Si se utiliza la transformada de Laplace resolver la ecuación diferencial X"+ 5X' + 6X = 3t – 3e-3t cos (2t) %3D con condiciones iniciales X(0) = 0 y X' (0) = -1, entonces la solución de la ecuación algebraica asociada es:
Si se utiliza la transformada de Laplace resolver la ecuación diferencial X"+ 5X' + 6X = 3t – 3e-3t cos (2t) %3D con condiciones iniciales X(0) = 0 y X' (0) = -1, entonces la solución de la ecuación algebraica asociada es:
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Using the Laplace transform solve the differential equation ? ′ ′ + 5? ′ + 6? = 3? - 3? −3? cos (2?) with initial conditions ? (0) = 0 and ? ′ (0) = −1, so the solution of the equation associated algebraic is:
![d)
+e
+e-3s
s3
4. Si se utiliza la transformada de Laplace resolver la ecuación diferencial
X" + 5X' + 6X = 3t – 3e-3t cos (2t)
con condiciones iniciales X(0) = 0 y X'(0) = -1, entonces la solución de la ecuación
algebraica asociada es:
%3D
-1
3.
3(s+3)
(s+3)2+4 s2((s+3)² +4) (s+3)2+4
3(s-3)
a) L[X] =
-1
3
b) L[X] =
(s+3)²+4' s²((s+3)²+4)
((s+3)²+4)((s-3)²+4)
3s2
3(s+3)
(s+3)2+4 ((s+3)2+4)2
3(s+3)
(s+3)2+4s2((s+3)²+4) ((s+3)2+4)2
-1
c) L[X] =
(s+3)2+4
-1
3
d) L[X] =
%3D
5. Para calcular la transformada inversa de Laplace de la función
3s2
puede
(s2+1)2
utilizarse el teorema de convolución, dando como resultado:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F23977c4d-31ec-49c4-91da-52b0435cc7cf%2F5d447ce0-eb0b-461d-ae62-eb639822feb4%2Fvnp88bc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:d)
+e
+e-3s
s3
4. Si se utiliza la transformada de Laplace resolver la ecuación diferencial
X" + 5X' + 6X = 3t – 3e-3t cos (2t)
con condiciones iniciales X(0) = 0 y X'(0) = -1, entonces la solución de la ecuación
algebraica asociada es:
%3D
-1
3.
3(s+3)
(s+3)2+4 s2((s+3)² +4) (s+3)2+4
3(s-3)
a) L[X] =
-1
3
b) L[X] =
(s+3)²+4' s²((s+3)²+4)
((s+3)²+4)((s-3)²+4)
3s2
3(s+3)
(s+3)2+4 ((s+3)2+4)2
3(s+3)
(s+3)2+4s2((s+3)²+4) ((s+3)2+4)2
-1
c) L[X] =
(s+3)2+4
-1
3
d) L[X] =
%3D
5. Para calcular la transformada inversa de Laplace de la función
3s2
puede
(s2+1)2
utilizarse el teorema de convolución, dando como resultado:
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