Formula

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)=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)`

---

Examples
1. Find y(0.2) for `y''=1+2xy-x^2z`, `x_0=0, y_0=1, z_0=0`, with step length 0.1 using Euler method (2nd order derivative)

Solution:
Given `y^('')=1+2xy-x^2z, y(0)=1, y'(0)=0, 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)=1+2xy-x^2z=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)`

---

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

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

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

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

`=1+(0)`

`=1`

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

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

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

`=0+(0.1)`

`=0.1`

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

---

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

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

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

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

`=1+(0.01)`

`=1.01`

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

`:.y(0.2)=1.01`

`n`	`x_n`	`y_n`	`z_n`	`x_(n+1)`	`y_(n+1)`	`z_(n+1)`
0	0	1	0	0.1	1	0.1
1	0.1	1	0.1	0.2	1.01

---

---