solution of differential equations I am looking for the solution of this  differential equations
$${\frac {d^{2}x}{d{t}^{2}}} +\gamma\, \left[  \left( {
\frac {dx}{dt}}\right) ^{2}+ \left( {\frac {dz}{dt}}
  \right) ^{2} \right]=0$$
$${\frac {d^{2}z}{d{t}^{2}}} +g+\gamma\, \left[ 
 \left( {\frac {dx}{dt}}  \right) ^{2}+ \left( {
\frac {dz}{dt}} \right) ^{2} \right] =0$$
 A: Let $X(t)=x'(t)$ and let $Z(t)=z'(t)$, then we get
$$
X'(t)+\gamma(X^2+Z^2)=0
$$
$$
Z'(t)+g+\gamma(X^2+Z^2)=0.
$$
Taking the difference of the two, we get
$$
(Z(t)-X(t))' = -g.
$$
This implies that
$$
Z(t)=X(t)-gt+C.
$$
Substituting this back to the first differential equation, we get
$$
X'(t)+\gamma(X^2+(X-gt+C)^2)=0.
$$
The only unknown in this last equation is $X$. This last equation is not separable or exact. I am not quite sure how to continue from here. Hope this helps a little.
A: Continuation of the EagleToLearn's partial answer : 
$$X'(t)+\gamma(X^2+(X-gt+C)^2)=0.$$
$$
X'(t)=-\gamma\left(2X^2+2(-gt+C)X+(-gt+C)^2\right).
$$
This is a Riccati ODE. The convenient change of function is : $\quad X(t)=\frac{Y'}{2\gamma Y}$
$\frac{Y''}{2\gamma Y}-\frac{(Y')^2}{2\gamma Y^2}= -2\gamma\left(\frac{Y'}{2\gamma Y}\right)^2+2\gamma(gt-C)\frac{Y'}{2\gamma Y}-\gamma(gt-C)^2$
$\frac{Y''}{2\gamma Y}= 2\gamma(gt-C)\frac{Y'}{2\gamma Y}-\gamma(gt-C)^2$
$$\frac{d^2Y}{dt^2} -2\gamma(gt-C)\frac{dY}{dt}+2\gamma^2(gt-C)^2Y(t)=0$$
Change of variable : $\quad \theta=(gt-C)^2\quad$ After transformation :
$$\frac{d^2Y}{d^2\theta} +(1-\frac{2\gamma}{g}) \frac{dY}{d\theta}+\frac{\gamma^2}{g^2}Y(\theta)=0$$
This is a linear ODE easy to solve :
$Y(\theta)=c_1\exp\left(\left(\frac{\gamma}{g}-\frac12+\sqrt{\frac12-\frac{\gamma}{g} } \right)\theta\right) +c_2\exp\left(\left(\frac{\gamma}{g}-\frac12-\sqrt{\frac12-\frac{\gamma}{g} } \right)\theta\right)$
$$Y(t)=c_1\exp\left(\pm\left(\frac{\gamma}{g}-\frac12+\sqrt{\frac12-\frac{\gamma}{g} } \right) \sqrt{gt-C} \right) \\+c_2\exp\left(\pm\left(\frac{\gamma}{g}-\frac12-\sqrt{\frac12-\frac{\gamma}{g} } \right)\sqrt{gt-C}\right)$$
$$x'(t)=X(t)=\frac{1}{2\gamma}\frac{1}{Y}\frac{dY}{dt} \quad\implies\quad x(t)=\frac{1}{2\gamma}\ln|c_3Y|$$
With $C_1=c_1c_3$ and $C_2=c_2c_3$ , the result is :
$$x(t)=\frac{1}{2\gamma}\ln\left| C_1\exp\left(\pm\left(\frac{\gamma}{g}-\frac12+\sqrt{\frac12-\frac{\gamma}{g} } \right) \sqrt{gt-C} \right) \\+C_2\exp\left(\pm\left(\frac{\gamma}{g}-\frac12-\sqrt{\frac12-\frac{\gamma}{g} } \right)\sqrt{gt-C}\right)\right|$$
The constants and the correct sign of $\pm$ have to be determined according to some conditions (not specified in the wording of the question).
Note (depending on the missing conditions) : For some conditions the above arduous calculus might become much simpler. 
