 s

E 3− 3 e−2 s


f(t) = e−4 t + e−2 t

A F (s) = −6+2 s(−4+s) (−2+s)

B F (s) = 2(−4+s) (−2+s)

C F (s) = 2(−4−s) (2+s)

D F (s) = −6−2 s(−4−s) (2+s)



Y (s) =A

(s+B) (s+ C)4


−16 y − 6 y′ + y′′ = t2 e8 t

Respuesta:



A F (s) = 2 s(4+s2) (64+s2)

B F (s) = 636+s2 + 10

100+s2

C F (s) = 12 ( 6

36+s2 + 10100+s2 )

D F (s) = −636+s2 + 10

100+s2




2 + 7 s−8− 2 s+ s2

Respuesta:


f(t) = sen(6 t) e−4 t

A F (s) = 4+s52+8 s+s2

B F (s) = 4+s52−8 s+s2

C F (s) = 652+8 s+s2

D F (s) = 652−8 s+s2



A F (s) = s(

125+s2 + 1

36+s2

)B F (s) = 1

2 s(

11+s2 −

1121+s2

)C F (s) = s2

(25+s2) (36+s2)

D F (s) = s(

11+s2 −

1121+s2

)20. Indique cual de las opciones siguientes es la ecuacion subsidiaria del problema con condiciones iniciales y(0) = 0 y y′(0) = 0

con ecuacion:16 y + 10 y′ + y′′ = U4(t) + U8(t)


A Y (s) = e4 s+e8 s

s (2+s) (8+s)

B Y (s) = e−8 s+e−4 s

s (2+s) (8+s)

C Y (s) = e−4 s + 1s (2+s) (8+s) e

−8 s

D Y (s) = e4 s + 1s (2+s) (8+s) e

8 s


f(t) =114t2 sen(7 t)

A F (s) = −49+s2

(49+s2)3

B F (s) = −49+3 s2

(7+s2)3

C F (s) = 49+3 s2

(49+s2)3

D F (s) = −49+3 s2

(49+s2)3


F (s) =4

−2 + s

A 2 e2 t

B 8 e−2 t

C 4 e2 t

D 4 e−2 t


f(t) =

0 si 0 ≤ t < 32 t si 3 ≤ t < 70 si 7 ≤ t

A −(e−7 s + e−3 s + 7 s e−7 s + 3 s e−3 s

)2s2

B −(e−7 s − e−3 s + 7 s e−7 s − 3 s e−3 s

)2s2

C(e−7 s − e−3 s + 7 s e−7 s − 3 s e−3 s

)2s2

D −(e−7 s + e−3 s + 7 s e−7 s + 3 s e−3 s

)2s


F (s) =A

s2+ (

B

s+C

s2) eD s


f(t) ={t si 0 < t ≤ 30 si t > 3

Respuesta:





f(t) =112t sen(6 t)

A F (s) = 1(36+s2)2

B F (s) = s(−36+s2)2

C F (s) = 1(−36+s2)2

D F (s) = s(36+s2)2


f(t) =

0 si 0 ≤ t < 23 t si 2 ≤ t < 40 si 4 ≤ t

A −(e−4 s + e−2 s + 4 s e−4 s + 2 s e−2 s

)3s

B(e−4 s − e−2 s + 4 s e−4 s − 2 s e−2 s

)3s2

C −(e−4 s − e−2 s + 4 s e−4 s − 2 s e−2 s

)3s2

D −(e−4 s + e−2 s + 4 s e−4 s + 2 s e−2 s

)3s2


F (s) =4 + 6 s

−6 + s+ s2

A f(t) = − 145 e−3 t + 16

5 e−2 t

B f(t) = 145 e−3 t + 16

5 e−2 t

C f(t) = 145 e−3 t + 16

5 e2 t

D f(t) = 165 e2 t + 14

5 e3 t



A F (s) = 12 ( 3

9+s2 + 13169+s2 )

B F (s) = 39+s2 + 13

169+s2

C F (s) = −39+s2 + 13

169+s2

D F (s) = 5 s(25+s2) (64+s2)


40 y + 13 y′ + y′′ = U8(t)

A y(t) = − 83 e−8 t + 8

3 e−5 t − 1

40 U8(t)− 124 e

(64−8 t) U8(t)− 115 e

(40−5 t) U8(t)

B y(t) = − 83 e−8 t + 8

3 e−5 t + 1

40 U8(t) + 124 e

(64−8 t) U8(t)− 115 e

(40−5 t) U8(t)


C y(t) = − 83 e−8 t + 8

3 e−5 t + 1

40 U8(t)− 115 e

(−40+5 t) U8(t) + 124 e

(−64+8 t) U8(t)

D y(t) = 83 e−8 t − 8

3 e−5 t + 1

40 U8(t) + 124 e

(64−8 t) U8(t)− 115 e

(40−5 t) U8(t)



A F (s) = − 1(5−s) (1+s2)

B F (s) = 1(5+s) (1+s2)

C F (s) = 4−s−s2(5−s) (1+s2)

D F (s) = 6+s+s2

(5+s) (1+s2)


F (s) =A

s2+ (

B

s+C

s2) eD s


f(t) ={t si 0 < t ≤ 50 si t > 5

Respuesta:


f(t) = −5 e2 t + e3 t

A F (s) = 17−6 s(3−s) (−2+s)

B F (s) = 17+6 s(2+s) (3+s)

C F (s) = −13−4 s(2+s) (3+s)

D F (s) = −13+4 s(3−s) (−2+s)



A F (s) = 9 s(−3+s) (3+s) (9+s2)

B F (s) =−3 (54−3 s2)

(18−6 s+s2) (18+6 s+s2)

C F (s) = − 1(6−s) (6+s)

D F (s) = 9(6−s) (6+s)

E F (s) = −9 s(−3+s)2 (3+s)2


f(t) ={

2 t si 0 ≤ t < 40 si 4 ≤ t


A 2s2 −

1+4 ss2 e−4 s

B 2s2 −

2s2 e−4 s

C 2s2 −

2 (1+4 s)s2 e−4 s

D − 2s2 + 2 (1+4 s)

s2 e−4 s


16 y + y′′ = 16 t

A Y (s) = 16+4 s2+7 s3

s2 (16+s2)

B Y (s) = 16+4 s2+7 s3

16+s2

C Y (s) = 28+4 s2+7 s3

s2 (16+s2)

D Y (s) = 16−4 s2+7 s3

s2 (16+s2)


L(√t) =

12

√π

s32


32

A F (s) = 3√π

s32

B F (s) = 3√π

s52

C F (s) = 34

√π

s32

D F (s) = 34

√π

s52


f(t) = − cos(3 t) + 2 sen(3 t)

A F (s) = 6−s9+s2

B F (s) = 6+s9+s2

C F (s) = −3−2 s9+s2

D F (s) = −3+2 s9+s2


Y (s) =As3 +B s2 + C s+D

(s2 + E)2


4 y + y′′ = 2 cos(2 t)

Respuesta:


Y (s) =As3 +B s2 + C s+D

s2 (s2 + E)


64 y + y′′ = 64 t

Respuesta:


16. Indique cual de las siguientes opciones contiene la forma adecuada para la transformada inversa de la siguente funcion siendoH(s) un polinomio de grado 1.

