2. 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=hf(x_0,y_0,z_0)=(0.2)*f(0,1,0)=(0.2)*(0)=0`
`l_1=hg(x_0,y_0,z_0)=(0.2)*g(0,1,0)=(0.2)*(-1)=-0.2`
`k_2=hf(x_0+h/2,y_0+k_1/2,z_0+l_1/2)=(0.2)*f(0.1,1,-0.1)=(0.2)*(-0.1)=-0.02`
`l_2=hg(x_0+h/2,y_0+k_1/2,z_0+l_1/2)=(0.2)*g(0.1,1,-0.1)=(0.2)*(-0.999)=-0.1998`
`k_3=hf(x_0+h,y_0+2k_2-k_1,z_0+2l_2-l_1)=(0.2)*f(0.2,0.96,-0.1996)=(0.2)*(-0.1996)=-0.03992`
`l_3=hg(x_0+h,y_0+2k_2-k_1,z_0+2l_2-l_1)=(0.2)*g(0.2,0.96,-0.1996)=(0.2)*(-0.91363)=-0.18273`
Now,
`y_1=y_0+1/6(k_1+4k_2+k_3)`
`y_1=1+1/6[0+4(-0.02)+(-0.03992)]`
`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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