3. Runge-Kutta 3 method (first order differential equation) example ( Enter your problem )
  1. Formula-1 & Example-1 : `y'=(x-y)/2`
  2. Example-2 : `y'=-2x-y`
  3. Example-3 : `y'=-y`
  4. Formula-2 & Example-1 : `y'=(x-y)/2`
  5. Example-2 : `y'=-2x-y`
  6. Example-3 : `y'=-y`
Other related methods
  1. Euler method (first order differential equation)
  2. Runge-Kutta 2 method (first order differential equation)
  3. Runge-Kutta 3 method (first order differential equation)
  4. Runge-Kutta 4 method (first order differential equation)
  5. Improved Euler method / Modified Euler method (first order differential equation)
  6. Midpoint Euler method (first order differential equation)
  7. Taylor Series method (first order differential equation)
  8. Euler method (second order differential equation)
  9. Runge-Kutta 2 method (second order differential equation)
  10. Runge-Kutta 3 method (second order differential equation)
  11. Runge-Kutta 4 method (second order differential equation)
  12. Improved Euler method / Modified Euler method (second order differential equation)
  13. Midpoint Euler method (second order differential equation)
  14. Taylor Series method (second order differential equation)

4. Formula-2 & Example-1 : `y'=(x-y)/2`
(Previous example)
6. Example-3 : `y'=-y`
(Next example)

5. Example-2 : `y'=-2x-y`





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

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

Third order Runge-Kutta (RK3) method formula
`k_1=f(x_n,y_n)`

`k_2=f(x_n+h/2,y_n+(hk_1)/2)`

`k_3=f(x_n+h,y_n+2hk_2-hk_1)`

`y_(n+1)=y_n+h/6(k_1+4k_2+k_3)`



for `n=0,x_0=0,y_0=-1`

`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)`

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

`=-0.9145`

`x_1=x_0+h=0+0.1=0.1`



for `n=1,x_1=0.1,y_1=-0.9145`

`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)`

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

`=-0.8562`

`x_2=x_1+h=0.1+0.1=0.2`



for `n=2,x_2=0.2,y_2=-0.8562`

`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)`

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

`=-0.8224`

`x_3=x_2+h=0.2+0.1=0.3`



for `n=3,x_3=0.3,y_3=-0.8224`

`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)`

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

`=-0.8109`

`x_4=x_3+h=0.3+0.1=0.4`



for `n=4,x_4=0.4,y_4=-0.8109`

`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)`

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

`=-0.8196`

`x_5=x_4+h=0.4+0.1=0.5`

`:.y(0.5)=-0.8196`

`n``x_n``y_n``k_1``k_2``k_3``x_(n+1)``y_(n+1)`
00-110.850.730.1-0.9145
10.1-0.91450.71450.57880.47020.2-0.8562
20.2-0.85620.45620.33340.23510.3-0.8224
30.3-0.82240.22240.11130.02240.4-0.8109
40.4-0.81090.0109-0.0896-0.17010.5-0.8196





This material is intended as a summary. Use your textbook for detail explanation.
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4. Formula-2 & Example-1 : `y'=(x-y)/2`
(Previous example)
6. Example-3 : `y'=-y`
(Next example)





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