F (s) =H(s)

−12 + s+ s2

A f(t) = B e−3 t +Ae4 t

B f(t) = Ae−4 t +B e−3 t

C f(t) = B e3 t +Ae4 t

D f(t) = Ae−4 t +B e3 t


32 y − 12 y′ + y′′ = 8 e4 t

A Y (s) = −24+8 s(−4+s) (32−12 s+s2)

B Y (s) = 24+8 s(−4+s) (32−12 s+s2)

C Y (s) = 24+8 s32−12 s+s2

D Y (s) = −24+8 s(−4+s) (32+12 s+s2)


f(t) = (4 + 2 t) e5 t

A F (s) = −18+4 s(5−s)2

B F (s) = 22+4 s(5+s)2

C F (s) = 2+4 s(−5+s) s2

D F (s) = −18−4 s(5+s)2


f(t) ={

3 si 0 ≤ t < 20 si 2 ≤ t

A 3s −

3s e−2 s

B 3 s− 3 s e−2 s

C − 3s + 3

s e−2 s

D 3− 3 e−2 s

E −3 + 3 e−2 s


F (s) =4

52 + 12 s+ s2e−π s


B f(t) = e(−6π−6 t) sen(4 t)Uπ(t)

C f(t) = e(6π−6 t) sen(4 t)Uπ(t)



x′ − y′ = 4x+ 6 y

x′ + y′ = 6x+ 4 y



A {X(s) = −13+2 s−10−4 s+s2 , Y (s) = 17−3 s

−10−4 s+s2 }

B {X(s) = 13−2 s−10−4 s+s2 , Y (s) = −17+3 s

−10−4 s+s2 }

C {X(s) = 17+2 s10+4 s−s2 , Y (s) = 13−3 s

−10−4 s+s2 }

D {X(s) = 17+2 s−10−4 s+s2 , Y (s) = −13+3 s

−10−4 s+s2 }


F (s) =8 + s

(−7 + s) (−1 + s)

A − 52 e−7 t + 3

2 e−t

B 52 e−7 t − 3

2 e−t

C 43 e−7 t − 1

3 e−t

D 13 e

t + 43 e

7 t

E 13 e

t − 43 e

7 t

F − 32 e

t + 52 e

7 t


F (s) =5− 7 s16 + s2

A 54 cos(4 t)− 7 sen(4 t)

B 5 cos(4 t)− 7 sen(4 t)

C −7 cos(4 t) + 5 sen(4 t)

D −7 cos(4 t) + 54 sen(4 t)

E cos(4 t) + sen(4 t)

F −7 cos(4 t)− 54 sen(4 t)



A F (s) = s2

(16+s2) (36+s2)

B F (s) = 12 s(

14+s2 −

1100+s2

)C F (s) = s

(1

16+s2 + 136+s2

)D F (s) = s

(1

4+s2 −1

100+s2

)






A F (s) = −(1 + s2)−1 + 749+s2

B F (s) = 12 ((1 + s2)−1 + 7

49+s2 )

C F (s) = (1 + s2)−1 + 749+s2

D F (s) = 3 s(9+s2) (16+s2)


F (s) =A

s2+ (

B

s+C

s2) eD s


f(t) ={t si 0 < t ≤ 50 si t > 5

Respuesta:



A F (s) = 12 s(

14+s2 −

1196+s2

)B F (s) = s

(1

36+s2 + 164+s2

)C F (s) = s2

(36+s2) (64+s2)

D F (s) = s(

14+s2 −

1196+s2

)4. Indique cual de las opciones siguientes es la ecuacion subsidiaria del problema con condiciones iniciales y(0) = 6 y y′(0) = 4

con ecuacion:64 y + y′′ = 64 t

A Y (s) = 64−4 s2+6 s3

s2 (64+s2)

B Y (s) = 64+4 s2+6 s3

64+s2

C Y (s) = 64+4 s2+6 s3

s2 (64+s2)

D Y (s) = 48+4 s2+6 s3

s2 (64+s2)


F (s) =7 + s

(−8 + s) (−2 + s)


A 12 e

2 t + 32 e

8 t

B 12 e

2 t − 32 e

8 t

C 52 e−8 t − 3

2 e−2 t

D 32 e−8 t − 1

2 e−2 t

E − 52 e−8 t + 3

2 e−2 t

F − 32 e

2 t + 52 e

8 t


Y (s) =As3 +B s2 + C s+D

(s2 + E)2


49 y + y′′ = 4 cos(7 t)

Respuesta:


f(t) ={

2 si 0 ≤ t < 50 si 5 ≤ t

A − 2s + 2

s e−5 s

B 2− 2 e−5 s

C 2 s− 2 s e−5 s

D 2s −

2s e−5 s

E −2 + 2 e−5 s


x′ = x+ 7 y

y′ = 7x+ y


A {X(s) = 25−4 s−48−2 s+s2 , Y (s) = 31−3 s

48+2 s−s2 }

B {X(s) = −25+4 s−48−2 s+s2 , Y (s) = 31−3 s

−48−2 s+s2 }

C {X(s) = 17+4 s−48−2 s+s2 , Y (s) = 25+3 s

−48−2 s+s2 }

D {X(s) = 17+4 s48+2 s−s2 , Y (s) = 25+3 s

48+2 s−s2 }


f(t) = (2 + 4 t) e−3 t

A F (s) = −2+2 s(−3+s)2

B F (s) = 10+2 s(−3−s)2

C F (s) = 4+2 ss2 (3+s)


D F (s) = 10−2 s(−3+s)2


F (s) =4

41 + 10 s+ s2e−π s

A f(t) = e(5π−5 t) sen(4 t)Uπ(t)

B f(t) = e(−5π−5 t) sen(4 t)Uπ(t)

