problema 4

Prévia do material em texto

Solución: 
Dadoy′=4X-2y,y(0)=0,h=0.25,y(1)=? 
 
Método RK de cuarto orden 
k1=hF(X0,y0)=(0.25)F(0,0)=(0.25)⋅(0)=0 
 
k2=hF(X0+h2,y0+k12)=(0.25)F(0.125,0)=(0.25)⋅(0.5)=0.125 
 
k3=hF(X0+h2,y0+k22)=(0.25)F(0.125,0.0625)=(0.25)⋅(0.375)=0.0938 
 
k4=hF(X0+h,y0+k3)=(0.25)F(0.25,0.0938)=(0.25)⋅(0.8125)=0.2031 
 
y1=y0+16(k1+2k2+2k3+k4) 
 
y1=0+16[0+2(0.125)+2(0.0938)+(0.2031)] 
 
y1=0.1068 
 
∴y(0.25)=0.1068 
 
 
 
otra vez tomando(X1,y1)en lugar de(X0,y0)y repetir el proceso 
 
k1=hF(X1,y1)=(0.25)F(0.25,0.1068)=(0.25)⋅(0.7865)=0.1966 
 
k2=hF(X1+h2,y1+k12)=(0.25)F(0.375,0.2051)=(0.25)⋅(1.0898)=0.2725 
 
k3=hF(X1+h2,y1+k22)=(0.25)F(0.375,0.243)=(0.25)⋅(1.014)=0.2535 
 
k4=hF(X1+h,y1+k3)=(0.25)F(0.5,0.3603)=(0.25)⋅(1.2795)=0.3199 
 
y2=y1+16(k1+2k2+2k3+k4) 
 
y2=0.1068+16[0.1966+2(0.2725)+2(0.2535)+(0.3199)] 
 
y2=0.3682 
 
 ∴y(0.5)=0.3682 
 
 
 
otra vez tomando(X2,y2)en lugar de(X0,y0)y repetir el proceso 
 
k1=hF(X2,y2)=(0.25)F(0.5,0.3682)=(0.25)⋅(1.2637)=0.3159 
 
k2=hF(X2+h2,y2+k12)=(0.25)F(0.625,0.5261)=(0.25)⋅(1.4477)=0.3619 
 
k3=hF(X2+h2,y2+k22)=(0.25)F(0.625,0.5491)=(0.25)⋅(1.4017)=0.3504 
 
k4=hF(X2+h,y2+k3)=(0.25)F(0.75,0.7186)=(0.25)⋅(1.5628)=0.3907 
 
y3=y2+16(k1+2k2+2k3+k4) 
 
y3=0.3682+16[0.3159+2(0.3619)+2(0.3504)+(0.3907)] 
 
y3=0.7234 
 
∴y(0.75)=0.7234 
 
 
 
otra vez tomando(X3,y3)en lugar de(X0,y0)y repetir el proceso 
 
k1=hF(X3,y3)=(0.25)F(0.75,0.7234)=(0.25)⋅(1.5532)=0.3883 
 
k2=hF(X3+h2,y3+k12)=(0.25)F(0.875,0.9175)=(0.25)⋅(1.6649)=0.4162 
 
k3=hF(X3+h2,y3+k22)=(0.25)F(0.875,0.9315)=(0.25)⋅(1.637)=0.4092 
 
k4=hF(X3+h,y3+k3)=(0.25)F(1,1.1326)=(0.25)⋅(1.7347)=0.4337 
 
y4=y3+16(k1+2k2+2k3+k4) 
 
y4=0.7234+16[0.3883+2(0.4162)+2(0.4092)+(0.4337)] 
 
y4=1.1355 
 
∴y(1)=1.1355 
 
 
 
∴y(1)=1.1355