Runge-Kutta 4 method (2nd order derivative) Example-2

Find y(0.2) for y′′=xz2-y2 $y''=xz^2-y^2$ , x0=0,y0=1,z0=0 $x_0=0, y_0=1, z_0=0$ , with step length 0.2 using Runge-Kutta 4 method (2nd order derivative)

Solution:
Given

y′′=xz2-y2,y(0)=1,y′(0)=0,h=0.2,y(0.2)=? $y^('')=xz^2-y^2, y(0)=1, y'(0)=0, h=0.2, y(0.2)=?$

put

dydx=z $(dy)/(dx)=z$ and differentiate w.r.t. x, we obtain

d2ydx2=dzdx $(d^2y)/(dx^2)=(dz)/(dx)$

We have system of equations

dydx=z=f(x,y,z) $(dy)/(dx)=z=f(x,y,z)$

dzdx=xz2-y2=g(x,y,z) $(dz)/(dx)=xz^2-y^2=g(x,y,z)$

Forth order R-K method for second order differential equation

k1=f(x0,y0,z0)=f(0,1,0)=0 $k_1=f(x_0,y_0,z_0)=f(0,1,0)=0$

l1=g(x0,y0,z0)=g(0,1,0)=-1 $l_1=g(x_0,y_0,z_0)=g(0,1,0)=-1$

k2=f(x0+h2,y0+hk12,z0+hl12)=f(0.1,1,-0.1)=-0.1 $k_2=f(x_0+h/2,y_0+(hk_1)/2,z_0+(hl_1)/2)=f(0.1,1,-0.1)=-0.1$

l2=g(x0+h2,y0+hk12,z0+hl12)=g(0.1,1,-0.1)=-0.999 $l_2=g(x_0+h/2,y_0+(hk_1)/2,z_0+(hl_1)/2)=g(0.1,1,-0.1)=-0.999$

k3=f(x0+h2,y0+hk22,z0+hl22)=f(0.1,0.99,-0.0999)=-0.0999 $k_3=f(x_0+h/2,y_0+(hk_2)/2,z_0+(hl_2)/2)=f(0.1,0.99,-0.0999)=-0.0999$

l3=g(x0+h2,y0+hk22,z0+hl22)=g(0.1,0.99,-0.0999)=-0.9791 $l_3=g(x_0+h/2,y_0+(hk_2)/2,z_0+(hl_2)/2)=g(0.1,0.99,-0.0999)=-0.9791$

k4=f(x0+h,y0+hk3,z0+hl3)=f(0.2,0.98002,-0.19582)=-0.19582 $k_4=f(x_0+h,y_0+hk_3,z_0+hl_3)=f(0.2,0.98002,-0.19582)=-0.19582$

l4=g(x0+h,y0+hk3,z0+hl3)=g(0.2,0.98002,-0.19582)=-0.95277 $l_4=g(x_0+h,y_0+hk_3,z_0+hl_3)=g(0.2,0.98002,-0.19582)=-0.95277$

Now,

y1=y0+h6(k1+2k2+2k3+k4) $y_1=y_0+h/6(k_1+2k_2+2k_3+k_4)$

y1=1+0.26[0+2(-0.1)+2(-0.0999)+(-0.19582)] $y_1=1+0.2/6[0+2(-0.1)+2(-0.0999)+(-0.19582)]$

y1=0.98015 $y_1=0.98015$

$:.y(0.2)=0.98015$

This material is intended as a summary. Use your textbook for detail explanation.
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5. Example-2