C f(t) = e(5π−5 t) cos(4 t)Uπ(t)



f(t) =

0 si 0 ≤ t < 45 t si 4 ≤ t < 90 si 9 ≤ t

A(e−9 s − e−4 s + 9 s e−9 s − 4 s e−4 s

)5s2

B −(e−9 s − e−4 s + 9 s e−9 s − 4 s e−4 s

)5s2

C −(e−9 s + e−4 s + 9 s e−9 s + 4 s e−4 s

)5s2

D −(e−9 s + e−4 s + 9 s e−9 s + 4 s e−4 s

)5s


F (s) =4

−2 + s

A 8 e−2 t

B 4 e2 t

C 4 e−2 t

D 2 e2 t


f(t) ={

5 t si 0 ≤ t < 10 si 1 ≤ t

A − 5s2 + 5 (1+s)

s2 e−s

B 5s2 −

5 (1+s)s2 e−s

C 5s2 −

5s2 e−s

D 5s2 −

1+ss2 e−s


L(t32 ) =

34

√π

s52


52

A F (s) = 152

√π

s52

B F (s) = 158

√π

s72

C F (s) = 152

√π

s72

D F (s) = 158

√π

s52


20 y + 9 y′ + y′′ = U4(t)


A y(t) = 4 e−5 t − 4 e−4 t + 120 U4(t) + 1

5 e(20−5 t) U4(t)− 1

4 e(16−4 t) U4(t)

B y(t) = −4 e−5 t + 4 e−4 t + 120 U4(t)− 1

4 e(−16+4 t) U4(t) + 1

5 e(−20+5 t) U4(t)

C y(t) = −4 e−5 t + 4 e−4 t + 120 U4(t) + 1

5 e(20−5 t) U4(t)− 1

4 e(16−4 t) U4(t)

D y(t) = −4 e−5 t + 4 e−4 t − 120 U4(t) + 1

5 e(20−5 t) U4(t) + 1

4 e(16−4 t) U4(t)


F (s) =s

82 + 2 s+ s2

A f(t) = cos(9 t)− 19 sen(9 t)

B f(t) = (cos(9 t)− 19 sen(9 t)) e−t


D f(t) = (cos(9 t)− sen(9 t)) e−t


Y (s) =A

(s+B) (s+ C)5


−24 y − 5 y′ + y′′ = t3 e8 t

Respuesta:


f(t) = e−5 t + sen(5 t)

A F (s) = −5(−5−s) (25+s2)

B F (s) = 5(−5+s) (25+s2)

C F (s) = 5 s+s2

(−5+s) (25+s2)

D F (s) = −50−5 s−s2(−5−s) (25+s2)



A F (s) = 10(10−s) (10+s)

B F (s) = − 52

1(10−s) (10+s)

C F (s) = −10 s(−5+s)2 (5+s)2

D F (s) =−2 (250−5 s2)

(50−10 s+s2) (50+10 s+s2)

E F (s) = 10 s(−5+s) (5+s) (25+s2)


64 y + y′′ = 7 sen(8 t)

A Y (s) = 56+56 s+7 s2

(−64+s2)2

B Y (s) = 56+448 s+7 s3

(64+s2)2


C Y (s) = 56+56 s+7 s3

(64+s2)2

D Y (s) = 56+448 s+7 s3

(−64+s2)2


f(t) = e−2 t + 2 e−t

A F (s) = 3−s(−2+s) (−1+s)

B F (s) = −5−3 s(−2−s) (1+s)

C F (s) = 3+s(−2−s) (1+s)

D F (s) = −5+3 s(−2+s) (−1+s)


f(t) =114t2 sen(7 t)

A F (s) = −49+s2

(49+s2)3

B F (s) = 49+3 s2

(49+s2)3

C F (s) = −49+3 s2

(49+s2)3

D F (s) = −49+3 s2

(7+s2)3




−8 + 8 s−4 s+ s3

Respuesta:


f(t) = −2 cos(5 t) + sen(5 t)

A F (s) = −10+s25+s2

B F (s) = 5+2 s25+s2

C F (s) = 5−2 s25+s2

D F (s) = −10−s25+s2





36 y + y′′ = 3 sen(6 t)

A Y (s) = 18+108 s+3 s3

(36+s2)2

B Y (s) = 18+18 s+3 s3

(36+s2)2

C Y (s) = 18+18 s+3 s2

(−36+s2)2

D Y (s) = 18+108 s+3 s3

(−36+s2)2



A F (s) = −4 s(−2+s) (2+s) (4+s2)

B F (s) = 4 s(−2+s)2 (2+s)2

C F (s) =−2 (−16+2 s2)

(8−4 s+s2) (8+4 s+s2)

D F (s) = −4(−4−s) (−4+s)

E F (s) = 1(−4−s) (−4+s)


F (s) =8

(9 + s)3

A f(t) = 4 t2 e9 t

B f(t) = 8 t2 e9 t

C f(t) = 4 t2 e−9 t

D f(t) = 8 t2 e−9 t


x′ = 2 t+ 5x+ 9 y

y′ = 4 + 9x+ 5 y


A {X(s) =2 (−5+19 s+13 s2+s3)s2 (−56−10 s+s2) , Y (s) =

2 (9−10 s+s2+2 s3)s2 (−56−10 s+s2) }

B {X(s) =2 (−5+19 s+23 s2−s3)s2 (−56−10 s+s2) , Y (s) =

2 (9−10 s−17 s2+2 s3)s2 (−56−10 s+s2) }

C {X(s) =2 (−5+19 s−23 s2+s3)s2 (−56−10 s+s2) , Y (s) =

2 (9−10 s+21 s2−2 s3)s2 (−56−10 s+s2) }

D {X(s) =2 (−5+19 s−13 s2−s3)s2 (−56−10 s+s2) , Y (s) =

2 (9−10 s+3 s2−2 s3)s2 (−56−10 s+s2) }



f(t) ={

6 t si 0 ≤ t < 50 si 5 ≤ t

A 6s2 −

6 (1+5 s)s2 e−5 s

B − 6s2 + 6 (1+5 s)

s2 e−5 s

C 6s2 −

1+5 ss2 e−5 s

D 6s2 −

6s2 e−5 s


f(t) =114t sen(7 t)

A F (s) = 1(49+s2)2

B F (s) = 1(−49+s2)2

C F (s) = s(49+s2)2

D F (s) = s(−49+s2)2


Y (s) =As3 +B s2 + C s+D

(s2 + E)2


64 y + y′′ = 5 sen(8 t)

Respuesta:


3 y + y′ = t cos(2 t) + sen(2 t) e4 t

A Y (s) =6+ 4 s

(4+s2)2+ 2

20−8 s+s2

3+s

B Y (s) =6+ 2 s2

(4+s2)2−(4+s2)

−1+ 2

20+8 s+s2

3+s

C Y (s) =6+ 2 s2

(4+s2)2−(4+s2)

−1+ −4+s

20−8 s+s2

3+s

D Y (s) =6+ 2 s2

(4+s2)2−(4+s2)

