Post on 04-Dec-2015
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Ejercicio D
3y " + 14 y ' + 58y = 0;
Auxiliar :3m2+14m+58=0 ;Baskara :
[−14±√196−696 ]6
−14±10√56
;o :
(−73 )±( 53 )√5
y=e(−73 ) x
∗{C1 sen [( 53 )√5 ]x+C2cos [( 53 )√5] x};
Reemplazocon y (1)=1 ;
1=e−73 ∗{C 1 sen [(53 )√5]+C2cos [(53 )√5] ]; o :
{C1 sen [( 53 )√5 ]+C2cos [( 53 )√5 ]]=e73
Reemplazocon y ' (1 )=1 ; primeroderivo dydx:
y=e(−73 ) x
∗{C1 sen [( 53 )√5 ]x+C2cos [( 53 )√5] x}
y '=(−73 )∗e(−73 ) x∗{C1 sen [( 53 )√5 ]x+C2cos [( 53 )√5] x}+¿
e(−73 )x
∗{C1∗[( 53 )√5]∗cos [( 53 )√5] x−C2∗[( 53 )√5]∗sen [( 53 )√5] x}
1=(−73 )∗e−73 ∗{C1 sen [( 53 )√5]+C2cos [( 53 )√5]}+¿
e−73 ∗{C1∗[(53 )√5]∗cos [( 53 )√5]−C 2∗[( 53 )√5 ]∗sen [(53 )√5]}
1=e(−7/3)∗¿
1=e−73 ∗(((−73 ){C1 sen[( 53 )√5]+C 2cos[( 53 )√5]}+[(53 )√5]{C1∗cos [(53 )√5]−C2∗sen [( 53 )√5]}))
Como :{C1 sen [( 53 )√5]+C2cos [( 53 )√5]]=e73 ;reemplazo :
1=e−73 ∗(((−73 )e
73+[( 53 )√5]{C1∗cos [( 53 )√5]−C 2∗sen[(53 )√5]}))
También de :{C 1 sen [(53 )√5]+C2cos [(53 )√5] ]=e73 ;obtengo :
C2cos [(5 /3)√5 ]¿=e(7 /3 )−C1 sen [(5 /3)√5];reemplazo :
1=e−73 ∗(((−73 )e
73+[( 53 )√5]{C1∗cos [( 53 )√5]−e
73−C 1 sen [(53 )√5]})); intento despejarC 1:
e73=(−73 )e
73+[( 53 )√5]{C1∗cos [( 53 )√5]−e
73−C 1 sen [(53 )√5]};
e73+( 73 )∗e
73=[( 53 )√5 ]{C1∗cos [( 53 )√5 ]−e
73−C1 sen[( 53 )√5]}
2e73+( 73 )∗e
73=[( 53 )√5]{C1∗cos[( 53 )√5]−C1 sen [( 53 )√5]};
[e73∗(2+73 )]∗35
√5={C1∗cos[( 53 )√5]−C1 sen [( 53 )√5]};
e73∗135
√5=C1∗{cos[(53 )√5]−sen [( 53 )√5]};
C1={[ e73 ]∗(135 )}
(√5{cos [( 53 )√5]−sen [( 53 )√5 ]})Luego de realizar "todas las cuentas" y obtenido C1, reemplazamos en cualquiera de las dos fórmulas finales y obtenemos C2.
Ejercicio5 :
Encontrar eloperador diferencial queanule a :
a. x+3 xy e6 x
b. (x3−2 x ) (x2−1 )
c. xex
Solución:
El enunciadode a ,debe ser :
a. x+3 xe6x
El operador queanula a x esD2 , ya que :
D2 x=DD ( x )=D (1 )=0
El operador queanula a3 xe6x es (D−6)2 , yaque :
(D−6 )2 (3x e6x )= (D−6 ) (D−6 ) (3x e6x )=¿
(D−6 ) [D (3x e6x )−6 (3x e6x ) ]=(D−6 ) [3 x D (e6 x )+e6 xD (3 x )−18 xe6 x]=¿
(D−6 ) [ (3x ) (6e6 x)+e6 x (3 )−18x e6x ]= (D−6 ) (18 x e6 x+3e6 x−18x e6x )=¿
(D−6 ) (3e6 x )=D (3e6x )−6 (3e6 x )=3 (6 )e6x−18 e6 x=18e6 x−18e6x=0
Respuesta a :Por lo tanto , esclaro que D2(D−6)2anula ax+3 x e6 x
b. (x3−2 x ) (x2−1 )=x5−x3−2 x3+2x=x5−3x3−2x
D6 (x5−3 x3−2x )=D5D (x5−3 x3−2 x )=¿
D5 (5 x4−9 x2−2 )=D4D (5x 4−9 x2−2 )=¿
D4 (20 x3−18 x−0 )=D3D ¿
D2D (60x2−18 )=D2(120 x−0)=DD (120 x )=D (120 )=0
Respuesta b :Por lo tanto , esclaroque D6anulaa (x3−2 x ) (x2−1 )
c. xex
El operador queanula a xex es (D−1 )2 , yaque :
(D−1 )2 (xex )= (D−1 ) (D−1 ) (xex )=¿
(D−1 ) [D (x ex )−x ex ]=(D−1 ) [ x D (ex )+ex D ( x )−x ex ]=¿
(D−1 ) [ x ex+ex−x ex ]=(D−1 ) (ex )=D (ex )−ex=ex−e x=0
Respuesta c :Por lotanto , es claroque (D−1 )2anula a xex