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 2 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)`
Method-1 : Using formula `k_2=hf(x_0+h,y_0+k_1,z_0+l_1)`
Second 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,y_0+k_1,z_0+l_1)=(0.1)*f(0.1,0.1,0.6)=(0.1)*(0.6)=0.06`
`l_2=hg(x_0+h,y_0+k_1,z_0+l_1)=(0.1)*g(0.1,0.1,0.6)=(0.1)*(-2.8)=-0.28`
`y_1=y_0+(k_1+k_2)/2=0+0.08=0.08`
`:.y(0.1)=0.08`
`:.y(0.1)=0.08`
Method-2 : Using formula `k_2=hf(x_0+h/2,y_0+k_1/2,z_0+l_1/2)`
Second 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`
`y_1=y_0+k_2=0+0.08=0.08`
`:.y(0.1)=0.08`
`:.y(0.1)=0.08`
This material is intended as a summary. Use your textbook for detail explanation.
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