−1+ 2

20−8 s+s2

3+s


f(t) = 3 cos(t) + sen(t)

A F (s) = 3+s1+s2

B F (s) = 1−3 s1+s2

C F (s) = 1+3 s1+s2

D F (s) = 3−s1+s2


f(t) = e−3 t + 5 e−2 t


A F (s) = 13−4 s(−3+s) (−2+s)

B F (s) = −17−6 s(−3−s) (2+s)

C F (s) = −17+6 s(−3+s) (−2+s)

D F (s) = 13+4 s(−3−s) (2+s)


4 y + y′′ = f(t)


f(t) =

0 si 0 ≤ t < π

1 si π ≤ t < 2π0 si 2π ≤ t

A y(t) = 7 cos(2 t) + 14 (1− cos(2 t)) Uπ(t)− 1

4 (1− cos(2 t)) U2π(t)

B y(t) = 7 cos(2 t) + 14 (1 + cos(2 t)) Uπ(t)− 1

4 (1 + cos(2 t)) U2π(t)

C y(t) = 7 cos(2 t) + 14 (1− cos(4 t)) Uπ(t)− 1

4 (1− cos(4 t)) U2π(t)

D y(t) = 72 sen(2 t) + 1

4 (1− cos(2 t)) Uπ(t)− 14 (1− cos(2 t)) U2π(t)

E y(t) = 7 cos(2 t) + 12 (1− cos(2 t)) Uπ(t)− 1

2 (1− cos(2 t)) U2π(t)


F (s) =s

100 + 12 s+ s2e−π s

A f(t) = e(6π−6 t) (cos(8 t)− 6 sen(8 t)) Uπ(t)

B f(t) = e(6π−6 t)(cos(8 t)− 3

4 sen(8 t))Uπ(t)

C f(t) = cos(8 t)Uπ(t) e−6 t



f(t) =

0 si 0 ≤ t < 22 t si 2 ≤ t < 50 si 5 ≤ t

A −(e−5 s − e−2 s + 5 s e−5 s − 2 s e−2 s

)2s2

B(e−5 s − e−2 s + 5 s e−5 s − 2 s e−2 s

)2s2

C −(e−5 s + e−2 s + 5 s e−5 s + 2 s e−2 s

)2s2

D −(e−5 s + e−2 s + 5 s e−5 s + 2 s e−2 s

)2s


F (s) =2

25 + s2

A 2 sen(5 t)

B 25 sen(5 t)

C 25 cos(25 t)

D 25 cos(5 t)

E 2 cos(5 t)

F 25 sen(25 t)



Y (s) =A

(s+B) (s+ C)7


−30 y − y′ + y′′ = t5 e6 t

Respuesta:


f(t) ={

6 si 0 ≤ t < 20 si 2 ≤ t

A 6− 6 e−2 s

B − 6s + 6

s e−2 s

C 6 s− 6 s e−2 s

D 6s −

6s e−2 s

E −6 + 6 e−2 s


F (s) =A

s2+ (

B

s+C

s2) eD s


f(t) ={t si 0 < t ≤ 60 si t > 6

Respuesta:


f(t) =sen(5 t)

t

A F (s) = 15 Atan(s)

B F (s) = Atan( 5s )

C F (s) = Atan(s)

D F (s) = Atan( 15 s)



A F (s) = 39+s2 + 9

81+s2

B F (s) = 3 s(9+s2) (36+s2)

C F (s) = −39+s2 + 9

81+s2

D F (s) = 12 ( 3

9+s2 + 981+s2 )




A F (s) = 1(4+s) (1+s2)

B F (s) = − 1(4−s) (1+s2)

C F (s) = 5+s+s2

(4+s) (1+s2)

D F (s) = 3−s−s2(4−s) (1+s2)


f(t) = t cosh(4 t)

A F (s) = 16+s2

−16+s2

B F (s) = (16 + s2)−1

C F (s) = 8 s(−16+s2)2

D F (s) = 16+s2

(−16+s2)2



s

17 + 2 s+ s2

Respuesta:


F (s) =5 + 5 s+ 6 s2

s6

A 32 t

4 + t5 + 56 t

6

B 6 t3 + 5 t4 + 5 t5

C 2 t3 + 54 t

4 + t5

D 6 t4 + 5 t5 + 5 t6

E t3 + 524 t

4 + 124 t

5

F 14 t

4 + 124 t

5 + 1144 t

6



A F (s) = 12 s(

14+s2 −

1100+s2

)B F (s) = s

(1

16+s2 + 136+s2

)C F (s) = s

(1

4+s2 −1

100+s2

)D F (s) = s2

(16+s2) (36+s2)





f(t) = cos(t) + 2 sen(t)

A F (s) = 2−s1+s2

B F (s) = 1−2 s1+s2

C F (s) = 2+s1+s2

D F (s) = 1+2 s1+s2


18 y + 9 y′ + y′′ = U2(t)

A y(t) = − 23 e−6 t + 2

3 e−3 t − 1

18 U2(t)− 118 e

(12−6 t) U2(t)− 19 e

(6−3 t) U2(t)

B y(t) = − 23 e−6 t + 2

3 e−3 t + 1

18 U2(t)− 19 e

(−6+3 t) U2(t) + 118 e

(−12+6 t) U2(t)

C y(t) = − 23 e−6 t + 2

3 e−3 t + 1

18 U2(t) + 118 e

(12−6 t) U2(t)− 19 e

(6−3 t) U2(t)

D y(t) = 23 e−6 t − 2

3 e−3 t + 1

18 U2(t) + 118 e

(12−6 t) U2(t)− 19 e

(6−3 t) U2(t)


f(t) =12t2 cos(4 t)

A F (s) = 48 s+s3

(16+s2)3

B F (s) = −12 s+s3

(16+s2)3

C F (s) = 12 s+s3

(16+s2)3

D F (s) = −48 s+s3

(16+s2)3


f(t) = sen(7 t) e−4 t

A F (s) = 765+8 s+s2

B F (s) = 4+s65−8 s+s2

C F (s) = 765−8 s+s2

D F (s) = 4+s65+8 s+s2


F (s) =9

3 + s

A 9 e3 t

B −3 e−3 t

C 9 e−3 t

D −27 e3 t



f(t) = et + e2 t

A F (s) = 3+2 s(1+s) (2+s)

B F (s) = 1(1+s) (2+s)

C F (s) = 3−2 s(1−s) (−2+s)

D F (s) = 1(1−s) (−2+s)


f(t) ={

2 t si 0 ≤ t < 10 si 1 ≤ t

A 2s2 −

1+ss2 e−s

B 2s2 −

2 (1+s)s2 e−s

C 2s2 −

2s2 e−s

D − 2s2 + 2 (1+s)

s2 e−s


Y (s) =A

(s+B) (s+ C)5


−42 y − y′ + y′′ = t3 e7 t

Respuesta:


