Ecuaciones Diferenciales
Enviado por fer9426 • 30 de Mayo de 2013 • 383 Palabras (2 Páginas) • 488 Visitas
Fernando Espinosa Sanchez 2cv11 2013300118
Variacion de Parametros
1.-
xy^'+4y=x^3-x
x dy/dx+4y=x^3-x
dy/dx+4y/x=x^2-1
P(x)=4x
Factor integrante
e^∫▒〖4/x dx〗=e^(4 ln(x) )=e^ln(x^4 ) = x^4
x^4 [dy/dx+4 y/x=x^2-1]
x^4 y^'+4x^3 y=x^6-x^4
d/dx [x^4 y]=x^6-x^4
∫▒〖d(x^4 y)=∫▒〖x^6-x^4 〗 dx〗
x^4 y=x^7/7-x^5/5+c
y=(x^7/7-x^5/5)/(x^4/1) + c
y=x^7/(7x^4 )-x^5/(5x^4 ) +c
y=x^3/7-x/5+ c
2.-
y’’ + 3y’ + 2y = sen(ex)y’’ + 3y’ + 2y = 0
m2 + 3m + 2 = 0 (m + 2)(m + 1) = 0 m1 = -2; m2 = -1
yh = C1 e-2x + C2 e-xç y1 y2Y
W(y1; y2)W(y1; y2) = e-2xe-x = -e-3x + 2e-3x = e-3x -2e-2x -e-x
U’1 = -y2f(x) = -e-x sen (ex) = -e2x sen (ex) W(y1; y2) e-3xU’2 = y1f(x) = e-2xsen (ex) = exsen (ex)W(y1; y2) e-3x
u1 = ∫ u’1 dx y u2 = ∫ u’2 dxu1 =∫ u’1 dx= ∫-e2x sen (ex) dx z= ex
dz = ex dx dx = dz/z
= -∫z2sen(z) dz/z= -∫z sen(z)
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)
u2 =∫u’2 dx = ∫ex sen (ex) dx =∫z sen z dz/z = ∫senz dz= -cos z = -cos(ex)
solucion particular
yp = u1y1 + u2y2yp = u1y1 + u2y2 = [ex cos(ex) - sen (ex)] e-2x -e-x cos(ex) = -e-2xsen (ex)
solucion general
y = yh + yp = C1y1 + C2y2 + u1y1 + u2y2y = yh + yp = C1e-2x + C2e-x – e-2x sen (ex)
coeficientes indeterminados
1.-
y′′−3y′+2y=0; (D2−3D+2) y=0; r2−3r + 2=0; r1=2; r2=1
yh = C1⋅e2x + C2⋅ex
yp = A⋅sin2x + B⋅cos2x
derivando dos veces:
y′=2A⋅cos2x−2B⋅sin2x;y′′=−4A⋅sin2x−4B⋅cos2x
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