Find y(0.2) for `y''=-4z-4y`, `x_0=0, y_0=0, z_0=1`, with step length 0.1 using Improved Euler / Modified 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`

Improved Euler method / Modified Euler method for second order differential equation formula
`y_(n+1)=y_n+h/2 [k_(1y) + k_(2y)]`

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

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

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

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

`k_(2z)=g(x_n+h,y_n+hk_(1y),z_n+hk_(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,y_0+hk_(1y),z_0+hk_(1z))`

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

`=f(0.1,0.1,0.6)`

`=0.6`

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

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

`=g(0.1,0.1,0.6)`

`=-2.8`

`y_1=y_0+h/2 [k_(1y) + k_(2y)]`

`=0+0.1/2 [1 + 0.6]`

`=0.08`

`z_1=z_0+h/2 [k_(1z) + k_(2z)]`

`=1+0.1/2 [-4 + -2.8]`

`=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,y_1+hk_(1y),z_1+hk_(1z))`

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

`=f(0.2,0.146,0.364)`

`=0.364`

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

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

`=g(0.2,0.146,0.364)`

`=-2.04`

`y_2=y_1+h/2 [k_(1y) + k_(2y)]`

`=0.08+0.1/2 [0.66 + 0.364]`

`=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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