9 y + y′′ = 5 cos(3 t)

A Y (s) = 50+5 s2

(9+s2)2

B Y (s) =s (45+5 s2)

(9+s2)2

C Y (s) =s (50−5 s2)

(9+s2)2

D Y (s) =s (50+5 s2)

(9+s2)2


f(t) =e−8 t − e−7 t

t

A F (s) = ln( 7+s8+s )

B F (s) = − 17+s + 1

8+s

C F (s) = 1−8+s −

1−7+s

D F (s) = 7+s8+s



F (s) =A

s2+ (

B

s+C

s2) eD s


f(t) ={t si 0 < t ≤ 30 si t > 3

Respuesta:


f(t) = e−4 t + sen(2 t)

A F (s) = −12−2 s−s2(−4−s) (4+s2)

B F (s) = −2(−4−s) (4+s2)

C F (s) = 2(−4+s) (4+s2)

D F (s) = −4+2 s+s2

(−4+s) (4+s2)



s

−35 + 2 s+ s2

Respuesta:




2 + 3 s−27 + 6 s+ s2

Respuesta:


f(t) =

0 si 0 ≤ t < 46 t si 4 ≤ t < 90 si 9 ≤ t

A −(e−9 s + e−4 s + 9 s e−9 s + 4 s e−4 s

)6s

B(e−9 s − e−4 s + 9 s e−9 s − 4 s e−4 s

)6s2

C −(e−9 s − e−4 s + 9 s e−9 s − 4 s e−4 s

)6s2

D −(e−9 s + e−4 s + 9 s e−9 s + 4 s e−4 s

)6s2



A F (s) = s2

(4+s2) (49+s2)


B F (s) = s(

14+s2 + 1

49+s2

)C F (s) = 1

2 s(

125+s2 −

181+s2

)D F (s) = s

(1

25+s2 −1

81+s2

)17. Cual de las siguientes opciones representa la transformada de Laplace de la funcion: (Sugerencia: use las definiciones de

las funciones de las funciones hiperbolicas mediante exponenciales)


A F (s) = 54

1(−10−s) (−10+s)

B F (s) =−4 (−250+5 s2)

(50−10 s+s2) (50+10 s+s2)

C F (s) = 20 s(−5+s)2 (5+s)2

D F (s) = −20(−10−s) (−10+s)

E F (s) = −20 s(−5+s) (5+s) (25+s2)


x′ = 5 t+ x+ 9 y

y′ = 4 + 9x+ y


A {X(s) = −5+41 s+41 s2−5 s3

s2 (−80−2 s+s2) , Y (s) = 45−4 s−45 s2+4 s3

s2 (−80−2 s+s2) }

B {X(s) = −5+41 s−41 s2+5 s3

s2 (−80−2 s+s2) , Y (s) = 45−4 s+53 s2−4 s3

s2 (−80−2 s+s2) }

C {X(s) = −5+41 s−31 s2−5 s3

s2 (−80−2 s+s2) , Y (s) = 45−4 s−37 s2−4 s3

s2 (−80−2 s+s2) }

D {X(s) = −5+41 s+31 s2+5 s3

s2 (−80−2 s+s2) , Y (s) = 45−4 s+45 s2+4 s3

s2 (−80−2 s+s2) }



A F (s) = −24+s2 + 10

100+s2

B F (s) = 6 s(16+s2) (36+s2)

C F (s) = 12 ( −2

4+s2 + 10100+s2 )

D F (s) = 24+s2 + 10

100+s2


f(t) ={

3 si 0 ≤ t < 30 si 3 ≤ t

A 3s −

3s e−3 s

B −3 + 3 e−3 s


C − 3s + 3

s e−3 s

D 3− 3 e−3 s

E 3 s− 3 s e−3 s


−56 y + y′ + y′′ = t5 e7 t

A Y (s) = 5(−7+s)7 (8+s)

B Y (s) = 120(−7+s)6 (8+s)

C Y (s) = 120(−7+s)7 (8+s)

D Y (s) = 120(−8+s) (7+s)7


Y (s) =As3 +B s2 + C s+D

(s2 + E)2


9 y + y′′ = 8 sen(3 t)

Respuesta:


F (s) =8

100 + 12 s+ s2e−π s

A f(t) = e(6π−6 t) cos(8 t)Uπ(t)


C f(t) = e(−6π−6 t) sen(8 t)Uπ(t)



F (s) =3 + 4 s+ 4 s2

s5

A 4 t2 + 4 t3 + 3 t4

B 23 t

3 + 16 t

4 + 140 t

5

C 4 t3 + 4 t4 + 3 t5

D 2 t2 + 43 t

3 + 34 t

4

E 43 t

3 + t4 + 35 t

5

F 2 t2 + 23 t

3 + 18 t

4





F (s) =9

25 + s2

A 95 sen(25 t)

B 9 sen(5 t)

C 9 cos(5 t)

D 95 cos(5 t)

E 95 sen(5 t)

F 95 cos(25 t)


F (s) =3

4 + s2

A 3 cos(2 t)

B 3 sen(2 t)

C 32 sen(4 t)

D 32 cos(4 t)

E 32 cos(2 t)

F 32 sen(2 t)


Y (s) =As+B

(s2 + C s+D) (s+ E)


6 y − 5 y′ + y′′ = 2 e3 t

Respuesta:


f(t) ={

2 si 0 ≤ t < 20 si 2 ≤ t

A 2s −

2s e−2 s

B −2 + 2 e−2 s

C − 2s + 2

s e−2 s

D 2− 2 e−2 s

E 2 s− 2 s e−2 s


f(t) = e−4 t − sen(3 t)

A F (s) = −3(−4+s) (9+s2)

B F (s) = 3(−4−s) (9+s2)


C F (s) = 3+3 s−s2(−4−s) (9+s2)

D F (s) = 21−3 s+s2

(−4+s) (9+s2)


x′ = 5 + 5x+ 10 y

y′ = 4 t+ 10x+ 5 y


A {X(s) = 40−25 s+75 s2−4 s3

s2 (−75−10 s+s2) , Y (s) = 20−54 s+65 s2−5 s3

s2 (75+10 s−s2) }

B {X(s) = 40−25 s+35 s2+4 s3

s2 (−75−10 s+s2) , Y (s) = 20−54 s−15 s2−5 s3

s2 (75+10 s−s2) }

C {X(s) = 40−25 s−65 s2+4 s3

s2 (−75−10 s+s2) , Y (s) = 20−54 s−65 s2+5 s3

s2 (75+10 s−s2) }

D {X(s) = 40−25 s−25 s2−4 s3

s2 (−75−10 s+s2) , Y (s) = 20−54 s+15 s2+5 s3

s2 (75+10 s−s2) }



A F (s) = 589+16 s+s2

B F (s) = 539−16 s+s2

C F (s) = 8+s89+16 s+s2

D F (s) = −8+s39−16 s+s2


f(t) = e−3 t + 5 et

A F (s) = −14−6 s(−3−s) (−1+s)

