ecuaciones diferenciales - cb.mty.itesm.mxcb.mty.itesm.mx/ma2001/alumno/tareas/ma2001-hw-7a.pdf ·...

Ecuaciones DiferencialesTarea No 7: Transformada de Laplace

Maestra Marıa Graciela Trevino, Agosto-Diciembre 2018

Grupo: Matrıcula: Nombre: Tipo:-1

1. Cual de las siguientes opciones representa la transformada

de Laplace de la funcion:

f(t) = −2 cos(5 t) − 4 sen(5 t)

A F (s) = −10−4 s25+s2

B F (s) = −20−2 s25+s2

C F (s) = −10+4 s25+s2

D F (s) = −20+2 s25+s2

2. Cual de las siguientes opciones representa la transforma-

da de Laplace de la funcion: (Sugerencia:Use identidades

trigonometricas para convertir un producto en una suma)

f(t) = sen(3 t) sen(8 t)

A F (s) = s(

125+s2 − 1

121+s2

)B F (s) = s

(1

9+s2 + 164+s2

)C F (s) = 1

2 s(

125+s2 − 1

121+s2

)D F (s) = s2

(9+s2) (64+s2)


de Laplace Inversa de la funcion:

F (s) =4 + 5 s + 9 s2

s6

A 9 t3 + 5 t4 + 4 t5

B 94 t

4 + t5 + 23 t

6

C 38 t

4 + 124 t

5 + 1180 t

6

D 3 t3 + 54 t

4 + 45 t

5

E 32 t

3 + 524 t

4 + 130 t

5

F 9 t4 + 5 t5 + 4 t6


Inversa de Laplace de la funcion:

F (s) =2 + s

(−8 + s) (−4 + s)

A − 52 e−8 t + 3

2 e−4 t

B 94 e−8 t − 5

4 e−4 t

C 54 e

4 t − 94 e

8 t

D 52 e−8 t − 3

2 e−4 t

E − 32 e

4 t + 52 e

8 t

F 54 e

4 t + 94 e

8 t

5. Indique cual de las siguientes opciones representa la trans-

formada inversa de la funcion:

F (s) =−8 + 8 s

−4 s + s3

A f(t) = 2 − e−2 t + 3 e2 t

B f(t) = 2 + 3 e−2 t − e2 t

C f(t) = 2 − 3 e−2 t + e2 t

D f(t) = 2 + e−2 t − 3 e2 t



F (s) =7 s

(9 + s2) (16 + s2)

A f(t) = sen(3 t) − sen(4 t)

B f(t) = sen(3 t) + sen(4 t)

C f(t) = cos(3 t) − cos(4 t)

D f(t) = cos(3 t) + cos(4 t)

7. Utilizando Transformada de Laplace, resuelva el problema

con condiciones iniciales y(0) = 0 y y′(0) = 0 y ecuacion

diferencial:

49 y + y′′ = sen(3 t)

A y(t) = 140 sen(3 t) − 3

40 sen(7 t)

B y(t) = − 140 sen(3 t) + 3

280 sen(7 t)

C y(t) = 3280 sen(3 t) − 1

40 s sen(7 t)

D y(t) = 140 sen(3 t) − 3

280 sen(7 t)

8. Indique cual de las opciones siguientes es la ecuacion sub-

sidiaria del problema con condiciones iniciales y(0) = 6 y

y′(0) = 0 con ecuacion:

16 y + y′′ = 6 sen(4 t)

Ecuaciones Diferenciales, Tarea No 7: Transformada de Laplace, Tipo: -1 2

A Y (s) = 24+96 s+6 s3

(16+s2)2

B Y (s) = 24+24 s+6 s2

(−16+s2)2

C Y (s) = 24+24 s+6 s3

(16+s2)2

D Y (s) = 24+96 s+6 s3

(−16+s2)2



Grupo: Matrıcula: Nombre: Tipo:0



f(t) = −2 e−4 t + e−t

A F (s) = −6+3 s(−4+s) (−1+s)

B F (s) = −2−s(−4+s) (−1+s)

C F (s) = −6−3 s(−1−s) (4+s)

D F (s) = −2+s(−1−s) (4+s)




f(t) = cos(5 t) sen(7 t)

