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-3ss34. Si se utiliza la transformada de Laplace resolver la ecuación diferencialX" + 5X' + 6X = 3t – 3e-3t cos (2t)con condiciones iniciales X(0) = 0 y X'(0) = -1, entonces la solución de la ecuaciónalgebraica asociada es:%3D-13.3(s+3)(s+3)2+4 s2((s+3)² +4) (s+3)2+43(s-3)a) L[X] =-13b) L[X] =(s+3)²+4' s²((s+3)²+4)((s+3)²+4)((s-3)²+4)3s23(s+3)(s+3)2+4 ((s+3)2+4)23(s+3)(s+3)2+4s2((s+3)²+4) ((s+3)2+4)2-1c) L[X] =(s+3)2+4-13d) L[X] =%3D5. Para calcular la transformada inversa de Laplace de la función3s2puede(s2+1)2utilizarse el teorema de convolución, dando como resultado:

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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:

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