B F (s) = −14+6 s(−3+s) (1+s)

C F (s) = 16+4 s(−3−s) (−1+s)

D F (s) = 16−4 s(−3+s) (1+s)


−12 y + 4 y′ + y′′ = t5 e2 t

A Y (s) = 120(−6+s) (2+s)7

B Y (s) = 5(−2+s)7 (6+s)

C Y (s) = 120(−2+s)6 (6+s)

D Y (s) = 120(−2+s)7 (6+s)


F (s) =−27− 21 s−9 s+ s3

A f(t) = 3 + 5 e−3 t − 2 e3 t

B f(t) = 3− 5 e−3 t + 2 e3 t

C f(t) = 3 + 2 e−3 t − 5 e3 t

D f(t) = 3− 2 e−3 t + 5 e3 t




A F (s) = 12 ( 5

25+s2 + 11121+s2 )

B F (s) = −525+s2 + 11

121+s2

C F (s) = 525+s2 + 11

121+s2

D F (s) = 3 s(9+s2) (64+s2)


49 y + y′′ = 49 t

A Y (s) = 49−2 s2+5 s3

s2 (49+s2)

B Y (s) = 49+2 s2+5 s3

49+s2

C Y (s) = 49+2 s2+5 s3

s2 (49+s2)

D Y (s) = 35+2 s2+5 s3

s2 (49+s2)



A F (s) = 4 s(−4+s) (4+s) (16+s2)

B F (s) = − 128−4 s2

(32−8 s+s2) (32+8 s+s2)

C F (s) = −4(8−s) (8+s)

D F (s) = −4 s(−4+s)2 (4+s)2

E F (s) = 4(8−s) (8+s)


f(t) ={

5 t si 0 ≤ t < 50 si 5 ≤ t

A 5s2 −

5s2 e−5 s

B 5s2 −

5 (1+5 s)s2 e−5 s

C 5s2 −

1+5 ss2 e−5 s

D − 5s2 + 5 (1+5 s)

s2 e−5 s


f(t) = 2 cos(2 t)− 3 sen(2 t)

A F (s) = −6−2 s4+s2

B F (s) = −6+2 s4+s2


C F (s) = 4+3 s4+s2

D F (s) = 4−3 s4+s2


f(t) =

0 si 0 ≤ t < 55 t si 5 ≤ t < 90 si 9 ≤ t

A −(e−9 s − e−5 s + 9 s e−9 s − 5 s e−5 s

)5s2

B −(e−9 s + e−5 s + 9 s e−9 s + 5 s e−5 s

)5s2

C(e−9 s − e−5 s + 9 s e−9 s − 5 s e−5 s

)5s2

D −(e−9 s + e−5 s + 9 s e−9 s + 5 s e−5 s

)5s


F (s) =A

s2+ (

B

s+C

s2) eD s


f(t) ={t si 0 < t ≤ 50 si t > 5

Respuesta:


f(t) =− cos(5 t) + cos(6 t)

t

A F (s) = − s25+s2 + s

36+s2

B F (s) = ln( 6+s5+s )

C F (s) = s25+s2 + s

36+s2

D F (s) = 12 ln( 25+s2

36+s2 )



A F (s) = s(

136+s2 + 1

64+s2

)B F (s) = s2

(36+s2) (64+s2)

C F (s) = 12 s(

14+s2 −

1196+s2

)D F (s) = s

(1

4+s2 −1

196+s2

)20. Indique cual opcion contiene la transformada inversa de Laplace de :

F (s) =s

65 + 8 s+ s2e−π s

A f(t) = − e(4π−4 t) (cos(7 t)− 4 sen(7 t)) Uπ(t)

B f(t) = − e(4π−4 t) cos(7 t)Uπ(t)

C f(t) = − cos(7 t)Uπ(t) e−4 t

D f(t) = − e(4π−4 t)(cos(7 t)− 4

7 sen(7 t))Uπ(t)



f(t) =12t2 cos(3 t)

A F (s) = 27 s+s3

(9+s2)3

B F (s) = −27 s+s3

(9+s2)3

C F (s) = −9 s+s3

(9+s2)3

D F (s) = 9 s+s3

(9+s2)3


Y (s) =As3 +B s2 + C s+D

(s2 + E)2


16 y + y′′ = 3 sen(4 t)

Respuesta:


F (s) =s

29 + 4 s+ s2

A f(t) = cos(5 t)− 25 sen(5 t)

B f(t) = cos(5 t)− 2 sen(5 t)

C f(t) = (cos(5 t)− 25 sen(5 t)) e−2 t

D f(t) = (cos(5 t)− 2 sen(5 t)) e−2 t


25 y + y′′ = sen(8 t)U2π(t)

A y(t) = 7 cos(5 t)− 8195 sen(5 t)U2π(t)− 1

39 sen(8 t)U2π(t)

B y(t) = 7 sen(5 t)− 8195 cos(5 t)U2π(t)− 1

39 cos(8 t)U2π(t)

C y(t) = 7 cos(5 t)− 8195 sen(5 t)U2π(t) + 1

39 sen(8 t)U2π(t)

D y(t) = 7 cos(5 t) + 8195 sen(5 t)U2π(t)− 1

39 sen(8 t)U2π(t)





2 y + y′ = f(t)

donde

f(t) =

0 t ≤ 02 + t t ≤ 10 1 < t

A Y (s) = 7+s−2+ 2s

2+s

B Y (s) = 72+s

C Y (s) = 7+ 1s−

1s e−s

2+s

D Y (s) = 7+s−2−s−2 e−s+ 2s−

3s e−s

2+s



A F (s) = s(

19+s2 −

181+s2

)B F (s) = s

(1

9+s2 + 136+s2

)C F (s) = s2

(9+s2) (36+s2)

D F (s) = 12 s(

19+s2 −

181+s2

)3. Cual es la transformada de Laplace de la funcion:


A F (s) = 9+s97+18 s+s2

B F (s) = 465−18 s+s2

C F (s) = 497+18 s+s2

D F (s) = −9+s65−18 s+s2


f(t) =

0 si 0 ≤ t < 36 t si 3 ≤ t < 60 si 6 ≤ t

A(e−6 s − e−3 s + 6 s e−6 s − 3 s e−3 s

)6s2

B −(e−6 s − e−3 s + 6 s e−6 s − 3 s e−3 s

)6s2

C −(e−6 s + e−3 s + 6 s e−6 s + 3 s e−3 s

)6s

D −(e−6 s + e−3 s + 6 s e−6 s + 3 s e−3 s

)6s2



F (s) =10

16 + s2

A 10 cos(4 t)

