3. 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=hf(x_0,y_0,z_0)=(0.1)*f(0,0,1)=(0.1)*(1)=0.1`
`l_1=hg(x_0,y_0,z_0)=(0.1)*g(0,0,1)=(0.1)*(-4)=-0.4`
`k_2=hf(x_0+h/2,y_0+k_1/2,z_0+l_1/2)=(0.1)*f(0.05,0.05,0.8)=(0.1)*(0.8)=0.08`
`l_2=hg(x_0+h/2,y_0+k_1/2,z_0+l_1/2)=(0.1)*g(0.05,0.05,0.8)=(0.1)*(-3.4)=-0.34`
`k_3=hf(x_0+h/2,y_0+k_2/2,z_0+l_2/2)=(0.1)*f(0.05,0.04,0.83)=(0.1)*(0.83)=0.083`
`l_3=hg(x_0+h/2,y_0+k_2/2,z_0+l_2/2)=(0.1)*g(0.05,0.04,0.83)=(0.1)*(-3.48)=-0.348`
`k_4=hf(x_0+h,y_0+k_3,z_0+l_3)=(0.1)*f(0.1,0.083,0.652)=(0.1)*(0.652)=0.0652`
`l_4=hg(x_0+h,y_0+k_3,z_0+l_3)=(0.1)*g(0.1,0.083,0.652)=(0.1)*(-2.94)=-0.294`
Now,
`y_1=y_0+1/6(k_1+2k_2+2k_3+k_4)`
`y_1=0+1/6[0.1+2(0.08)+2(0.083)+(0.0652)]`
`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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