A F (s) = 5 s(25+s2) (49+s2)

B F (s) = −24+s2 + 12

144+s2

C F (s) = 12 ( 2

4+s2 + 12144+s2 )

D F (s) = 24+s2 + 12

144+s2



F (s) =2 + 2 s

9 + s2

A 2 cos(3 t) + 23 sen(3 t)

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

C 2 cos(3 t) + 2 sen(3 t)

D 2 cos(3 t) − 23 sen(3 t)

E 23 cos(3 t) + 2 sen(3 t)

F 2 cos(3 t) + 2 sen(3 t)



F (s) =6 + s

(−7 + s) (−2 + s)

A 35 e

2 t − 85 e

7 t

B − 135 e−7 t + 8

5 e−2 t

C 35 e

2 t + 85 e

7 t

D 135 e−7 t − 8

5 e−2 t

E − 85 e

2 t + 135 e7 t

F 85 e−7 t − 3

5 e−2 t



F (s) =4

45 − 14 s + s2

A f(t) = −1 e5 t + e9 t

B f(t) = e−9 t + e−5 t

C f(t) = e−9 t − e−5 t

D f(t) = e5 t + e9 t



F (s) =12

(4 + s2) (16 + s2)

A f(t) = 12 cos(2 t) + 1

4 cos(4 t)

B f(t) = 12 sen(2 t) + 1

4 sen(4 t)

C f(t) = 12 sen(2 t) − 1

4 sen(4 t)

D f(t) = cos(2 t) + cos(4 t)



diferencial:

25 y + y′′ = sen(4 t)

A y(t) = − 19 sen(4 t) + 4

45 sen(5 t)

B y(t) = 445 sen(4 t) − 1

9 s sen(5 t)

C y(t) = 19 sen(4 t) − 4

9 sen(5 t)

D y(t) = 19 sen(4 t) − 4

45 sen(5 t)




10 y − 7 y′ + y′′ = 5 e2 t

Ecuaciones Diferenciales, Tarea No 7: Transformada de Laplace, Tipo: 0 2

A Y (s) = 5+5 s10−7 s+s2

B Y (s) = −5+5 s(−2+s) (10−7 s+s2)

C Y (s) = 5+5 s(−2+s) (10−7 s+s2)

D Y (s) = −5+5 s(−2+s) (10+7 s+s2)






f(t) = e−t − sen(5 t)

A F (s) = 5(−1−s) (25+s2)

B F (s) = −20+5 s−s2(−1−s) (25+s2)

C F (s) = 30−5 s+s2

(−1+s) (25+s2)

D F (s) = −5(−1+s) (25+s2)



trigonometricas para eliminar el cuadrado)

f(t) = 3 sen2(5 t)

A F (s) = 3(

1s − s

25+s2

)B F (s) = 3

2 ( 1s + s

25+s2 )

C F (s) = 16 ( 1

s − s100+s2 )

D F (s) = 3(

1s − s

100+s2

)E F (s) = 3

2 ( 1s − s

100+s2 )



F (s) =1 + 6 s + 3 s2

s7

A 140 t

5 + 1120 t

6 + 15040 t

7

B 18 t

4 + 120 t

5 + 1720 t

6

C 34 t

4 + 65 t

5 + 16 t

6

D 35 t

5 + t6 + 17 t

7

E 3 t4 + 6 t5 + t6

F 3 t5 + 6 t6 + t7



F (s) =4 + s

(−8 + s) (−4 + s)

A 94 e−8 t − 5

4 e−4 t

B −3 e−8 t + 2 e−4 t

C 3 e−8 t − 2 e−4 t

D 54 e

4 t + 94 e

8 t

E 54 e

4 t − 94 e

8 t

F −2 e4 t + 3 e8 t



F (s) =−8 − 8 s

−4 s + s3

A f(t) = 2 + 3 e−2 t − e2 t

B f(t) = 2 + e−2 t − 3 e2 t

C f(t) = 2 − 3 e−2 t + e2 t

D f(t) = 2 − e−2 t + 3 e2 t



F (s) =55 s

(9 + s2) (64 + s2)

A f(t) = sen(3 t) + sen(8 t)

