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 Euler method (second order differential equation)

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)`

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=1,z_0=0`

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

`=1+(0.2)*f(0,1,0)`

`=1+(0.2)*(0)`

`=1+(0)`

`=1`

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

`=0+(0.2)*g(0,1,0)`

`=0+(0.2)*(-1)`

`=0+(-0.2)`

`=-0.2`

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

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

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

`=1+(0.2)*f(0.2,1,-0.2)`

`=1+(0.2)*(-0.2)`

`=1+(-0.04)`

`=0.96`

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

`:.y(0.4)=0.96`

`n`	`x_n`	`y_n`	`z_n`	`x_(n+1)`	`y_(n+1)`	`z_(n+1)`
0	0	1	0	0.2	1	-0.2
1	0.2	1	-0.2	0.4	0.96

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