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 ]