variacion de parametros
TRANSCRIPT
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Ecuaciónes Diferenciales
VARIACIÓN DE PARÁMETROS
Salvador Solis Valdez
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Para explicar este tema comenzare con un ejemplo:
Solucion de y’’ + 3y’ + 2y = sen(ex)y’’ + 3y’ + 2y = 0
1. Hallamos y1 y y2 soluciones linealmente independientes de la homogenea asociada:
m2 + 3m + 2 = 0
(m + 2)(m + 1) = 0
m1 = -2; m2 = -1
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yh = C1 e-2x + C2 e-xç
y1 y2
Y ENCONTRAMOS Y1 Y Y2
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2.- Por cramer hallamos W(y1; y2)
W(y1; y2) = e-2x e-x = -e-3x + 2e-3x = e-3x
-2e-2x -e-x
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3.- Hallamos
U’1 = -y2f(x) = -e-x sen (ex) = -e2x sen (ex)
W(y1; y2) e-3x
U’2 = y1f(x) = e-2x sen (ex) = ex sen (ex)
W(y1; y2) e-3x
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4.Integramos u1 = ∫ u’1 dx y u2 = ∫ u’2 dx
• u1 =∫ u’1 dx
• = ∫-e2x sen (ex) dx
z= ex
haciendo dz = ex dx
dx = dz/z
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= -∫z2 sen(z) dz/z
= -∫z sen(z) dz
integrando por partes v = z dv = dz
dw = -sen zdz w = cos z
= z cos z -∫cos z dz
= z cos z - sen z
= ex cos(ex) - sen (ex)
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u2 =∫u’2 dx = ∫ex sen (ex) dx
=∫z sen z dz/z = ∫senz dz
= -cos z = -cos(ex)
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5. La solucion particular yp = u1y1 + u2y2
yp = u1y1 + u2y2
= [ex cos(ex) - sen (ex)] e-2x -e-x cos(ex)
= -e-2x sen (ex)
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6. La solucion general y = yh + yp = C1y1 + C2y2 + u1y1 + u2y2
y = yh + yp
= C1e-2x + C2e-x – e-2x sen (ex)
Espero que este ejemplo les haya ayudado