B f(t) = sen(3 t) − sen(8 t)

C f(t) = cos(3 t) + cos(8 t)

D f(t) = cos(3 t) − cos(8 t)



diferencial:

16 y + y′′ = sen(3 t)

A y(t) = 17 sen(3 t) − 3

7 sen(4 t)

B y(t) = 328 sen(3 t) − 1

7 s sen(4 t)

C y(t) = 17 sen(3 t) − 3

28 sen(4 t)

D y(t) = − 17 sen(3 t) + 3

28 sen(4 t)





49 y + y′′ = 7 sen(7 t)

A Y (s) = 49+343 s+7 s3

(−49+s2)2

B Y (s) = 49+49 s+7 s3

(49+s2)2

C Y (s) = 49+343 s+7 s3

(49+s2)2

D Y (s) = 49+49 s+7 s2

(−49+s2)2






f(t) = et − sen(t)

A F (s) = −s+s2

(1+s) (1+s2)

B F (s) = −2+s−s2(1−s) (1+s2)

C F (s) = 1(1−s) (1+s2)

D F (s) = − 1(1+s) (1+s2)


de Laplace de la funcion: (Sugerencia: use las definicio-

nes de las funciones de las funciones hiperbolicas mediante

exponenciales)

f(t) = 5 cosh(2 t) senh(2 t)

A F (s) =5 (16−2 s2)

(8−4 s+s2) (8+4 s+s2)

B F (s) = 25

1(4−s) (4+s)

C F (s) = −10(4−s) (4+s)

D F (s) = 10 s(−2+s)2 (2+s)2

E F (s) = −10 s(−2+s) (2+s) (4+s2)



F (s) =2 + 5 s

36 + s2

A 5 cos(6 t) + 2 sen(6 t)

B 2 cos(6 t) + 5 sen(6 t)

C 13 cos(6 t) + 5 sen(6 t)

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

E 5 cos(6 t) + 13 sen(6 t)

F 5 cos(6 t) − 13 sen(6 t)



F (s) =7 + s

(−6 + s) (−4 + s)

A 132 e−6 t − 11

2 e−4 t

B 52 e

4 t − 72 e

6 t

C 72 e−6 t − 5

2 e−4 t

D − 112 e4 t + 13

2 e6 t

E 52 e

4 t + 72 e

6 t

F − 132 e−6 t + 11

2 e−4 t



F (s) =4

32 − 12 s + s2

A f(t) = e4 t + e8 t

B f(t) = −1 e4 t + e8 t

C f(t) = e−8 t − e−4 t

D f(t) = e−8 t + e−4 t



F (s) =−17 s

(64 + s2) (81 + s2)

A f(t) = cos(8 t) + cos(9 t)

B f(t) = − cos(8 t) + cos(9 t)

C f(t) = −sen(8 t) + sen(9 t)

D f(t) = sen(8 t) + sen(9 t)



diferencial:

64 y + y′′ = sen(4 t)

A y(t) = 148 sen(4 t) − 1

12 sen(8 t)

B y(t) = 148 sen(4 t) − 1

96 sen(8 t)

C y(t) = − 148 sen(4 t) + 1

96 sen(8 t)

D y(t) = 196 sen(4 t) − 1

48 s sen(8 t)





16 y + y′′ = 5 sen(4 t)

A Y (s) = 20+20 s+5 s2

(−16+s2)2

B Y (s) = 20+80 s+5 s3

(−16+s2)2

C Y (s) = 20+80 s+5 s3

(16+s2)2

D Y (s) = 20+20 s+5 s3

(16+s2)2






f(t) = 4 + e2 t + 3 t

A F (s) = −6−5 s+5 s2

(−2+s) s2

B F (s) = 6+11 s+5 s2

s2 (2+s)

C F (s) = 12(−2+s) s3

D F (s) = 12s3 (2+s)




f(t) = 7 sen2(8 t)

A F (s) = 7(

1s − s

64+s2

)B F (s) = 1

14 ( 1s − s

256+s2 )

C F (s) = 72 ( 1

s + s64+s2 )

D F (s) = 72 ( 1

s − s256+s2 )

