6. Example-3
Find y(0.1) for `y''=-4z-4y`, `x_0=0, y_0=0, z_0=1`, with step length 0.1 using Runge-Kutta 4 method (2nd order derivative)
Solution:
Given `y^('')=-4z-4y, y(0)=0, y'(0)=1, h=0.1, y(0.1)=?`
put `(dy)/(dx)=z` and differentiate w.r.t. x, we obtain `(d^2y)/(dx^2)=(dz)/(dx)`
We have system of equations
`(dy)/(dx)=z=f(x,y,z)`
`(dz)/(dx)=-4z-4y=g(x,y,z)`
Forth order R-K method for second order differential equation
`k_1=f(x_0,y_0,z_0)=f(0,0,1)=1`
`l_1=g(x_0,y_0,z_0)=g(0,0,1)=-4`
`k_2=f(x_0+h/2,y_0+(hk_1)/2,z_0+(hl_1)/2)=f(0.05,0.05,0.8)=0.8`
`l_2=g(x_0+h/2,y_0+(hk_1)/2,z_0+(hl_1)/2)=g(0.05,0.05,0.8)=-3.4`
`k_3=f(x_0+h/2,y_0+(hk_2)/2,z_0+(hl_2)/2)=f(0.05,0.04,0.83)=0.83`
`l_3=g(x_0+h/2,y_0+(hk_2)/2,z_0+(hl_2)/2)=g(0.05,0.04,0.83)=-3.48`
`k_4=f(x_0+h,y_0+hk_3,z_0+hl_3)=f(0.1,0.083,0.652)=0.652`
`l_4=g(x_0+h,y_0+hk_3,z_0+hl_3)=g(0.1,0.083,0.652)=-2.94`
Now,
`y_1=y_0+h/6(k_1+2k_2+2k_3+k_4)`
`y_1=0+0.1/6[1+2(0.8)+2(0.83)+(0.652)]`
`y_1=0.08187`
`:.y(0.1)=0.08187`
`:.y(0.1)=0.08187`
This material is intended as a summary. Use your textbook for detail explanation.
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