Soluciones edo

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E.D.O.

1) Dadas las siguientes EDOs con condiciones iníciales y(0)=1 , y'(0)=0
Considerando un paso de integración h= 0.2 calcular el valor de la variable dependiente para los valores de la variable independiente 0.2  0.4  0.6  0.8  1.0 utilizando los métodos de:
a) y' + y = x

Y'=X-Y
Xn+1=Xn+h
Yn+1=Yn+hf(Xn,Yn)

fXn,Yn=Y'=X-Y
h=0.2

X0=0
Y0=1
fX0,Y0=f0,1=-1

X1=X0+h=0+0.2=0.2Y1=Y0+hfX0,Y0=1+0.2-1=1-0.2=0.8
fX1,Y1=f(0.2 , 0.8)=0.2-0.8=-0.6

X2=X1+h=0.2+0.2=0.4
Y2=Y1+hfX1,Y1=0.8+0.2-0.6=0.8-0.12=0.68
fX2,Y2=f(0.2 , 0.68)=0.2-0.68=-0.48

X3=X2+h=0.4+0.2=0.6
Y3=Y2+hfX2,Y2=0.68+0.2-0.48=0.68-0.096=0.584
fX3,Y3=f(0.6 , 0.584)=0.6-0.584=0.016

X4=X3+h=0.6+0.2=0.8
Y4=Y3+hfX3,Y3=0.584+0.20.016=0.584-0.0032=0.5808
fX4,Y4=f(0.8 , 0.5808)=0.8-0.5808=0.2192X5=X4+h=0.8+0.2=1
Y5=Y4+hfX4,Y4=0.5808+0.20.2192=0.5808-0.04384=0.53696
fX5,Y5=f(1 , 0.53696)=1-0.53696=0.46304

b XY'+Y=e2X

Y'=e2X-YX
Xn+1=Xn+h
Yn+1=Yn+hf(Xn,Yn)

fXn,Yn=Y'=e2X-YX
h=0.2

X0=0
Y0=1
fX0,Y0=f0,1=∞

En este caso Y’ no está definida para (X0 , Y0) lo cual se comprueba al desarrollar la ecuación diferencial lineal XY'+Y=e2X cuya solución es Y=e2X-12X la cualpresenta una asíntota vertical en X= 0, por esta razón comenzaré en X1= 0.2 , Y1=1.22956 (estos valores iníciales son los reales)
fXn,Yn=Y'=e2X-YX

X1=X0+h=0+0.2=0.2
Y1=Y0+hfX0,Y0=1.22956≈1.23
fX1,Y1=f(0.2 , 1.23)=e0.4-1.230.2=1.30912≈1.31

X2=X1+h=0.2+0.2=0.4
Y2=Y1+hfX1,Y1=1.23+0.21.31=1.492
fX2,Y2=f(0.4 , 1.492)=e0.8-1.4920.4=1.834

X3=X2+h=0.4+0.2=0.6
Y3=Y2+hfX2,Y2=1.492+0.21.834=1.8588≈1.86
fX3,Y3=f(0.6 , 1.86)=e1.2-1.860.6=2.433

X4=X3+h=0.6+0.2=0.8
Y4 =Y3+hfX3,Y3=1.86+0.22.433=2.3466≈2.35
fX4,Y4=f(0.8 , 2.35)=e1.6-2.350.8=3.254

X5=X4+h=0.8+0.2=1
Y5 =Y4+hfX4,Y4=2.35+0.23.254=3
fX5,Y5=f(1 , 3)=e2-31=4.389≈4.4

c) y'' + y' - 6y = 0
En este caso
y’’= – y’ + 6y
z = y’ y’(0) = 0
z’ = y’’

x0 = 0
y0 = 1
z0 = 0
k1 = f(x0 ,y0 ,z0) =f(0,1,0) = z = 0
K1= F(x0 ,y0 ,z0) = F(0,1,0) = 6y – z = 6
Y1 = y0 + hk1 = 1 + 0.2(0) = 1
Z1 = z0 +hK1 = 0 + 0.2(6) = 1.2
X1 = x0 + h = 0.2