E F (s) = 7(

1s − s

256+s2

)3. Cual de las siguientes opciones representa la transformada


F (s) =5 + 4 s

4 + s2

A 52 cos(2 t) + 4 sen(2 t)

B 4 cos(2 t) − 52 sen(2 t)

C cos(2 t) + sen(2 t)

D 5 cos(2 t) + 4 sen(2 t)

E 4 cos(2 t) + 5 sen(2 t)

F 4 cos(2 t) + 52 sen(2 t)



F (s) =8 + s

(−7 + s) (−1 + s)

A 43 e−7 t − 1

3 e−t

B 52 e−7 t − 3

2 e−t

C 13 e

t − 43 e

7 t

D − 52 e−7 t + 3

2 e−t

E 13 e

t + 43 e

7 t

F − 32 e

t + 52 e

7 t



F (s) =6 + 3 s

−30 + s + s2

A f(t) = − 1211 e−6 t + 21

11 e−5 t

B f(t) = 1211 e−6 t + 21

11 e5 t

C f(t) = 1211 e−6 t + 21

11 e−5 t

D f(t) = 2111 e

5 t + 1211 e

6 t



F (s) =32

(4 + s2) (36 + s2)

A f(t) = 12 cos(2 t) + 1

6 cos(6 t)

B f(t) = cos(2 t) + cos(6 t)

C f(t) = 12 sen(2 t) − 1

6 sen(6 t)

D f(t) = 12 sen(2 t) + 1

6 sen(6 t)



diferencial:

16 y + y′′ = sen(2 t)

A y(t) = 112 sen(2 t) − 1

6 sen(4 t)

B y(t) = 124 sen(2 t) − 1

12 s sen(4 t)

C y(t) = − 112 sen(2 t) + 1

24 sen(4 t)

D y(t) = 112 sen(2 t) − 1

24 sen(4 t)





4 y + y′′ = 8 sen(2 t)

A Y (s) = 16+32 s+8 s3

(−4+s2)2

B Y (s) = 16+16 s+8 s2

(−4+s2)2

C Y (s) = 16+32 s+8 s3

(4+s2)2

D Y (s) = 16+16 s+8 s3

(4+s2)2






f(t) = 3 cos(3 t) + 4 sen(3 t)

A F (s) = 12−3 s9+s2

B F (s) = 9+4 s9+s2

C F (s) = 9−4 s9+s2

D F (s) = 12+3 s9+s2





A F (s) = 39+s2 + 11

121+s2

B F (s) = −39+s2 + 11

121+s2

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

D F (s) = 12 ( 3

9+s2 + 11121+s2 )



F (s) =4 − 3 s

25 + s2

A cos(5 t) + sen(5 t)

B −3 cos(5 t) + 45 sen(5 t)

C 4 cos(5 t) − 3 sen(5 t)

D −3 cos(5 t) + 4 sen(5 t)

E −3 cos(5 t) − 45 sen(5 t)

F 45 cos(5 t) − 3 sen(5 t)



F (s) =5 + s

(−7 + s) (−3 + s)

A −3 e−7 t + 2 e−3 t

B 3 e−7 t − 2 e−3 t

C e3 t + 2 e7 t

D e3 t − 2 e7 t

E 2 e−7 t − e−3 t

F −2 e3 t + 3 e7 t



F (s) =7 + 4 s

−40 − 3 s + s2

A f(t) = e5 t + 3 e8 t

B f(t) = e−5 t + 3 e8 t

C f(t) = 3 e−8 t + e−5 t

D f(t) = 3 e−8 t − e−5 t



F (s) =−77 s

(4 + s2) (81 + s2)

A f(t) = cos(2 t) + cos(9 t)

B f(t) = − cos(2 t) + cos(9 t)

C f(t) = sen(2 t) + sen(9 t)

D f(t) = −sen(2 t) + sen(9 t)



diferencial:

9 y + y′′ = sen(4 t)

A y(t) = 421 sen(3 t) − 1

7 sen(4 t)

B y(t) = − 421 sen(3 t) + 1

7 sen(4 t)

C y(t) = 47 sen(3 t) − 1

7 sen(4 t)

D y(t) = 17 s sen(3 t) − 4

21 sen(4 t)




4 y + y′′ = 3 cos(2 t)


