5. Example-2
Find y(0.2) for `y''=xz^2-y^2`, `x_0=0, y_0=1, z_0=0`, with step length 0.2 using Runge-Kutta 4 method (2nd order derivative)
Solution: Given `y^('')=xz^2-y^2, y(0)=1, y'(0)=0, h=0.2, y(0.2)=?`
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)=xz^2-y^2=g(x,y,z)`
Forth order R-K method for second order differential equation `k_1=f(x_0,y_0,z_0)=f(0,1,0)=0`
`l_1=g(x_0,y_0,z_0)=g(0,1,0)=-1`
`k_2=f(x_0+h/2,y_0+(hk_1)/2,z_0+(hl_1)/2)=f(0.1,1,-0.1)=-0.1`
`l_2=g(x_0+h/2,y_0+(hk_1)/2,z_0+(hl_1)/2)=g(0.1,1,-0.1)=-0.999`
`k_3=f(x_0+h/2,y_0+(hk_2)/2,z_0+(hl_2)/2)=f(0.1,0.99,-0.0999)=-0.0999`
`l_3=g(x_0+h/2,y_0+(hk_2)/2,z_0+(hl_2)/2)=g(0.1,0.99,-0.0999)=-0.9791`
`k_4=f(x_0+h,y_0+hk_3,z_0+hl_3)=f(0.2,0.98002,-0.19582)=-0.19582`
`l_4=g(x_0+h,y_0+hk_3,z_0+hl_3)=g(0.2,0.98002,-0.19582)=-0.95277`
Now, `y_1=y_0+h/6(k_1+2k_2+2k_3+k_4)`
`y_1=1+0.2/6[0+2(-0.1)+2(-0.0999)+(-0.19582)]`
`y_1=0.98015`
`:.y(0.2)=0.98015`
`:.y(0.2)=0.98015`
This material is intended as a summary. Use your textbook for detail explanation. Any bug, improvement, feedback then
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