Find y(0.4) for `y''=xz^2-y^2`, `x_0=0, y_0=1, z_0=0`, with step length 0.2 using Midpoint Euler method (2nd order derivative)

Solution:
Given `y^('')=xz^2-y^2, y(0)=1, y'(0)=0, h=0.2, y(0.4)=?`

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)`

Here, `x_0=0,y_0=1,z_0=0,h=0.2,x_n=0.4`

Midpoint Euler method for second order differential equation formula
`y_(n+1)=y_n+h k_(2y)`

`k_(1y)=f(x_n,y_n,z_n)`

`k_(2y)=f(x_n+h/2,y_n+h/2 k_(1y),z_n+h/2 k_(1z))`

`z_(n+1)=z_n+h k_(2z)`

`k_(1z)=g(x_n,y_n,z_n)`

`k_(2z)=g(x_n+h/2,y_n+h/2 k_(1y),z_n+h/2 k_(1z))`

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for `n=0,x_0=0,y_0=1,z_0=0`

`k_(1y)=f(x_0,y_0,z_0)`

`=f(0,1,0)`

`=0`

`k_(1z)=g(x_0,y_0,z_0)`

`=g(0,1,0)`

`=-1`

`k_(2y)=f(x_0+h/2,y_0+h/2 k_(1y),z_0+h/2 k_(1z))`

`=f(0+0.2/2,1+0.2/2 *0,0+0.2/2 *-1)`

`=f(0.1,1,-0.1)`

`=-0.1`

`k_(2z)=g(x_0+h/2,y_0+h/2 k_(1y),z_0+h/2 k_(1z))`

`=g(0+0.2/2,1+0.2/2 *0,0+0.2/2 *-1)`

`=g(0.1,1,-0.1)`

`=-0.999`

`y_(1)=y_0+h k_(2y)`

`=1+0.2*-0.1`

`=0.98`

`z_(1)=z_0+h k_(2z)`

`=0+0.2*-0.999`

`=-0.1998`

`x_1=x_0+h=0+0.2=0.2`

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for `n=1,x_1=0.2,y_1=0.98,z_1=-0.1998`

`k_(1y)=f(x_1,y_1,z_1)`

`=f(0.2,0.98,-0.1998)`

`=-0.1998`

`k_(1z)=g(x_1,y_1,z_1)`

`=g(0.2,0.98,-0.1998)`

`=-0.9524`

`k_(2y)=f(x_1+h/2,y_1+h/2 k_(1y),z_1+h/2 k_(1z))`

`=f(0.2+0.2/2,0.98+0.2/2 *-0.1998,-0.1998+0.2/2 *-0.9524)`

`=f(0.3,0.96,-0.295)`

`=-0.295`

`k_(2z)=g(x_1+h/2,y_1+h/2 k_(1y),z_1+h/2 k_(1z))`

`=g(0.2+0.2/2,0.98+0.2/2 *-0.1998,-0.1998+0.2/2 *-0.9524)`

`=g(0.3,0.96,-0.295)`

`=-0.8955`

`y_(2)=y_1+h k_(2y)`

`=0.98+0.2*-0.295`

`=0.921`

`x_2=x_1+h=0.2+0.2=0.4`

`:.y(0.4)=0.921`

`n`	`x_n`	`y_n`	`z_n`	`x_(n+1)`	`y_(n+1)`	`z_(n+1)`
0	0	1	0	0.2	0.98	-0.1998
1	0.2	0.98	-0.1998	0.4	0.921

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