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