Find y(0.2) for `y''=-4z-4y`, `x_0=0, y_0=0, z_0=1`, with step length 0.1 using Euler method (second order differential equation)

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)`

Euler method for second order differential equation formula
`y_(n+1)=y_n+hf(x_n,y_n,z_n)`

`z_(n+1)=z_n+hg(x_n,y_n,z_n)`

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

`y_1=y_0+hf(x_0,y_0,z_0)`

`=0+(0.1)*f(0,0,1)`

`=0+(0.1)*(1)`

`=0+(0.1)`

`=0.1`

`z_1=z_0+hg(x_0,y_0,z_0)`

`=1+(0.1)*g(0,0,1)`

`=1+(0.1)*(-4)`

`=1+(-0.4)`

`=0.6`

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

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for `n=1,x_1=0.1,y_1=0.1,z_1=0.6`

`y_2=y_1+hf(x_1,y_1,z_1)`

`=0.1+(0.1)*f(0.1,0.1,0.6)`

`=0.1+(0.1)*(0.6)`

`=0.1+(0.06)`

`=0.16`

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

`:.y(0.2)=0.16`

`n`	`x_n`	`y_n`	`z_n`	`x_(n+1)`	`y_(n+1)`	`z_(n+1)`
0	0	0	1	0.1	0.1	0.6
1	0.1	0.1	0.6	0.2	0.16

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