X1 = 0.2
Y1 = 1
Z1 = 1.2
k2 = f(x1 ,y1 ,z1) = f(0.2 ,1 ,1.2) = z = 1.2
K2= F(x1 ,y1 ,z1) = F(0.2 ,1 ,1.2) = 6y – z = 4.8
Y2 = y1 + hk1 = 1 + 0.2(1.2) = 1.24
Z2 = z1 +hK1 = 1.2 + 0.2(4.8) = 2.16
X2 = x1 + h = 0.4

X2 = 0.4
Y2 = 1.24
Z2 = 2.16k3 = f(x2 ,y2 ,z2) = f(0.4 , 1.24 , 2.16) = z = 2.16
K3= F(x2 ,y2 ,z2) = F(0.4 , 1.24 , 2.16) = 6y – z = 5.28
Y3 = y2 + hk3 = 1.24 + 0.2(2.16) = 1.672
Z3 = z2 +hK3 = 2.16 + 0.2(5.28) = 3.216
X3 = x2 + h = 0.6

X3 = 0.6
Y3 = 1.672
Z3 = 3.216
k4 = f(x3 ,y3 ,z3) = f(0.6 , 1.672 , 3.216) = z = 3.216
K4= F(x3 ,y3 ,z3) = F(0.6 , 1.672 , 3.216) = 6y – z = 6.816
Y4 = y3 + hk4 = 1.672 +0.2(3.216) = 2.3152
Z4 = z3 +hK4 = 3.216 + 0.2(6.816) = 4.5792
X4 = x3 + h = 0.8

X4 = 0.8
Y4 = 2.3152
Z4 = 4.5792
k5 = f(x4 ,y4 ,z4) = f(0.8 , 2.3152 , 4.5792) = z = 4.5792
K5= F(x4 ,y4 ,z4) = F(0.8 , 2.3152 , 4.5792) = 6y – z = 9.312
Y5 = y4 + hk5 = 2.3152 + 0.2(4.5792) = 3.23104
Z5 = z4 +hK5 = 4.5792 + 0.2(9.312) = 6.4416
X5 = x4 + h = 1

X5 = 1
Y5 = 3.23104
Z5 = 6.4416
k6 = f(x5,y5 ,z5) = f(1 , 3.23104 , 6.4416) = z = 6.4416
K6= F(x5 ,y5 ,z5) = F(1 , 3.23104 , 6.4416) = 6y – z = 12.94464
Y6 = y5 + hk6 = 3.23104 + 0.2(6.4416) = 4.51936
Z6 = z5 +hK6 = 6.4416 + 0.2(12.94464) = 9.030528

d)

Y''+2Y'+Y=0 Y''=-2Y'-Y
Z=Y' → Y'=fX,Y,Z=Z
Z'=Y'' → Z'=FX,Y,Z=-2Z-Y

Sustituyendo

Z'=-2Z-Y
Xn+1=Xn+h
Yn+1=Yn+hkn+1Zn+1=Zn+hKn+1

kn+1=f(Xn,Yn,Zn)
Kn+1=F(Xn,Yn,Zn)

X0=0
Y0=1
Z0=0

k1=fX0,Y0,Z0=Z0=0
K1=FX0,Y0,Z0=-2Z0-Y0=-1
X1=X0+h=0.2
Y1=Y0+hk1=1+0.20=1
Z1=Z0+hK1=0+0.2-1=-0.2

k2=fX1,Y1,Z1=Z1=-0.2
K2=FX1,Y1,Z1=-2Z1-Y1=0.4-1=-0.6
X2=X1+h=0.2+0.2=0.4
Y2=Y1+hk2=1+0.2-0.2=0.96
Z2=Z1+hK2=-0.2+0.2-0.6=-0.32

k3=fX2,Y2,Z2=Z2=-0.32
K3=FX2,Y2,Z2=-2Z2-Y2=0.64-0.96=-0.32
X3=X2+h=0.4+0.2=0.6...
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