B 10 sen(4 t)

C 52 sen(16 t)

D 52 cos(16 t)

E 52 sen(4 t)

F 52 cos(4 t)


f(t) ={

5 t si 0 ≤ t < 20 si 2 ≤ t

A 5s2 −

1+2 ss2 e−2 s

B 5s2 −

5s2 e−2 s

C 5s2 −

5 (1+2 s)s2 e−2 s

D − 5s2 + 5 (1+2 s)

s2 e−2 s



A F (s) = 12 ( −2

4+s2 + 14196+s2 )

B F (s) = 8 s(36+s2) (64+s2)

C F (s) = 24+s2 + 14

196+s2

D F (s) = −24+s2 + 14

196+s2


Y (s) =A

(s+B) (s+ C)7


−15 y + 2 y′ + y′′ = t5 e3 t

Respuesta:


L(t52 ) =

158

√π

s72


72

A F (s) = 10516

√π

s72

B F (s) = 10516

√π

s92

C F (s) = 1054

√π

s92

D F (s) = 1054

√π

s72


6 y + y′ = t cos(2 t) + sen(2 t) e5 t


A Y (s) =2+ 2 s2

(4+s2)2−(4+s2)

−1+ 2

29+10 s+s2

6+s

B Y (s) =2+ 4 s

(4+s2)2+ 2

29−10 s+s2

6+s

C Y (s) =2+ 2 s2

(4+s2)2−(4+s2)

−1+ 2

29−10 s+s2

6+s

D Y (s) =2+ 2 s2

(4+s2)2−(4+s2)

−1+ −5+s

29−10 s+s2

6+s


F (s) =5− 5 s9 + s2

A −5 cos(3 t) + 5 sen(3 t)

B cos(3 t) + sen(3 t)

C −5 cos(3 t) + 53 sen(3 t)

D −5 cos(3 t)− 53 sen(3 t)

E 53 cos(3 t)− 5 sen(3 t)

F 5 cos(3 t)− 5 sen(3 t)


F (s) =A

s2+ (

B

s+C

s2) eD s


f(t) ={t si 0 < t ≤ 20 si t > 2

Respuesta:




−64− 8 s−16 s+ s3

Respuesta:


F (s) =s

26 + 2 s+ s2

A f(t) = cos(5 t)− 15 sen(5 t)

B f(t) = (cos(5 t)− sen(5 t)) e−t


D f(t) = (cos(5 t)− 15 sen(5 t)) e−t


Y (s) =As+B

(s2 + C s+D) (s+ E)


8 y − 6 y′ + y′′ = 4 e2 t

Respuesta:



f(t) = 3 cos(4 t)− 2 sen(4 t)

A F (s) = −8+3 s16+s2

B F (s) = 12+2 s16+s2

C F (s) = −8−3 s16+s2

D F (s) = 12−2 s16+s2


f(t) = e−3 t − 2 e4 t

A F (s) = 10+s(−3−s) (−4+s)

B F (s) = 10−s(−3+s) (4+s)

C F (s) = −2+3 s(−3+s) (4+s)

D F (s) = −2−3 s(−3−s) (−4+s)


f(t) =110t2 sen(5 t)

A F (s) = −25+s2

(25+s2)3

B F (s) = −25+3 s2

(25+s2)3

C F (s) = −25+3 s2

(5+s2)3

D F (s) = 25+3 s2

(25+s2)3


x′ − y′ = x+ 7 y

x′ + y′ = 7x+ y


A {X(s) = 25+3 s−24−s+s2 , Y (s) = −7+4 s

−24−s+s2 }

B {X(s) = −7+3 s−24−s+s2 , Y (s) = 25−4 s

−24−s+s2 }

C {X(s) = −7+3 s24+s−s2 , Y (s) = −25+4 s

−24−s+s2 }

D {X(s) = 25+3 s24+s−s2 , Y (s) = 7−4 s

−24−s+s2 }


F (s) =5

34 + 6 s+ s2e−π s

A f(t) = − e(−3π−3 t) sen(5 t)Uπ(t)

B f(t) = − e(3π−3 t) cos(5 t)Uπ(t)

C f(t) = − e(3π+3 t) sen(5 t)Uπ(t)

D f(t) = − e(3π−3 t) sen(5 t)Uπ(t)


24 y + 10 y′ + y′′ = U7(t)


A y(t) = − 72 e−6 t + 7

2 e−4 t + 1

24 U7(t) + 112 e

(42−6 t) U7(t)− 18 e

(28−4 t) U7(t)

B y(t) = − 72 e−6 t + 7

2 e−4 t + 1

24 U7(t)− 18 e

(−28+4 t) U7(t) + 112 e

(−42+6 t) U7(t)

C y(t) = 72 e−6 t − 7

2 e−4 t + 1

24 U7(t) + 112 e

(42−6 t) U7(t)− 18 e

(28−4 t) U7(t)

D y(t) = − 72 e−6 t + 7

2 e−4 t − 1

24 U7(t) + 112 e

(42−6 t) U7(t) + 18 e

(28−4 t) U7(t)


f(t) ={

3 si 0 ≤ t < 30 si 3 ≤ t

A 3 s− 3 s e−3 s

B 3− 3 e−3 s

C − 3s + 3

s e−3 s

D −3 + 3 e−3 s

E 3s −

3s e−3 s


f(t) = e−3 t + sen(t)

A F (s) = − 1(−3−s) (1+s2)

B F (s) = 1(−3+s) (1+s2)

C F (s) = −4−s−s2(−3−s) (1+s2)

D F (s) = −2+s+s2

(−3+s) (1+s2)



A F (s) = −128+4 s2

(32−8 s+s2) (32+8 s+s2)

B F (s) = 4(−8−s) (−8+s)

C F (s) = −4 s(−4+s)2 (4+s)2

D F (s) = 4 s(−4+s) (4+s) (16+s2)

E F (s) = −4(−8−s) (−8+s)





f(t) =

0 si 0 ≤ t < 54 t si 5 ≤ t < 80 si 8 ≤ t

A −(e−8 s − e−5 s + 8 s e−8 s − 5 s e−5 s

)4s2

B −(e−8 s + e−5 s + 8 s e−8 s + 5 s e−5 s

)4s

C(e−8 s − e−5 s + 8 s e−8 s − 5 s e−5 s

)4s2

D −(e−8 s + e−5 s + 8 s e−8 s + 5 s e−5 s

)4s2


f(t) =112t2 sen(6 t)