A Y (s) = 15+3 s2

(4+s2)2

B Y (s) =s (12+3 s2)(4+s2)2

C Y (s) =s (15−3 s2)(4+s2)2

D Y (s) =s (15+3 s2)(4+s2)2






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

A F (s) = 3(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) = −6−3 s−s2(1−s) (9+s2)




f(t) = cos(4 t) cos(6 t)

A F (s) = s2

(16+s2) (36+s2)

B F (s) = s(

14+s2 + 1

100+s2

)C F (s) = 1

2 s(

14+s2 + 1

100+s2

)D F (s) = s

(1

16+s2 + 136+s2



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

s7

A 6 t5 + 6 t6 + 4 t7

B 14 t

4 + 120 t

5 + 1180 t

6

C 120 t

5 + 1120 t

6 + 11260 t

7

D 32 t

4 + 65 t

5 + 23 t

6

E 65 t

5 + t6 + 47 t

7

F 6 t4 + 6 t5 + 4 t6



F (s) =3 + s

(−7 + s) (−3 + s)

A − 52 e−7 t + 3

2 e−3 t

B e3 t − 2 e7 t

C 2 e−7 t − e−3 t

D 52 e−7 t − 3

2 e−3 t

E − 32 e

3 t + 52 e

7 t

F e3 t + 2 e7 t



F (s) =7 + 4 s

−18 + 7 s + s2

A f(t) = − 2911 e−9 t + 15

11 e−2 t

B f(t) = 1511 e

2 t + 2911 e

9 t

C f(t) = 2911 e−9 t + 15

11 e−2 t

D f(t) = 2911 e−9 t + 15

11 e2 t



F (s) =−60 s

(4 + s2) (64 + s2)

A f(t) = cos(2 t) + cos(8 t)

B f(t) = −sen(2 t) + sen(8 t)

C f(t) = − cos(2 t) + cos(8 t)

D f(t) = sen(2 t) + sen(8 t)



diferencial:

16 y + y′′ = sen(8 t)

A y(t) = 124 sen(4 t) − 1

48 sen(8 t)

B y(t) = − 124 sen(4 t) + 1

48 sen(8 t)

C y(t) = 16 sen(4 t) − 1

48 sen(8 t)

D y(t) = 148 s sen(4 t) − 1

24 sen(8 t)




36 y + y′′ = 7 cos(6 t)


A Y (s) = 259+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) =s (252+7 s2)(36+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) = 1+3 s1+s2

D F (s) = −3−s1+s2




f(t) = 8 sen2(3 t)

A F (s) = 116 ( 1

s − s36+s2 )

B F (s) = 4(

1s + s

9+s2

)C F (s) = 8

(1s − s

36+s2

)D F (s) = 8

(1s − s

9+s2

)E F (s) = 4

(1s − s

36+s2



F (s) =1 + 4 s + 6 s2

s5

A 6 t3 + 4 t4 + t5

B 2 t3 + t4 + 15 t

5

C 6 t2 + 4 t3 + t4

D 3 t2 + 43 t

3 + 14 t

4

E t3 + 16 t

4 + 1120 t

5

F 3 t2 + 23 t

3 + 124 t

4



F (s) =5 + s

(−6 + s) (−1 + s)

A 75 e−6 t − 2

5 e−t

B 115 e−6 t − 6

5 e−t

C 25 e

t − 75 e

6 t

D − 115 e−6 t + 6

5 e−t

E 25 e

t + 75 e

6 t

F − 65 e

t + 115 e6 t



F (s) =6

27 − 12 s + s2

A f(t) = e3 t + e9 t

B f(t) = e−9 t − e−3 t

C f(t) = −1 e3 t + e9 t

D f(t) = e−9 t + e−3 t



F (s) =56

(25 + s2) (81 + s2)

A f(t) = 15 sen(5 t) + 1

9 sen(9 t)

B f(t) = cos(5 t) + cos(9 t)

C f(t) = 15 cos(5 t) + 1

9 cos(9 t)

D f(t) = 15 sen(5 t) − 1

9 sen(9 t)



diferencial:

25 y + y′′ = sen(2 t)

A y(t) = 121 sen(2 t) − 2

105 sen(5 t)

