Apply Rungee Kutta method of fourth order to find an approximate Value of y when x=0.4 given that `dy/dx=(y-x)/(y+x),y=1` π=π ππππ π=π Taking h=0.2.
Solution
(1) `dy/dx=(y-x)/(y+x) x_0=y_0=1, h=0.2`
π(π,π)=` (y-x)/(y+x)`
ππ=π.π(ππ,ππ)=π.ππ(π,π)=π.π
`k_2=h.f(x_0+h/2,y_0+k_1/2)=0.2 f (0.1,1.1)=0.1661`
`k_3=h.f (x_0+h/2,y_0+k_2/2)=0.2.f(0.1.1.0833)=0.1661`
`k_4=h.f (x_0+h,y_0+k_3)=0.2f(0.2,.1.1661)=0.1414`
`k= (k_1+2k_2+2k_3+k_4)/6=(0.2+2(0.1666)+2(0.1661)+0.1414)/6=0.1678`
∴` y(0.2)=y_0+k=1+0.1678=1.1678`
(II) `x_1=0.2,y_2=1.1678,h=0.2`
`k_5=h.f(x_1,y_1)=0.2f(0.2,1.1678)=0.1415`
`k_6=h.f(x_1+h/2,y_1 ++k_5/2)=0.2 f(0.3,1.2285)=0.1220`
`k_7=h.f(x_1+h/2,y_1 ++k_6/2)=0.2f(0.3,1.2285)=0.1214`
`k_8=h.f(x_1+h,y_1+k_7)=0.2f(0.4,1.2892)=0.10`
`k* =( k_5+2k_6+2k_7+k_8)/6 = (0.1415+2(0.1215)+2(0.1215)+0.1052)/6=0.1222`
π(π.π)=ππ+π∗=π.ππππ+π.ππππ=π.πππ