A F (s) = −36+3 s2

(36+s2)3

B F (s) = −36+s2

(36+s2)3

C F (s) = −36+3 s2

(6+s2)3

D F (s) = 36+3 s2

(36+s2)3


F (s) =A

s2+ (

B

s+C

s2) eD s


f(t) ={t si 0 < t ≤ 40 si t > 4

Respuesta:


f(t) = cos(3 t) e−9 t

A F (s) = 390+18 s+s2

B F (s) = 9+s90+18 s+s2

C F (s) = −9+s90−18 s+s2

D F (s) = 390−18 s+s2



A F (s) = s2

(16+s2) (36+s2)


B F (s) = s(

14+s2 −

1100+s2

)C F (s) = s

(1

16+s2 + 136+s2

)D F (s) = 1

2 s(

14+s2 −

1100+s2


f(t) ={

2 si 0 ≤ t < 20 si 2 ≤ t

A − 2s + 2

s e−2 s

B 2− 2 e−2 s

C −2 + 2 e−2 s

D 2 s− 2 s e−2 s

E 2s −

2s e−2 s


F (s) =2 + s

(−6 + s) (−5 + s)

A −7 e5 t + 8 e6 t

B −8 e−6 t + 7 e−5 t

C 6 e5 t + 7 e6 t

D 6 e5 t − 7 e6 t

E 8 e−6 t − 7 e−5 t

F 7 e−6 t − 6 e−5 t


Y (s) =As3 +B s2 + C s+D

(s2 + E)2


25 y + y′′ = 7 cos(5 t)

Respuesta:


6 y + y′ = t cos(6 t) + sen(6 t) e5 t

A Y (s) =5+ 2 s2

(36+s2)2−(36+s2)

−1+ 6

61+10 s+s2

6+s

B Y (s) =5+ 2 s2

(36+s2)2−(36+s2)

−1+ −5+s

61−10 s+s2

6+s

C Y (s) =5+ 2 s2

(36+s2)2−(36+s2)

−1+ 6

61−10 s+s2

6+s

D Y (s) =5+ 12 s

(36+s2)2+ 6

61−10 s+s2

6+s



F (s) =2

(4 + s)3

A f(t) = t2 e4 t

B f(t) = 2 t2 e−4 t

C f(t) = 2 t2 e4 t

D f(t) = t2 e−4 t



A F (s) = 2 s(4+s2) (36+s2)

B F (s) = −416+s2 + 8

64+s2

C F (s) = 416+s2 + 8

64+s2

D F (s) = 12 ( 4

16+s2 + 864+s2 )



A F (s) = −6(4−s) (4+s)

B F (s) = 6 s(−2+s)2 (2+s)2

C F (s) = −6 s(−2+s) (2+s) (4+s2)

D F (s) =3 (16−2 s2)

(8−4 s+s2) (8+4 s+s2)

E F (s) = 23

1(4−s) (4+s)


f(t) =sen(2 t)

t

A F (s) = Atan( 12 s)

B F (s) = Atan(s)

C F (s) = 12 Atan(s)



F (s) =s

13− 4 s+ s2

A f(t) = (cos(3 t) + 23 sen(3 t)) e2 t

B f(t) = cos(3 t)− 23 sen(3 t)

C f(t) = cos(3 t) + 23 sen(3 t)

D f(t) = (cos(3 t) + 2 sen(3 t)) e2 t


x′ = −3 + x+ 10 y

y′ = t+ 10x+ y



A {X(s) = 10+3 s+28 s2−s3s2 (−99−2 s+s2) , Y (s) = 1+29 s+13 s2−3 s3

s2 (99+2 s−s2) }

B {X(s) = 10+3 s−34 s2+s3

s2 (−99−2 s+s2) , Y (s) = 1+29 s−13 s2+3 s3

s2 (99+2 s−s2) }

C {X(s) = 10+3 s+26 s2+s3

s2 (−99−2 s+s2) , Y (s) = 1+29 s−7 s2−3 s3

s2 (99+2 s−s2) }

D {X(s) = 10+3 s−32 s2−s3s2 (−99−2 s+s2) , Y (s) = 1+29 s+7 s2+3 s3

s2 (99+2 s−s2) }


f(t) = −5 cos(t) + 3 sen(t)

A F (s) = 3+5 s1+s2

B F (s) = −5−3 s1+s2

C F (s) = −5+3 s1+s2

D F (s) = 3−5 s1+s2


f(t) = 3 e−2 t + e2 t

A F (s) = 4−4 s(2−s) (2+s)

B F (s) = 4+4 s(−2+s) (2+s)

C F (s) = −8−2 s(−2+s) (2+s)

D F (s) = −8+2 s(2−s) (2+s)


F (s) =7

58 + 6 s+ s2e−π s

A f(t) = − e(3π+3 t) sen(7 t)Uπ(t)

B f(t) = − e(3π−3 t) cos(7 t)Uπ(t)

C f(t) = − e(−3π−3 t) sen(7 t)Uπ(t)

D f(t) = − e(3π−3 t) sen(7 t)Uπ(t)


f(t) ={

6 t si 0 ≤ t < 50 si 5 ≤ t

A − 6s2 + 6 (1+5 s)

s2 e−5 s

B 6s2 −

6 (1+5 s)s2 e−5 s

C 6s2 −

6s2 e−5 s

D 6s2 −

1+5 ss2 e−5 s


24 y + 11 y′ + y′′ = U4(t) + U8(t)

A Y (s) = e4 s+e8 s

s (3+s) (8+s)


B Y (s) = e−8 s+e−4 s

s (3+s) (8+s)

C Y (s) = e−4 s + 1s (3+s) (8+s) e

−8 s

D Y (s) = e4 s + 1s (3+s) (8+s) e

8 s


−8 y − 2 y′ + y′′ = t5 e4 t

A Y (s) = 120(−2+s) (4+s)7

B Y (s) = 120(−4+s)7 (2+s)

C Y (s) = 120(−4+s)6 (2+s)

D Y (s) = 5(−4+s)7 (2+s)


f(t) = e−4 t − sen(2 t)

A F (s) = 4+2 s−s2(−4−s) (4+s2)

B F (s) = 2(−4−s) (4+s2)

C F (s) = −2(−4+s) (4+s2)

D F (s) = 12−2 s+s2

(−4+s) (4+s2)


F (s) =10−3 + s

A 10 e3 t

B 10 e−3 t

C 103 e3 t

D 30 e−3 t


Y (s) =A

(s+B) (s+ C)7


−18 y + 3 y′ + y′′ = t5 e3 t

Respuesta:

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