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:

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Chapter1: Functions And Models
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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:
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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