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 Improved Euler / Modified 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)`

Here, `x_0=0,y_0=1,z_0=0,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))`

---

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

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

`=f(0,1,0)`

`=0`

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

`=g(0,1,0)`

`=1`

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

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

`=f(0.1,1,0.1)`

`=0.1`

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

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

`=g(0.1,1,0.1)`

`=1.199`

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

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

`=1.005`

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

`=0+0.1/2 [1 + 1.199]`

`=0.11`

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

---

for `n=1,x_1=0.1,y_1=1.005,z_1=0.11`

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

`=f(0.1,1.005,0.11)`

`=0.11`

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

`=g(0.1,1.005,0.11)`

`=1.1999`

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

`=f(0.1+0.1,1.005+0.1*0.11,0.11+0.1*1.1999)`

`=f(0.2,1.016,0.2299)`

`=0.2299`

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

`=g(0.1+0.1,1.005+0.1*0.11,0.11+0.1*1.1999)`

`=g(0.2,1.016,0.2299)`

`=1.3972`

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

`=1.005+0.1/2 [0.11 + 0.2299]`

`=1.022`

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

`:.y(0.2)=1.022`

`n`	`x_n`	`y_n`	`z_n`	`x_(n+1)`	`y_(n+1)`	`z_(n+1)`
0	0	1	0	0.1	1.005	0.11
1	0.1	1.005	0.11	0.2	1.022