Solve nonlinear ODE's System $x_1'(t)=-x_1(t)x_3(t)$, $x_2'(t)=-x_2(t)x_3(t)$, $x_3'(t)=x_1^2(t)+x_2^2(t)$ I need to study the flow generated by the vector field $X(x,y,z)=(-xz,-yz,x^2+y^2)$. Therefore, I need to solve the system:
$$ \left\{  \begin{array}{ccc}  x_1'(t)&=&-x_1(t)x_3(t)\\
x_2'(t) &=& -x_2(t)x_3(t)\\
x_3'(t) &=&x_1^2(t)+x_2^2(t)
\end{array}\right. $$
I don't know how to solve this. I would appretiate any reference or help to solve this kind of equations.
 A: Put it into cylindrical coordinates
$$
x_{\,1} (t) = r(t)\cos \left( {\alpha (t)} \right)\quad x_{\,2} (t) = r(t)\sin \left( {\alpha (t)} \right)
$$
so
$$
\left\{ \matrix{
  x_{\,3} '(t) = r(t)^{\,2} \quad  \to \quad x_{\,3} (t) = \int_0^t {r(\tau )^{\,2} d\tau}  + c_{\,3}  = I(t) + c_{\,3}  \hfill \cr 
  x_{\,1} '(t) = r'(t)\cos \left( {\alpha (t)} \right) - r(t)\alpha '(t)\sin \left( {\alpha (t)} \right) =  \hfill \cr 
   =  - \left( {I(t) + c_{\,3} } \right)r(t)\cos \left( {\alpha (t)} \right) \hfill \cr 
  x_{\,2} '(t) = r'(t)\sin \left( {\alpha (t)} \right) + r(t)\alpha '(t)\cos \left( {\alpha (t)} \right) =  \hfill \cr 
   =  - \left( {I(t) + c_{\,3} } \right)r(t)\sin \left( {\alpha (t)} \right) \hfill \cr}  \right.
$$
which, omitting the time dependency to get a clearer view, becomes
$$
\left\{ \matrix{
  r'\cos \alpha  - r\alpha '\sin \alpha  =  - \left( {I + c_{\,3} } \right)r\cos \alpha  \hfill \cr 
  r'\sin \alpha  + r\alpha '\cos \alpha  =  - \left( {I + c_{\,3} } \right)r\sin \alpha  \hfill \cr}  \right.
$$
Multiplying the first by $\cos\alpha$ and the second by $\sin\alpha$ and summing gives
$$
r' =  - \left( {I + c_{\,3} } \right)r\quad  \to \quad {{d^{\,2} } \over {dt^{\,2} }}\ln \left( r \right) =  - r^{\,2} \quad  \to \quad {{d^{\,2} } \over {dt^{\,2} }}\rho  = e^{\, - \,2\rho } 
$$
while instead multiplying by $-\sin\alpha$ and $\cos\alpha$ and summing we get
$$
r\alpha ' = 0
$$
A: Hint
This might help
$$
\left\{  \begin{array}{ccc}  x_1'(t)&=&-x_1(t)x_3(t)\\
x_2'(t) &=& -x_2(t)x_3(t)
\end{array}\right.  \implies \frac {dx_1}{dx_2}=\frac {x_1}{x_2} \implies x_1=Kx_2$$
$$
\left\{  \begin{array}{ccc}  x_1'(t)&=&-x_1(t)x_3(t)\\
x_2'(t) &=& -x_2(t)x_3(t)\\
x_3'(t) &=&x_1^2(t)+x_2^2(t)
\end{array}\right.$$
$$x_3'(t) =x_1^2(t)+x_2^2(t)$$
$$\frac {dx_3}{dx_2} =\frac {x_1^2(t)+x_2^2(t)}{-x_2(t)x_3(t)}$$
$$\frac {dx_3}{dx_2} =\frac {K^2x_2(t)+x_2(t)}{-x_3(t)}$$
$$-\int x_3{dx_3} =(K^2+1)\int x_2dx_2$$
$$x^2_3=-(K^2+1)x^2_2+C \implies x_2=h(x_3)$$
You can substitute this in the third equation and try to solve it 
$$x_3'(t) =(K^2+1)x_2^2(t)$$
$$x_3'(t) =C-x_3^2(t)$$
$$\int \frac {dx_3}{x_3^2-C} =-t+R$$
$$....$$
