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Find Runge-Kutta method y'=-2x-y, y(0)=1, xn=0.5, h=0.1

Solution:
Your problem `->` Runge-Kutta method y'=-2x-y, y(0)=1, xn=0.5, h=0.1


Given `y'=-2x-y, y(0)=1, h=0.1, y(0.5)=?`

Second order R-K method
`k_1=hf(x_0,y_0)=(0.1)f(0,1)=(0.1)*(-1)=-0.1`

`k_2=hf(x_0+h/2,y_0+k_1/2)=(0.1)f(0.05,0.95)=(0.1)*(-1.05)=-0.105`

`y_1=y_0+k_2=1-0.105=0.895`

`:.y(0.1)=0.895`

Again taking `(x_1,y_1)` in place of `(x_0,y_0)` repeat the process

`k_1=hf(x_1,y_1)=(0.1)f(0.1,0.895)=(0.1)*(-1.095)=-0.1095`

`k_2=hf(x_1+h/2,y_1+k_1/2)=(0.1)f(0.15,0.8402)=(0.1)*(-1.1402)=-0.114`

`y_2=y_1+k_2=0.895-0.114=0.781`

`:.y(0.2)=0.781`

Again taking `(x_2,y_2)` in place of `(x_0,y_0)` repeat the process

`k_1=hf(x_2,y_2)=(0.1)f(0.2,0.781)=(0.1)*(-1.181)=-0.1181`

`k_2=hf(x_2+h/2,y_2+k_1/2)=(0.1)f(0.25,0.7219)=(0.1)*(-1.2219)=-0.1222`

`y_3=y_2+k_2=0.781-0.1222=0.6588`

`:.y(0.3)=0.6588`

Again taking `(x_3,y_3)` in place of `(x_0,y_0)` repeat the process

`k_1=hf(x_3,y_3)=(0.1)f(0.3,0.6588)=(0.1)*(-1.2588)=-0.1259`

`k_2=hf(x_3+h/2,y_3+k_1/2)=(0.1)f(0.35,0.5958)=(0.1)*(-1.2958)=-0.1296`

`y_4=y_3+k_2=0.6588-0.1296=0.5292`

`:.y(0.4)=0.5292`

Again taking `(x_4,y_4)` in place of `(x_0,y_0)` repeat the process

`k_1=hf(x_4,y_4)=(0.1)f(0.4,0.5292)=(0.1)*(-1.3292)=-0.1329`

`k_2=hf(x_4+h/2,y_4+k_1/2)=(0.1)f(0.45,0.4627)=(0.1)*(-1.3627)=-0.1363`

`y_5=y_4+k_2=0.5292-0.1363=0.3929`

`:.y(0.5)=0.3929`



Third order R-K method





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