How to solve multiple dependent differential equation? I have next differential equation:
$$ (x_0x_3)'=-(x_1x_2)'+8\cdot(-x_2'-\dfrac{x_1'}{x_0}+\dfrac{x_1x_0'}{x_0^2})$$
where $x_0'=\dfrac{dx_0}{dz}$, and it means the same for every sign $'$. 
On the other side $x_0=(1+64(1-z))^{0.5}=f(z)$. 
I need to solve the first equation on that way that I need to find $x_3$. But because I have dependence $x_0=f(z)$, and because also $x_1=f(x_0)$ and $x_2=f(x_0)$, that means that they are all also dependent on $z$: $x_1=f(z)$ and $x_2=f(z)$, and it is impossible to integrate upper equation on this way:
$$\dfrac{d}{dz}(x_0x_3)=...  \,\,\,  / \cdot dz$$
$$d(x_0x_3)=... \,\,\, /integration$$
because I will lost some dependence between variables. 
I am trying to do all this in Mupad which is part of Matlab and it is according to my opinion appropriate for symbolical equations. So I have tried to tell Mupad that I have $$x_0=(1+64(1-z))^{0.5}$$ $$x_1=8(\dfrac{1}{x_0}-1)$$ $$x_2=-\dfrac{x_1^2}{2x_0}+\dfrac{8}{x_0}(-x_1-ln(x_0))$$ and it accepted it. But as next step where I need to solve upper differential equation i tried with this:
ode::solve({-(x0*x3)'-(x1*x2)'+8*(-x2'-x1'/x0+x1*x0'/x0^2)=0},x3)

And also with different combinations with initial conditions
 ode::solve({-(x0*x3)'-(x1*x2)'+8*(-x2'-x1'/x0+x1*x0'/x0^2)=0,x3(0)=0},x3)

I always got:
Error: Invalid arguments 

Am I on the right way to solve this equation on this way and with Mupad and what would be incorrect here? Or would be correct way to solve this?
 A: You have $x_0=f(z)$ and $x_1=x_2=f(x_0)=f(f(z)),$ where 
$$f(z)=\sqrt{1+64(1-z)}.$$
Define $x_4=x_0\cdot x_3$. The original DE becomes
\begin{align*}
x_4'&=-(x_1x_2)'+8\cdot\left(-x_2'-\dfrac{x_1'}{x_0}+\dfrac{x_1x_0'}{x_0^2}\right) \\
x_4'&=-((x_1)^2)'+8\cdot\left(-x_1'-\dfrac{x_1'}{x_0}+\dfrac{x_1x_0'}{x_0^2}\right) \\
x_4'&=-((f(f(z)))^2)'+8\cdot\left(-(f(f(z)))'-\dfrac{(f(f(z)))'}{f(z)}+\dfrac{f(f(z))\,f'(z)}{f^2(z)}\right)\\
x_4'&=-2f(f(z))\cdot f'(f(z))\cdot f'(z)+8\cdot\left(\dfrac{-f^2(z)(f(f(z)))'-f(z)(f(f(z)))'+f(f(z))\,f'(z)}{f^2(z)}\right).
\end{align*}
We simplify what's in the parentheses:
\begin{align*}
()&=\dfrac{-f^2(z)(f(f(z)))'-f(z)(f(f(z)))'+f(f(z))\,f'(z)}{f^2(z)}\\
&=\dfrac{-f^2(z)f'(f(z))f'(z)-f(z)f'(f(z))f'(z)+f(f(z))\,f'(z)}{f^2(z)} \\
&=f'(z)\cdot\dfrac{-f^2(z)f'(f(z))-f(z)f'(f(z))+f(f(z))}{f^2(z)}.
\end{align*}
Plugging this back in:
$$x_4'=f'(z)\cdot\left(8\cdot\dfrac{-f^2(z)f'(f(z))-f(z)f'(f(z))+f(f(z))}{f^2(z)}-2f(f(z))\cdot f'(f(z))\right). $$
I would use Mathematica to define f[x_], and then simply integrate this expression with respect to $z$. That will give you $x_4$. Then you can solve for $x_3$.
A: Making substitutions and developing the DE gives
$$
x_3(z) x_0'(z)+\frac{8 x_1'(z)}{x_0(z)}+x_0(z)
   x_3'(z)+x_2(z) x_1'(z)+(x_1(z)+8) x_2'(z)-\frac{8x_1(z) x_0'(z)}{x_0(z)^2}=0
$$
Solving for $x_3(z)$ gives
$$
x_3'(z)= \frac{x_1(z) \left(8 x_0'(z)-x_0(z)^2 x_2'(z)\right)-x_0(z) \left(x_0(z)
   \left(x_3(z) x_0'(z)+x_2(z) x_1'(z)+8 x_2'(z)\right)+8 x_1'(z)\right)}{x_0(z)^3}
$$
and solving the $x_3(z)$ DE
$$
x_3(z) = \frac{x_0(z) \left(C_3-(x_1(z)+8) x_2(z)\right)-8 x_1(z)}{x_0(z)^2}
$$
All this was made with the help from a symbolic processor. I hope this helps.
Follows the MATHEMATICA script to obtain those results.

d[x1_, x2_, x3_, z_]:= D[x0[z] x3, z]+D[x1 x2, z]+8(D[x2, z]+D[x1, z]/x0[z]-x1 D[x0[z],z]/x0[z]^2)
dx30 = d[x1[z], x2[z], x3[z], z]
solx3 = Solve[dx30 == 0, x3'[z]][[1]]
equx3 = solx3 /. Rule -> Equal
solx3 = DSolve[equx3, x3, z][[1]]
x3z = x3[z] /. solx3
