Ecuaciones Diferenciales

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

Sustituyendo en la ecuación inicial:

(−4A+6B+2A)⋅sin2x + (−4B−6A+2B)⋅cos2x = 14⋅sin2x−18⋅cos2x

A = 2; B = 3

solución general:

yG = yh+yp

=C1⋅e2x + C2⋅ex + 2⋅sin2x+3⋅cos2x

2.-

p(x) = c0 + c1 cos x + c2 cos(2x) + c3 cos(3x)

p(xj) = yj

; 0 ≤ j ≤ 3 siendo x0 = 0; x1 = π/6; x2 = /π4; x3 = π/3 con y0 =0; y1 = 2 -√(3/2) ;

y2 = 1 + √(2/2); y3= 3/2

(█(■(1&1&1@1&√(3/2)&1/2@1&√(2/2)&0) ■(1@0@-√(2/2))@■(1&1/2 ) ■(-1/2&-1)))(█(■(co@c1)@c2@c3))=(█(█(0@2-√(3/2))@1+√(2/2)@3/2))

c0 = 1; c1 =-1;

c2 = 2; c3 = 2

p(x) = 1 cos(x) + 2 cos(2x) 2 cos(3x)

Circuito:

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