Find y(0.2) for `y''=xz^2-y^2`, `x_0=0, y_0=1, z_0=0`, with step length 0.2 using Taylor Series method (second order differential equation)

Solution:
Given `y^('')=xz^2-y^2, y(0)=1, y'(0)=0, h=0.2, 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)=xz^2-y^2=g(x,y,z)`

Here, `x_0=0,y_0=1,z_0=0,h=0.2,x_n=0.2`

Differentiating successively, we get

Derivative steps

`d/(dx)(xy'^2-y^2)`

`=d/(dx)(xy'^2)-d/(dx)(y^2)`

`d/(dx)(xy'^2)=y'^2+2xy'y''`

`d/(dx)(xy'^2)`

`=(d/(dx)(x))y'^2+x(d/(dx)(y'^2))`

`d/(dx)(y'^2)=2y'y''`

`d/(dx)(y'^2)`

`=2y'*d/(dx)(y')`

`=2y'*y''`

`=2y'y''`

`=1y'^2+x(2y'y'')`

`=y'^2+2xy'y''`

`d/(dx)(y^2)=2yy'`

`d/(dx)(y^2)`

`=2y*d/(dx)(y)`

`=2y*y'`

`=2yy'`

`=(y'^2+2xy'y'')-2yy'`

`=y'^2+2xy'y''-2yy'`

Now, `d^2/(dx^2)(xy'^2-y^2)=d/(dx)(y'^2+2xy'y''-2yy')`

`=d/(dx)(y'^2)+d/(dx)(2xy'y'')-d/(dx)(2yy')`

`d/(dx)(y'^2)=2y'y''`

`d/(dx)(y'^2)`

`=2y'*d/(dx)(y')`

`=2y'*y''`

`=2y'y''`

`d/(dx)(2xy'y'')=2y'y''+2xy''^2+2xy'y'''`

`d/(dx)(2xy'y'')`

`=2*(d/(dx)(x))y'y''+2x(d/(dx)(y'))y''+2xy'(d/(dx)(y''))`

`=2*1y'y''+2x(y'')y''+2xy'(y''')`

`=2y'y''+2xy''^2+2xy'y'''`

`d/(dx)(2yy')=2y'^2+2yy''`

`d/(dx)(2yy')`

`=2*(d/(dx)(y))y'+2y(d/(dx)(y'))`

`=2*(y')y'+2y(y'')`

`=2y'^2+2yy''`

`=2y'y''+(2y'y''+2xy''^2+2xy'y''')-(2y'^2+2yy'')`

`=2y'y''+2y'y''+2xy''^2+2xy'y'''-2y'^2-2yy''`

`=4y'y''+2xy''^2+2xy'y'''-2y'^2-2yy''`

`y'=xy'^2-y^2`

`y''=xy'^2-y^2`

`y'''=y'^2+2xy'y''-2yy'`

`y^(iv)=4y'y''+2xy''^2+2xy'y'''-2y'^2-2yy''`

Now substituting, we get
`y_0''=x_0y_0'^2-y_0^2=-1`

`y_0'''=y_0'^2+2x_0y_0'y_0''-2y_0y_0'=0`

`y_0^(iv)=4y_0'y_0''+2x_0y_0''^2+2x_0y_0'y_0'''-2y_0'^2-2y_0y_0''=2`

Putting these values in Taylor Series, we have
`y_1 = y_0 + hy_0' + h^2/(2!) y_0'' + h^3/(3!) y_0''' + h^4/(4!) y_0^(iv) + ...`

---

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

`=1+0.2*(0)+(0.2)^2/(2)*(-1)+(0.2)^3/(6)*(0)+(0.2)^4/(24)*(2)+...`

`=1+0-0.02+0+0+...`

`=0.9801`

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

`:.y(0.2)=0.9801`

`n`	`x_n`	`y_n`	`y_n'`	`y_n''`	`y_n'''`	`y_n^(iv)`	`x_(n+1)`	`y_(n+1)`
0	0	1	0	-1	0	2	0.2	0.9801

---

---