Ejercicios De Homogeneas Y No Homogeneas.docx
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Ejercicio 2 letra C. ο·
π¦ ´´ β 2π¦ Β΄ + 5π¦ = π π₯ cos 2π₯ π¦ = π¦π + π¦π π¦π" β 2π¦πβ² + 5π¦π = 0 π2 β 2π + 5 = 0
π=
βπ Β± βπ 2 β 4ππ 2π
π=
β(β2) Β± β(β2)2 β 4(1)(5) 2(1)
π=
2 Β± β4 β 20 2
π=
2 Β± ββ16 2
π1 = 1 + 2π π2 = 1 β 2π π¦π = πΆ1π βπ₯ cos(π½π₯) + πΆ2 π πΌπ₯ π ππ (π½π₯) π¦π = πΆ1π π₯ cos(2π₯) + πΆ2π π₯ π ππ (2π₯) Para especificar los coeficientes podemos utilizar las letras [A y B] π¦π = π΄π π₯ cos(2π₯) + π΅ π π₯ π ππ (2π₯) Podemos ver que los tΓ©rminos que se encontraron en yp son los mismos tΓ©rminos encontrados en yc. le agregamos un (x), a cada uno para diferenciarlos π¦π = π΄π₯ π π₯ cos(2π₯) + π΅π₯ π π₯ π ππ (2π₯)
Derivamos 2 veces (Yp), tomando en cuenta el grado de la ecuaciΓ³n ( y" ) π¦π = π΄π₯ π π₯ cos(2π₯) + π΅π₯ π π₯ π ππ (2π₯) π¦πβ² = π΄ π π₯ cos(2π₯) + π΄π₯ π π₯ cos(2π₯) β 2π΄π₯ π π₯ π ππ (2π₯) + π΅ π π₯ π ππ (2π₯) + π΅π₯ π π₯ π ππ (2π₯) + 2π΅π₯ π π₯ cos(2π₯) π¦π" = π΄ π π₯ cos (2π₯) β 2π΄ π π₯ (2π₯) + π΄ π π₯ cos (2π₯) + π΄π₯ π π₯ cos (2π₯) β 2π΄π₯ π π₯ π ππ (2π₯) β 2π΄ π π₯ π ππ (2π₯) β 2π΄π₯ π π₯ π ππ (2π₯) β 4π΄π₯ π π₯ cos (2π₯) + π΅ π π₯ π ππ (2π₯) + 2π΅ π π₯ cos (2π₯) + π΅ π π₯ π ππ (2π₯) + π΅π₯ π π₯ π ππ (2π₯) + 2π΅π₯ π π₯ cos (2π₯) + 2π΅ π π₯ cos (2π₯) + 2π΅π₯ π π₯ cos (2π₯) β 4π΅π₯ π π₯ π ππ (2π₯) Se realiza la bΓΊsqueda de los valores de (A y B), para que [Yp] sea una SoluciΓ³n: π¦" β 2π¦β² + 5π¦ π¦π" = π΄ π π₯ cos (2π₯) β 2π΄ π π₯ π ππ (2π₯) + π΄ π π₯ cos (2π₯) + π΄π₯ π π₯ cos (2π₯) β 2π΄π₯ π π₯ π ππ (2π₯) β 2π΄ π π₯ π ππ (2π₯) β 2π΄π₯ π π₯ π ππ (2π₯) β 4π΄π₯ π π₯ cos (2π₯) + π΅ π π₯ π ππ (2π₯) + 2π΅ π π₯ cos (2π₯) + π΅ π π₯ π ππ (2π₯) + π΅π₯ π π₯ π ππ (2π₯) + 2π΅π₯ π π₯ cos (2π₯) + 2π΅ π π₯ cos (2π₯) + 2π΅π₯ π π₯ cos (2π₯) β 4π΅π₯ π π₯ π ππ (2π₯) β2π¦πβ² = β2(π΄ π π₯ cos(2π₯) + π΄π₯ π π₯ cos(2π₯) β 2π΄π₯ π π₯ π ππ (2π₯) + π΅ π π₯ π ππ (2π₯) + π΅π₯ π π₯ π ππ (2π₯) + 2π΅π₯ π π₯ cos(2π₯)) 5π¦π = 5(π΄π₯ π π₯ cos(2π₯) + π΅π₯ π π₯ π ππ(2π₯) Agrupamos tΓ©rminos: π΄ π π₯ cos(2π₯) + π΄ π π₯ cos(2π₯) + 2π΅ π π₯ cos(2π₯) + 2π΅ π π₯ cos(2π₯) + 2π΅ π π₯ cos(2π₯) β 2π΄ π π₯ π ππ (2π₯) β 2π΄ π π₯ π ππ (2π₯) + π΅ π π₯ π ππ (2π₯) + π΅ π π₯ π ππ (2π₯) + π΄π₯ π π₯ cos(2π₯) β 4π΄π₯ π π₯ cos(2π₯) + 2π΅π₯ π π₯ cos(2π₯) + 2π΅π₯ π π₯ cos(2π₯) β 2π΄π₯ π π₯ π ππ (2π₯) β 2π΄π₯ π π₯ π ππ (2π₯) + π΅π₯ π π₯ π ππ (2π₯) β 4π΅π₯ π π₯ π ππ (2π₯) β 2π΄ π π₯ cos(2π₯) β 2π΄π₯ π π₯ cos(2π₯) β 4π΅π₯ π π₯ cos(2π₯) + 4π΄π₯ π π₯ π ππ (2π₯) β 2π΅π₯ π π₯ π ππ (2π₯) β 2π΅ π π₯ π ππ (2π₯)
5π΄π₯ π π₯ cos(2π₯) + 5π΅π₯ π π₯ π ππ (2π₯) Realizamos la SimplificaciΓ³n de TΓ©rminos 4π΅ π π₯ cos(2π₯) β4π΄ π π₯ π ππ (2π₯)
Formamos las siguientes ecuaciones y las igualamos: π¦" β 2π¦β² + 5π¦ = π π₯ cos (2π₯) 4π΅ π π₯ cos(2π₯) = π π₯ cos(2π₯) 4π΅ = 1 π΅ = 1/4 β4π΄ π π₯ π ππ (2π₯) β4π΄ = 0 π΄=0 Realizamos la sustituciΓ³n de los valores encontrados de A y B en la ecuaciΓ³n de yp: π¦π = π΄ π π₯ cos(2π₯) + π΅ π π₯ π ππ (2π₯) 1 π¦π = ( ) π π₯ π ππ (2π₯) 4
Buscamos la Y general: π¦π‘ = ππ + ππ 1 π¦π‘ = πΆ1 π π₯ cos(2π₯) + πΆ2 π π₯ π ππ (2π₯) + ( ) π π₯ π ππ (2π₯) 4
π¦ ´´ β 2π¦ Β΄ + 5π¦ = π π₯ cos 2π₯ π¦ = π¦π + π¦π π¦π" β 2π¦πβ² + 5π¦π = 0 π2 β 2π + 5 = 0
π=
βπ Β± βπ 2 β 4ππ 2π
π=
β(β2) Β± β(β2)2 β 4(1)(5) 2(1)
π=
2 Β± β4 β 20 2
π=
2 Β± ββ16 2
π1 = 1 + 2π π2 = 1 β 2π π¦π = πΆ1π βπ₯ cos(π½π₯) + πΆ2 π πΌπ₯ π ππ (π½π₯) π¦π = πΆ1π π₯ cos(2π₯) + πΆ2π π₯ π ππ (2π₯) Para especificar los coeficientes podemos utilizar las letras [A y B] π¦π = π΄π π₯ cos(2π₯) + π΅ π π₯ π ππ (2π₯) Podemos ver que los tΓ©rminos que se encontraron en yp son los mismos tΓ©rminos encontrados en yc. le agregamos un (x), a cada uno para diferenciarlos π¦π = π΄π₯ π π₯ cos(2π₯) + π΅π₯ π π₯ π ππ (2π₯)
Derivamos 2 veces (Yp), tomando en cuenta el grado de la ecuaciΓ³n ( y" ) π¦π = π΄π₯ π π₯ cos(2π₯) + π΅π₯ π π₯ π ππ (2π₯) π¦πβ² = π΄ π π₯ cos(2π₯) + π΄π₯ π π₯ cos(2π₯) β 2π΄π₯ π π₯ π ππ (2π₯) + π΅ π π₯ π ππ (2π₯) + π΅π₯ π π₯ π ππ (2π₯) + 2π΅π₯ π π₯ cos(2π₯) π¦π" = π΄ π π₯ cos (2π₯) β 2π΄ π π₯ (2π₯) + π΄ π π₯ cos (2π₯) + π΄π₯ π π₯ cos (2π₯) β 2π΄π₯ π π₯ π ππ (2π₯) β 2π΄ π π₯ π ππ (2π₯) β 2π΄π₯ π π₯ π ππ (2π₯) β 4π΄π₯ π π₯ cos (2π₯) + π΅ π π₯ π ππ (2π₯) + 2π΅ π π₯ cos (2π₯) + π΅ π π₯ π ππ (2π₯) + π΅π₯ π π₯ π ππ (2π₯) + 2π΅π₯ π π₯ cos (2π₯) + 2π΅ π π₯ cos (2π₯) + 2π΅π₯ π π₯ cos (2π₯) β 4π΅π₯ π π₯ π ππ (2π₯) Se realiza la bΓΊsqueda de los valores de (A y B), para que [Yp] sea una SoluciΓ³n: π¦" β 2π¦β² + 5π¦ π¦π" = π΄ π π₯ cos (2π₯) β 2π΄ π π₯ π ππ (2π₯) + π΄ π π₯ cos (2π₯) + π΄π₯ π π₯ cos (2π₯) β 2π΄π₯ π π₯ π ππ (2π₯) β 2π΄ π π₯ π ππ (2π₯) β 2π΄π₯ π π₯ π ππ (2π₯) β 4π΄π₯ π π₯ cos (2π₯) + π΅ π π₯ π ππ (2π₯) + 2π΅ π π₯ cos (2π₯) + π΅ π π₯ π ππ (2π₯) + π΅π₯ π π₯ π ππ (2π₯) + 2π΅π₯ π π₯ cos (2π₯) + 2π΅ π π₯ cos (2π₯) + 2π΅π₯ π π₯ cos (2π₯) β 4π΅π₯ π π₯ π ππ (2π₯) β2π¦πβ² = β2(π΄ π π₯ cos(2π₯) + π΄π₯ π π₯ cos(2π₯) β 2π΄π₯ π π₯ π ππ (2π₯) + π΅ π π₯ π ππ (2π₯) + π΅π₯ π π₯ π ππ (2π₯) + 2π΅π₯ π π₯ cos(2π₯)) 5π¦π = 5(π΄π₯ π π₯ cos(2π₯) + π΅π₯ π π₯ π ππ(2π₯) Agrupamos tΓ©rminos: π΄ π π₯ cos(2π₯) + π΄ π π₯ cos(2π₯) + 2π΅ π π₯ cos(2π₯) + 2π΅ π π₯ cos(2π₯) + 2π΅ π π₯ cos(2π₯) β 2π΄ π π₯ π ππ (2π₯) β 2π΄ π π₯ π ππ (2π₯) + π΅ π π₯ π ππ (2π₯) + π΅ π π₯ π ππ (2π₯) + π΄π₯ π π₯ cos(2π₯) β 4π΄π₯ π π₯ cos(2π₯) + 2π΅π₯ π π₯ cos(2π₯) + 2π΅π₯ π π₯ cos(2π₯) β 2π΄π₯ π π₯ π ππ (2π₯) β 2π΄π₯ π π₯ π ππ (2π₯) + π΅π₯ π π₯ π ππ (2π₯) β 4π΅π₯ π π₯ π ππ (2π₯) β 2π΄ π π₯ cos(2π₯) β 2π΄π₯ π π₯ cos(2π₯) β 4π΅π₯ π π₯ cos(2π₯) + 4π΄π₯ π π₯ π ππ (2π₯) β 2π΅π₯ π π₯ π ππ (2π₯) β 2π΅ π π₯ π ππ (2π₯)
5π΄π₯ π π₯ cos(2π₯) + 5π΅π₯ π π₯ π ππ (2π₯) Realizamos la SimplificaciΓ³n de TΓ©rminos 4π΅ π π₯ cos(2π₯) β4π΄ π π₯ π ππ (2π₯)
Formamos las siguientes ecuaciones y las igualamos: π¦" β 2π¦β² + 5π¦ = π π₯ cos (2π₯) 4π΅ π π₯ cos(2π₯) = π π₯ cos(2π₯) 4π΅ = 1 π΅ = 1/4 β4π΄ π π₯ π ππ (2π₯) β4π΄ = 0 π΄=0 Realizamos la sustituciΓ³n de los valores encontrados de A y B en la ecuaciΓ³n de yp: π¦π = π΄ π π₯ cos(2π₯) + π΅ π π₯ π ππ (2π₯) 1 π¦π = ( ) π π₯ π ππ (2π₯) 4
Buscamos la Y general: π¦π‘ = ππ + ππ 1 π¦π‘ = πΆ1 π π₯ cos(2π₯) + πΆ2 π π₯ π ππ (2π₯) + ( ) π π₯ π ππ (2π₯) 4
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