Find y(0.5) for `y'=-2x-y`, `x_0=0, y_0=-1`, with step length 0.1 using Runge-Kutta 3 method (1st order derivative)

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

Third order R-K method
`k_1=f(x_0,y_0)=f(0,-1)=1`

`k_2=f(x_0+h/2,y_0+(hk_1)/2)=f(0.05,-0.95)=0.85`

`k_3=f(x_0+h,y_0+2hk_2-hk_1)=f(0.1,-0.93)=0.73`

`y_1=y_0+h/6(k_1+4k_2+k_3)`

`y_1=-1+(0.1)/6[1+4(0.85)+(0.73)]`

`y_1=-0.9145`

`:.y(0.1)=-0.9145`

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

`k_1=f(x_1,y_1)=f(0.1,-0.9145)=0.7145`

`k_2=f(x_1+h/2,y_1+(hk_1)/2)=f(0.15,-0.8788)=0.5788`

`k_3=f(x_1+h,y_1+2hk_2-hk_1)=f(0.2,-0.8702)=0.4702`

`y_2=y_1+h/6(k_1+4k_2+k_3)`

`y_2=-0.9145+(0.1)/6[0.7145+4(0.5788)+(0.4702)]`

`y_2=-0.8562`

`:.y(0.2)=-0.8562`

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Again taking `(x_2,y_2)` in place of `(x_1,y_1)` and repeat the process

`k_1=f(x_2,y_2)=f(0.2,-0.8562)=0.4562`

`k_2=f(x_2+h/2,y_2+(hk_1)/2)=f(0.25,-0.8334)=0.3334`

`k_3=f(x_2+h,y_2+2hk_2-hk_1)=f(0.3,-0.8351)=0.2351`

`y_3=y_2+h/6(k_1+4k_2+k_3)`

`y_3=-0.8562+(0.1)/6[0.4562+4(0.3334)+(0.2351)]`

`y_3=-0.8224`

`:.y(0.3)=-0.8224`

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Again taking `(x_3,y_3)` in place of `(x_2,y_2)` and repeat the process

`k_1=f(x_3,y_3)=f(0.3,-0.8224)=0.2224`

`k_2=f(x_3+h/2,y_3+(hk_1)/2)=f(0.35,-0.8113)=0.1113`

`k_3=f(x_3+h,y_3+2hk_2-hk_1)=f(0.4,-0.8224)=0.0224`

`y_4=y_3+h/6(k_1+4k_2+k_3)`

`y_4=-0.8224+(0.1)/6[0.2224+4(0.1113)+(0.0224)]`

`y_4=-0.8109`

`:.y(0.4)=-0.8109`

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Again taking `(x_4,y_4)` in place of `(x_3,y_3)` and repeat the process

`k_1=f(x_4,y_4)=f(0.4,-0.8109)=0.0109`

`k_2=f(x_4+h/2,y_4+(hk_1)/2)=f(0.45,-0.8104)=-0.0896`

`k_3=f(x_4+h,y_4+2hk_2-hk_1)=f(0.5,-0.8299)=-0.1701`

`y_5=y_4+h/6(k_1+4k_2+k_3)`

`y_5=-0.8109+(0.1)/6[0.0109+4(-0.0896)+(-0.1701)]`

`y_5=-0.8196`

`:.y(0.5)=-0.8196`

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`:.y(0.5)=-0.8196`

`n`	`x_n`	`y_n`	`k_1`	`k_2`	`k_3`	`x_(n+1)`	`y_(n+1)`
0	0	-1	1	0.85	0.73	0.1	-0.9145
1	0.1	-0.9145	0.7145	0.5788	0.4702	0.2	-0.8562
2	0.2	-0.8562	0.4562	0.3334	0.2351	0.3	-0.8224
3	0.3	-0.8224	0.2224	0.1113	0.0224	0.4	-0.8109
4	0.4	-0.8109	0.0109	-0.0896	-0.1701	0.5	-0.8196

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