Find y(0.2) for `y''=-4z-4y`, `x_0=0, y_0=0, z_0=1`, with step length 0.1 using Midpoint Euler method (2nd order derivative)

Solution:
Given `y^('')=-4z-4y, y(0)=0, y'(0)=1, h=0.1, 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)=-4z-4y=g(x,y,z)`

Here, `x_0=0,y_0=0,z_0=1,h=0.1,x_n=0.2`

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=0,z_0=1`

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

`=f(0,0,1)`

`=1`

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

`=g(0,0,1)`

`=-4`

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

`=f(0+0.1/2,0+0.1/2 *1,1+0.1/2 *-4)`

`=f(0.05,0.05,0.8)`

`=0.8`

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

`=g(0+0.1/2,0+0.1/2 *1,1+0.1/2 *-4)`

`=g(0.05,0.05,0.8)`

`=-3.4`

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

`=0+0.1*0.8`

`=0.08`

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

`=1+0.1*-3.4`

`=0.66`

`x_1=x_0+h=0+0.1=0.1`

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for `n=1,x_1=0.1,y_1=0.08,z_1=0.66`

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

`=f(0.1,0.08,0.66)`

`=0.66`

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

`=g(0.1,0.08,0.66)`

`=-2.96`

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

`=f(0.1+0.1/2,0.08+0.1/2 *0.66,0.66+0.1/2 *-2.96)`

`=f(0.15,0.113,0.512)`

`=0.512`

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

`=g(0.1+0.1/2,0.08+0.1/2 *0.66,0.66+0.1/2 *-2.96)`

`=g(0.15,0.113,0.512)`

`=-2.5`

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

`=0.08+0.1*0.512`

`=0.1312`

`x_2=x_1+h=0.1+0.1=0.2`

`:.y(0.2)=0.1312`

`n`	`x_n`	`y_n`	`z_n`	`x_(n+1)`	`y_(n+1)`	`z_(n+1)`
0	0	0	1	0.1	0.08	0.66
1	0.1	0.08	0.66	0.2	0.1312

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