B y(t) = 121 sen(2 t) − 2

21 sen(5 t)

C y(t) = 2105 sen(2 t) − 1

21 s sen(5 t)

D y(t) = − 121 sen(2 t) + 2

105 sen(5 t)





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 s(−6+s) (18−9 s+s2)

C Y (s) = 15+3 s(−6+s) (18−9 s+s2)

D Y (s) = 15+3 s18−9 s+s2






f(t) = e−4 t − 3 e2 t

A F (s) = 14−2 s(−4+s) (2+s)

B F (s) = 14+2 s(−4−s) (−2+s)

C F (s) = −10+4 s(−4+s) (2+s)

D F (s) = −10−4 s(−4−s) (−2+s)


de Laplace de la funcion: (Sugerencia: use las definicio-

nes de las funciones de las funciones hiperbolicas mediante

exponenciales)

f(t) = − cosh(2 t) senh(2 t)

A F (s) = − 16−2 s2

(8−4 s+s2) (8+4 s+s2)

B F (s) = 2(4−s) (4+s)

C F (s) = 2 s(−2+s) (2+s) (4+s2)

D F (s) = −2 s(−2+s)2 (2+s)2

E F (s) = −2(4−s) (4+s)



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

s7

A t4 + 65 t

5 + 12 t

6

B 4 t5 + 6 t6 + 3 t7

C 16 t

4 + 120 t

5 + 1240 t

6

D 130 t

5 + 1120 t

6 + 11680 t

7

E 45 t

5 + t6 + 37 t

7

F 4 t4 + 6 t5 + 3 t6



F (s) =4 + s

(−8 + s) (−3 + s)

A − 125 e−8 t + 7

5 e−3 t

B 95 e−8 t − 4

5 e−3 t

C 125 e−8 t − 7

5 e−3 t

D 45 e

3 t + 95 e

8 t

E 45 e

3 t − 95 e

8 t

F − 75 e

3 t + 125 e8 t



F (s) =4

32 − 12 s + s2

A f(t) = e4 t + e8 t

B f(t) = e−8 t − e−4 t

C f(t) = −1 e4 t + e8 t

D f(t) = e−8 t + e−4 t



F (s) =−17

(64 + s2) (81 + s2)

A f(t) = 18 cos(8 t) + 1

9 cos(9 t)

B f(t) = cos(8 t) + cos(9 t)

C f(t) = 18 sen(8 t) + 1

9 sen(9 t)

D f(t) = − 18 sen(8 t) + 1

9 sen(9 t)



diferencial:

4 y + y′′ = sen(5 t)

A y(t) = 121 s sen(2 t) − 5

42 sen(5 t)

B y(t) = − 542 sen(2 t) + 1

21 sen(5 t)

C y(t) = 521 sen(2 t) − 1

21 sen(5 t)

D y(t) = 542 sen(2 t) − 1

21 sen(5 t)





14 y − 9 y′ + y′′ = 7 e2 t

A Y (s) = 7+7 s14−9 s+s2

B Y (s) = 7+7 s(−2+s) (14−9 s+s2)

C Y (s) = −7+7 s(−2+s) (14−9 s+s2)

D Y (s) = −7+7 s(−2+s) (14+9 s+s2)






f(t) = e2 t − sen(4 t)

A F (s) = 4(2−s) (16+s2)

B F (s) = −4(2+s) (16+s2)

C F (s) = 8−4 s+s2

(2+s) (16+s2)

D F (s) = −24+4 s−s2(2−s) (16+s2)





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

B F (s) = 525+s2 + 11

121+s2

C F (s) = −525+s2 + 11

121+s2

D F (s) = 12 ( 5

25+s2 + 11121+s2 )



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

s6

A 6 t4 + 5 t5 + 5 t6

B 6 t3 + 5 t4 + 5 t5

C 14 t

4 + 124 t

5 + 1144 t

6

D 32 t

4 + t5 + 56 t

6

E 2 t3 + 54 t

4 + t5

F t3 + 524 t

4 + 124 t

5



F (s) =5 + s

(−7 + s) (−4 + s)

