Pagina Pajua

1.-Dada la siguiente ecuación diferencial.

dydx=yx²-1.2y Donde:y0=1 Desde x=0 hasta x=3

A) Use el método de Euler con h=0.25 para resolver la ecu. Diferencial:

Xo;Yo, 0;1

* Iteración 1:
y1=yo+hfxo-yo
fxo,yo=1×0-1.2×1
fxo,yo=-1.2
y1=1+0.25×-1.2
y1=0.7
x1=xi+h
x1=0+0.25
x1=0.25
x1;y1=0.25;0.7

* Iteración 2:

y2=y1+hfx1,y1
fx1,y1=0.7×0.25²-1.2×0.7fx1,y1=-0.79625
y2=0.7+0.25×-0.79625
y2=0.5009

x2=x1+h
x2=0.25+0.25
x2=0.50
x2;y2=0.50;0.5009

* Iteración 3:

y3=y2+hf(x2,y2)
y3=0.5009+0.25(0.5009×0.52-1.2×0.5009)
y3=0.3819
x3=x2+h
x3=0.50+0.25
x3=0.75
x3;y3=0.75;0.3819

* Iteración 4:

y4=0.3819+0.250.3819×0.752-1.2×0.3819
y4=0.3210
x4=0.75+0.25=1
x4;y4=1;0.3210

* Iteración 5:y5=0.3210+0.2503210×12-1.2×0.3210
y5=0.3050
x5=1.25
x5;y5=(1.25;0.3050)

* Iteración 6:

y6=0.3050+0.250.3050×1.252-1.2×0.3050
y6=0.3326
x6=1.5
x6;y6=(1.5;0.3326)

* Iteración 7:

y7=0.3326+0.250.3326×1.52-1.2×0.3326
y7=0.4199
x7=1.75
x7;y7=(1.75;0.4199)

* Iteración 8:

y8=0.4199+0.250.4199×1.752-1.2×0.4199
y8=0.6154
x8=2
x8;y8=2;0.6154

* Iteración 9:y9=0.61540+0.25(0.6154×22-1.2×0.6154)
y9=1.0462
x9=2.25
x9;y9=2.25;1.0462

* Iteración 10:

y10=1.0462+0.25(1.0462×2.252-1.2×1.0462)
y10=2.0564
x10=2.5
x10;y10=2.5;2.0564

* Iteración 11:

y11=2.0564+0.252.0564×2.52-1.2×2.0564
y11=4.6526
x11=2.75
x11;y11=2.75;4.6526

* Iteración 12:

y12=4.6526+0.254.65×2.752-1.2×4.6526
y12=12.0531
x12=3

c) Use el método de punto medi con h=0.25 pararesolver el problema

dydx=yx2-1.2y

Donde:y0=1 h=0.25

* Formulas:

yi+1=y1+k2×h
k1=fxi,yi
k2=f(x1+12×h;y1+12k1×h)

* Iteración 1:

k1=f(xo,yo)=f=0.1
k1=1×0²-1.2×1
k1=-1.2
k2=fxo+12×h,yo+12ki×h
k2=f(0+12×0.25;1+12×-1.2×0.25
k2=f0.125;0.85
k2=0.85×0.1252-1.2×0.85
k2=-1.0067
y1=1-1.006719×0.25
y1=0.748320

* Iteración 2:

x1=xo+h
x1=0.25
k1=fx1,y1=f=(0.25;0.748320)k1=0.748320×0.252-1.2×0.748320
k1=-0.8514352
k2=f(0.25+12×0.25;0.748320+12×-0.851432×0.25
k2=f0.375;0.641891
k2=0.641891×0.375²-1.2×0.641891
k2=-0.680003
y2=0.748320-0.680003×0.25
y2=0.578319

* Iteración 3:

x2=x1+h
x2=0.5
k1=fx2,y2=f0.5;0.578319
k1=-0.549403
k2=f(0.5+12×0.25;0.578319+12×-0.549403×0.25)
k2=f0.625;0.509643
k2=0.509643×0.6252-1.2×0.50964
k2=-0.4125y3=0.573319-0.4125×0.25
y3=0.4752

* Iteración 4:

x3=x2+h
x3=0.75
k1=fx3,y3=f=(0.75;0.4752)
k1=0.4752×0.752-1.2×0.4752
k1=-0.3029
k2=f(0.75+12×0.25;0.4752+12×-0.3029×0.25
k2=f0.875;0.4373
k2=0.4373×0.8752-1.2×0.4373
k2=-0.1900
y4=0.4752-0.1900×0.25
y4=0.4277
x4=0.75+0.25
x4=1

* Iteración 5:

k1=fx4,y4=f=1;1,4277
k1=0.4277×12-1.2×0.4277
k1=-0.0855k2=f1+12×0.25;0.4277+12×-0.0855×0.25
k2=f1.125;0.4170
k2=0.4170×1.1252-1.2×0.4170
k2=0.0274
y5=0.4277+0.0274×0.25
y5=0.4346
x5=x4+h
x5=1.25

* Iteración 6:

k1=fx5;y5=f1.25;0.4346
k1=0.4346×1.25²-1.2×0.4346
k1=0.1575
k2=f1.25+12×0.25;0.4346+12×0.1575×0.25
k2=f1.375;0.4543
k2=0.4543×1.3752-1.2×0.4543
k2=0.3138
y6=0.4346+0.3138×0.25
y6=0.5131
x6=x5+h
x6=1.5

* Iteración 7:

k1=fx6;y6=f1.5;0.5131k1=0.5131×1.52-1.2×0.5131
k1=0.53876
k2=f1.5+12×0.25;0.5131+12×0.5388×0.25
k2=f1.625;0.5805
k2=0.5805×1.6252-1.2×0.5805
k2=0.8363
y7=0.5131+0.8363×0.25
y7=0.7222
x7=x6+h
x7=1.75

d) Use el método clásico de RK de tercer orden con h=0,25.
dydx=yx2-1,2y
y(0)= 1 , h=0,25

Solución: Formulas
yi+1=y1+16k1+4k2+k3h
k1=f(xi,yi)k2=fxi+12h, yi+12k1h
k3=f(xi+h, yi-k1h+2k2h)
* Iteración 1:
k1=fx0, y0=f(0, 1)
k1=1×02-1,2×1=-1,2

k2=f0+12×0.25 ;1+12×-1,2×0.25
k2=f0.125 ;0.85
k2=0.85×0.1252-1.2×0.85=-1.0067

k3=f(x0+h, y0-k1h+2k2h)
k3=f(0+0.25 ; 1+1.2×0.25+2×(-1.0067)×0.25)
k3=f(0.25 ; 0.7966)
k3=0.7966×0.1252-1.2×0.7966= -0.9062

y1=y0+16k1+4k2+k3h=0.7445
x1=x0+h=0.25
* Iteración 2:
k1=fx1,...