4.2.1 Metodo de Los Operadores
1. Dn (eax
f (x)) = eax (D + a)nf (x)
2. Dn (eax ) = eax an
#|R{ze}al
2. Dn (eax ) = eax an
#|R{ze}al
Demostración
Veamos 1. Por inducción:
n = 1 D(eax f (x)) = eaxDf (x) + f (x)aeax = eax(D + a)f (x) Supongamos que se cumple para n = k:
Dk (eax f (x)) = eax (D + a)k f (x)
y veamos que se cumple para n = k + 1. En efecto, teniendo en cuenta las hipótesis de inducción para n = 1 y para n = k, se tiene
Dk+1(eax f (x)) = Dk D(eaxf (x)) = Dk (eax(D + a)f (x))
= eax(D + a)k (D + a)f (x) = eax (D + a)k+1f (x) Veamos 2. Por inducción:
n = 1 D(eax ) = aeax
Supongamos que se cumple para n = k:
Dk (eax ) = ak eax
y veamos que se cumple para n = k + 1. En efecto, teniendo en cuenta las hipótesis de inducción para n = 1 y para n = k, se tiene
Dk+1(eax ) = Dk D(eax) = Dk (aeax)
= a(Dk eax ) = a(ak eax) = ak+1eax
El siguiente Teorema, llamado teorema básico de los operadores, nos permite sacar una exponencial que esta dentro de un operador.
Teorema Básico
de Operadores
1. Si f ∈ C n (I ) y L(D) es un operador diferencial lineal y a ∈ ℜ, entonces:
L(D)(eaxf (x)) = eaxL(D + a)f (x)
2. L(D)eax = L(a)eax
Demostraci´on 1:
L(D)(eaxf (x)) = (an (x)Dn + an
1(x)Dn−1 + . . . + a (x)D + a (x)D0 )(eaxf (x)) =
− 1 0
= an (x)Dn (eaxf (x)) + an
1(x)Dn−1(eax f (x)) + . . . + a (x)D(eax f (x)) + a (x)eaxf (x)
− 1 0
= an (x)eax(D + a)n f (x) + an
+ a0 (x)eaxf (x)
1(x)eax(D + a)n−1f (x) + . . . + a1(x)eax (D + a)f (x)
= eax(an (x)(D + a)n + an
1 (x)(D + a)n−1 + . . . + a (x)(D + a) + a (x))f (x)
− 1 0
= eaxL(D + a)f (x)
1. Si f ∈ C n (I ) y L(D) es un operador diferencial lineal y a ∈ ℜ, entonces:
L(D)(eaxf (x)) = eaxL(D + a)f (x)
2. L(D)eax = L(a)eax
Demostraci´on 1:
L(D)(eaxf (x)) = (an (x)Dn + an
1(x)Dn−1 + . . . + a (x)D + a (x)D0 )(eaxf (x)) =
− 1 0
= an (x)Dn (eaxf (x)) + an
1(x)Dn−1(eax f (x)) + . . . + a (x)D(eax f (x)) + a (x)eaxf (x)
− 1 0
= an (x)eax(D + a)n f (x) + an
+ a0 (x)eaxf (x)
1(x)eax(D + a)n−1f (x) + . . . + a1(x)eax (D + a)f (x)
= eax(an (x)(D + a)n + an
1 (x)(D + a)n−1 + . . . + a (x)(D + a) + a (x))f (x)
− 1 0
= eaxL(D + a)f (x)
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