<s>[INST] Solve the ODe with five variables How to solve $$\frac{dx}{2p}=\frac{dy}{2q}=\frac{du}{2(p^2+q^2}=\frac{dp}{2up}=\frac{dq}{2uq}=dt$$
as functions $$x=x(t), y=y(t), u=u(t), p=p(t), q=q(t)$$
My method is use the last three equalities to deduce $$\frac{d^2u}{dt^2}+u\frac{du}{dt}=0$$
But this nonlinearity troubles me...
 [/INST] Based on your last line:
Consider $F = u^2$
$$ \frac{df}{dt} = 2 u \frac{du}{dt} $$
Thus suppose we re-arrange
$$ \frac{d^2u}{dt^2} + u \frac{du}{dt} = 0 $$
Into
$$  \frac{d^2u}{dt^2} = - u \frac{du}{dt} $$
Then we can re-write it as
$$ \frac{d}{dt} \left[ \frac{du}{dt} \right] =  \frac{d}{dt}\left[ - \frac{1}{2}u^2\right] $$
Thus we have
$$ \frac{du}{dt} = C_1-\frac{1}{2}u^2 $$
From here we will use the standard algorithm for solving linear ODEs
$$ \frac{1}{C_1 - \frac{1}{2}u^2} du  = 1 dt$$
Since the choice of C is arbitrary this can be re-written as
$$2 \frac{1}{C_1 - u^2} du  = 1 dt$$
We can integrate both sides:
$$2 \int \frac{1}{C_1 - u^2} du  = t + C_2$$
The left hand side can be attacked with Partial Fractions: We decompose the integrand into:
$$ \frac{1}{C_1 - u^2} = \frac{1}{(\sqrt{C_1} - u)(\sqrt{C_1} + u)} = \frac{A}{\sqrt{C_1} - u} + \frac{B}{\sqrt{C_1} + u}$$
We find the constants A,B
$$ B(\sqrt{C_1} - u) + A(\sqrt{C_1} + u) = 0 $$
Which (based on grouping like terms of $C_1$ and $u$ tells us
$$ \begin{pmatrix} Au - Bu = 0 \\ \sqrt{C_1}(A + B) = 1 \end{pmatrix} \rightarrow \begin{pmatrix} A = B \\ (A + B) = \frac{1}{\sqrt{C_1}} \end{pmatrix} $$
Giving us
$$ A = B = \frac{1}{2 \sqrt{C_1}} $$
So our integral is now
$$2 \int \left[ \frac{\frac{1}{2 \sqrt{C_1}}}{\sqrt{C_1} -u} + \frac{\frac{1}{2 \sqrt{C_1}}}{\sqrt{C_1} + u} \right] du  $$
We factor the top parts giving us
$$ \frac{1}{\sqrt{C_1}} \int \left[ \frac{1}{\sqrt{C_1} -u} + \frac{1}{\sqrt{C_1} + u}  \right] du $$
Which integrates to
$$ \frac{1}{\sqrt{C_1}} \left( \ln(\sqrt{C_1} + u) - \ln(\sqrt{C_1} - u) \right) $$
So we have
$$2 \int \frac{1}{C_1 - u^2} du  = t + C_2$$
Gives rise to
$$ \frac{1}{\sqrt{C_1}} \left( \ln(\sqrt{C_1} + u) - \ln(\sqrt{C_1} - u) \right)  = t + C_2$$
Which resolves to
$$ \ln \left( \frac{\sqrt{C_1} + u}{\sqrt{C_1} - u} \right)   = C_1 t + C_2 $$
We exponentiate both sides to find
$$ \left( \frac{\sqrt{C_1} + u}{\sqrt{C_1} - u} \right) = C_2 e^{C_1t} $$
And now solve for $u$ as 
$$ \sqrt{C_1} + u = (C_2 e^{C_1t})(\sqrt{C_1} - u) \rightarrow $$
$$\left(1 - C_2 e^{C_1t} \right)u= (C_2 e^{C_1t}-1)\sqrt{C_1} $$
$$ u = \frac{(C_2 e^{C_1t}-1)\sqrt{C_1}}{\left(1 - C_2 e^{C_1t} \right)} $$
Alternate Trig Approach
$$ \int \frac{2}{C_1 - u^2} du = \frac{2}{\sqrt{C_1}} arctanh^{-1} \left( \frac{u}{\sqrt{C_1}} \right)  $$
So our solution then becomes
$$\frac{2}{\sqrt{C_1}} arctanh^{-1} \left( \frac{u}{\sqrt{C_1}} \right) = t + C_2 $$
$$ arctanh^{-1} \left( \frac{u}{\sqrt{C_1}} \right) = C_1t + C_2$$
$$ u = \sqrt{C_1}tanh\left( C_1t + C_2\right) $$
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