Post on 29-Jan-2016
description
DESARROLLAR:
x ´+6 x−2 y=0
6 x+ y ´−2 y=0
Dx+6 x−2 y=0
6 x+Dy−2 y=0
[ (D+6 ) x−2 y=0 ] por−6
[6 x+(D−2 ) y=0 ] por (D+6)
(D2+4D ) y=0
m (m+4 )=0
m=0m=−4
y (t )=C1+C2e−4 t
y ´ ( t )=−4C2 e−4 t
Reemplazando y (t )=C1+C2e−4 t
En la Ecuación 6 x+ y ´−2 y=0
x=2 y− y ´6
x (t )=2 [C1+C2 e−4 t ]−[−4C2 e
−4 t ]6
x (t )=C13
+C2e−4 t
x ´+ y=3
−x+ y ´=−2
x(o )= y(o )=1
Aplicando Laplace:
x ´+ y=3
L [x ´ ]+L [ y ]=3 L[1]
sX−x (o)+Y=3s
sX+Y=3+ss
−x+ y ´=−2
−L [x ]+L [ y ´ ]=−2L[1]
−X+sY− y (o )=−2s
−X+sY=−2+ss
Aplicando Cramer:
| s 1−1 s||XY |=| 3+ss−2+s
s|
X=| 3+ss 1
−2+ss
s|| s 1−1 s|
X=3+s−(−2+s
s)
s2−(−1)
X=
s2+2 s+2ss2+1
X= s2+2 s+2s (s2+1)
Aplicando Inversa de Laplace:
L−1 [X ]=L−1[ s2+2 s+2s (s2+1 ) ]x (t )=L−1[ As +Bs+c
s2+1 ]x (t )=L−1[ A ( s2+1 )+s (Bs+c )
s ( s2+1 ) ]x (t )=L−1[ A ( s2+1 )+s (Bs+c )
s ( s2+1 ) ]=L−1[ s2+2 s+2s (s2+1 ) ]x (t )=L−1[ ( A+B ) s2+Cs+A
s ( s2+1 ) ]=L−1[ s2+2 s+2s ( s2+1 ) ]A=2 B=-1 C=2
x (t )=L−1[ 2s+−s+2s2+1 ]
x (t )=L−1[ 2s ]+L−1[−s+2s2+1 ]x (t )=+2−L−1[ s
s2+1 ]+2 L−1[ 1
s2+1 ]x (t )=+2−cos (t)+2 sen (t)
Y=| s 3+s
s
−1−2+ss
|| s 1−1 s|
Y=−2+s−( 3+s
s)
s2−(−1)
Y=
s2−3 s−3ss2+1
Y= s2−3 s−3s(s¿¿2+1)¿
Aplicando Inversa de Laplace:
L−1 [Y ]=L−1¿
y (t )=L−1[ As +Bs+Cs2+1 ]
y (t )=L−1[ ( A+B ) s2+Cs+As ( s2+1 ) ]=L−1¿
A=-3 B=4 C=-3
y (t )=−L−1[ 3s ]+L−1[ 4 s−3s2+1 ]y (t )=−3+4 L−1[ s
s2+1 ]−3 L−1[ 1
s2+1 ]y (t )=−3+4 cos (t)−3 sen( t)
y ´ ´−2 y ´+ y=et .cos (t) y (o )=3 y ´ (o )=2
[ s2−2 s+1 ]Y−sY (o )−Y ´ (o )+2Y (o )= s−1s2−2 s+2
[ s2−2 s+1 ]Y−3 s−2+6= s−1s2−2 s+2
[ s2−2 s+1 ]Y−3 s+4= s−1s2−2 s+2
[ s2−2 s+1 ]Y= s−1s2−2 s+2
+3 s−4
Y= s−1
(s2−2 s+2)(s¿¿2−2 s+1)+3 s−4s2−2 s+1
¿
Y= s−1
(s2−2 s+2)(s¿¿2−2 s+1)+(3 s−4 )(s2−2 s+2)
(s¿¿2−2 s+1)(s2−2 s+2)¿¿
Y= 3 s3−10 s2+15 s−9(s2−2 s+2)(s¿¿2−2 s+1)¿
Aplicando la inversa de Laplace
L−1 [Y ]=L−1¿
y (t )=L−1[ 1−ss2−2 s+2 ]+L−1[ 4
s−1 ]+L−1[ 1
(s−1)2 ]y (t )=L−1[ 1−s
s2−2 s+2 ]+L−1[ 4s−1 ]+L−1[ 1
(s−1)2 ]