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 3 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)`
Third order R-K method
`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,y_0+2hk_2-hk_1,z_0+2hl_2-hl_1)=f(0.2,0.96,-0.1996)=-0.1996`
`l_3=g(x_0+h,y_0+2hk_2-hk_1,z_0+2hl_2-hl_1)=g(0.2,0.96,-0.1996)=-0.91363`
Now,
`y_1=y_0+h/6(k_1+4k_2+k_3)`
`y_1=1+0.2/6[0+4(-0.1)+(-0.1996)]`
`y_1=0.98001`
`:.y(0.2)=0.98001`
`:.y(0.2)=0.98001`
This material is intended as a summary. Use your textbook for detail explanation.
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