A 4 e−7 t − 3 e−4 t

B 53 e

4 t + 83 e

7 t

C −4 e−7 t + 3 e−4 t

D 53 e

4 t − 83 e

7 t

E −3 e4 t + 4 e7 t

F 83 e−7 t − 5

3 e−4 t



F (s) =2 + 4 s

−10 + 3 s + s2

A f(t) = 187 e−5 t + 10

7 e−2 t

B f(t) = 107 e2 t + 18

7 e5 t

C f(t) = − 187 e−5 t + 10

7 e−2 t

D f(t) = 187 e−5 t + 10

7 e2 t



F (s) =−20

(16 + s2) (36 + s2)

A f(t) = cos(4 t) + cos(6 t)

B f(t) = 14 cos(4 t) + 1

6 cos(6 t)

C f(t) = 14 sen(4 t) + 1

6 sen(6 t)

D f(t) = − 14 sen(4 t) + 1

6 sen(6 t)



diferencial:

9 y + y′′ = sen(4 t)

A y(t) = 47 sen(3 t) − 1

7 sen(4 t)

B y(t) = − 421 sen(3 t) + 1

7 sen(4 t)

C y(t) = 421 sen(3 t) − 1

7 sen(4 t)

D y(t) = 17 s sen(3 t) − 4

21 sen(4 t)




49 y + y′′ = 5 cos(7 t)


A Y (s) =s (245+5 s2)(49+s2)2

B Y (s) =s (250−5 s2)(49+s2)2

C Y (s) =s (250+5 s2)(49+s2)2

D Y (s) = 250+5 s2

(49+s2)2






f(t) = e2 t − sen(2 t)

A F (s) = −2 s+s2

(2+s) (4+s2)

B F (s) = 2(2−s) (4+s2)

C F (s) = −8+2 s−s2(2−s) (4+s2)

D F (s) = −2(2+s) (4+s2)




f(t) = 5 sen2(8 t)

A F (s) = 52 ( 1

s + s64+s2 )

B F (s) = 5(

1s − s

256+s2

)C F (s) = 5

(1s − s

64+s2

)D F (s) = 1

10 ( 1s − s

256+s2 )

E F (s) = 52 ( 1

s − s256+s2 )



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

s5

A 5 t3 + 4 t4 + 5 t5

B 5 t2 + 4 t3 + 5 t4

C 52 t

2 + 23 t

3 + 524 t

4

D 53 t

3 + t4 + t5

E 56 t

3 + 16 t

4 + 124 t

5

F 52 t

2 + 43 t

3 + 54 t

4



F (s) =5 + s

(−9 + s) (−5 + s)

A − 72 e−9 t + 5

2 e−5 t

B − 52 e

5 t + 72 e

9 t

C 72 e−9 t − 5

2 e−5 t

D 32 e

5 t − 52 e

9 t

E 52 e−9 t − 3

2 e−5 t

F 32 e

5 t + 52 e

9 t



F (s) =−2

15 − 8 s + s2

A f(t) = e−5 t + e−3 t

B f(t) = e3 t + e5 t

C f(t) = −1 e−5 t + e−3 t

D f(t) = e3 t − e5 t



F (s) =7 s

(9 + s2) (16 + s2)

A f(t) = cos(3 t) + cos(4 t)

B f(t) = sen(3 t) − sen(4 t)

C f(t) = cos(3 t) − cos(4 t)

D f(t) = sen(3 t) + sen(4 t)



diferencial:

36 y + y′′ = sen(8 t)

A y(t) = 27 sen(6 t) − 1

28 sen(8 t)

B y(t) = 121 sen(6 t) − 1

28 sen(8 t)

C y(t) = 128 s sen(6 t) − 1

21 sen(8 t)

D y(t) = − 121 sen(6 t) + 1

28 sen(8 t)





4 y + y′′ = 7 cos(2 t)

A Y (s) =s (35−7 s2)(4+s2)2

B Y (s) = 35+7 s2

(4+s2)2

C Y (s) =s (35+7 s2)(4+s2)2

D Y (s) =s (28+7 s2)(4+s2)2

ecuaciones diferenciales - cb.mty.itesm.mxcb.mty.itesm.mx/ma2001/alumno/tareas/ma2001-hw-7a.pdf ·...

Documents