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 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=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/2,y_0+k_2/2,z_0+l_2/2)=(0.2)*f(0.1,0.99,-0.0999)=(0.2)*(-0.0999)=-0.01998`
`l_3=hg(x_0+h/2,y_0+k_2/2,z_0+l_2/2)=(0.2)*g(0.1,0.99,-0.0999)=(0.2)*(-0.9791)=-0.19582`
`k_4=hf(x_0+h,y_0+k_3,z_0+l_3)=(0.2)*f(0.2,0.98002,-0.19582)=(0.2)*(-0.19582)=-0.03916`
`l_4=hg(x_0+h,y_0+k_3,z_0+l_3)=(0.2)*g(0.2,0.98002,-0.19582)=(0.2)*(-0.95277)=-0.19055`
Now,
`y_1=y_0+1/6(k_1+2k_2+2k_3+k_4)`
`y_1=1+1/6[0+2(-0.02)+2(-0.01998)+(-0.03916)]`
`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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